Can I set up a half squad, mmg and leader in a steeple?
EXPECTED
Answer: Yes. A steeple has a normal stacking limit of one HS, and a leader does not count against that limit; an MMG can be there if possessed by the HS, so a HS + leader + MMG setup in a steeple is legal.
References:
- (B23.?) Steeple stacking/occupancy — the Q&A notes a steeple has a normal stacking limit of one HS.
- (A5.5) Stacking — SMC do not count as squad-equivalents for normal stacking limits, so the leader can accompany the HS.
- (A4.43) Possession of SW — the MMG may be in the steeple if possessed by the HS/leader in that Location.
MODEL RESPONSE
**Answer:** No. A Steeple Location has a stacking limit of one HS-equivalent, and no Gun may be set up in a Steeple. Since an MMG is a Gun (SW) and the leader is an SMC, the total stacking (1 HS + 1 SMC + 1 Gun) exceeds the limit and the Gun is prohibited outright.
---
The relevant rules are:
- **(B31.21) Steeple Location Stacking**: A Steeple Location has a normal stacking limit (A5.1) of one HS-equivalent. No Gun may be set up in a Steeple Location.
- **(B31.2) Steeple Location**: Treated as a normal building Location except as stated otherwise (cannot be Fortified, has no Rooftop Location, and the +1 drm for stone rubble determination is NA).
Even if the stacking limit were not an issue, the prohibition on setting up any Gun in a Steeple Location (which includes an MMG as a Support Weapon/Gun) makes this setup illegal.
JUDGE REASONING
[Calculation] This is effectively a yes/no fact question, so the final answer must match exactly. The expected answer is Yes: a steeple can contain one HS, a leader does not count against stacking, and an MMG is allowed if possessed. The model answered No and incorrectly treated an MMG as a Gun prohibited from setup in a steeple. That directly contradicts the expected rule outcome.
A 4-6-7 squad declares a Dash from woods to woods via a road hex. A 4-4-7 squad fires at the 4-6-7 as it enters the road. What is the FP of the attack?
EXPECTED
2 FP. The 4 FP is halved for Dash (Area Fire). FFMO/FFNAM DRMs apply normally.
MODEL RESPONSE
**Answer:** 2 FP. The 4‑4‑7’s Fire Power is halved because all non-ordnance Defensive First Fire against a Dashing unit in the road Location is treated as Area Fire, and the road hex in a Dash is considered not Open Ground to the firer (otherwise the Dash would be illegal). The example confirms halving in this exact situation.
---
Step 1: A Dash (A4.63) requires the unit to move from a non-Open Ground Location through a road hex to another non-Open Ground Location. The road hex is not Open Ground to any firer for whom the Dash penalty applies; if it were Open Ground, the Dash would not be allowed as the starting/ending hexes must be non-Open Ground, and the road hex is the intervening hex.
Step 2: Per A4.63, all non-ordnance Defensive First Fire vs a Dashing unit in the road hex is considered Area Fire, halving the firer’s FP.
Step 3: The 4‑4‑7 has a normal FP of 4. Halved for Area Fire/Dash = 2 FP.
Calculation: 4 FP ÷ 2 = 2 FP.
References:
- (A4.63) Dash — Defensive First Fire vs a Dashing unit in the road is Area Fire, halving FP
- (A9.3) Area Fire halves FP
- Example in A4.63: 4‑4‑7 firing at Dashing 4‑6‑7 in road hex has its FP halved
JUDGE REASONING
[Calculation] This is a specific FP calculation, and the model’s final answer exactly matches the expected answer: 2 FP. It correctly states that the 4 FP attack is halved due to Dash being treated as Area Fire. The extra explanation is consistent and does not contradict the expected answer.
A building hex contains two squads and a crew manning a gun. How many MF does it cost for a third squad to enter this Location?
EXPECTED
3 MF. 2 MF to enter the building + 1 MF for overstacking (the Location already contains > 3 squad equivalents: 2 squads + 1 crew (counts as squad for stacking if manning Gun))
MODEL RESPONSE
**Answer:** It costs 4 MF for the third squad to enter the building hex (2 MF for the building + 2 MF for entering an overstacked Location).
---
Step 1: Base cost to enter a building hex is 2 MF (A4.34).
Step 2: Determine overstacking penalty. Normal stacking is 3 squad equivalents (5.1). Two squads + 1 crew = 2.5 squad equivalents (the crew counts as 0.5 squad equivalent per 5.5). The third squad entering makes it 3.5 squad equivalents, exceeding the limit by 0.5 squad equivalents.
Step 3: Overstacking penalty is +1 MF per squad equivalent (FRU) exceeding normal limits (5.11). 0.5 squad equivalents rounds up to 1, so +1 MF.
Step 4: However, the example in 5.11 shows that entering a Location already overstacked by 1½ squads costs +2 MF. Here, the Location is already at 2.5 squad equivalents (0.5 over the limit). The entering squad adds 1 squad equivalent, making the overstack 1.5 squad equivalents. Per the example, this incurs a +2 MF penalty.
Calculation: 2 (building) + 2 (overstack penalty) = 4 MF.
References:
- (A4.34) MF Cost for Buildings — entering a building hex costs 2 MF
- (5.1) Stacking Limits — normal limit is 3 squad equivalents per Location
- (5.5) Squad Equivalents — a crew is 0.5 squad equivalent
- (5.11) Movement Overstacking — +1 MF per squad equivalent exceeding normal limits; example shows +2 MF for entering a Location overstacked by 1½ squads
JUDGE REASONING
[Calculation] This is a specific MF-cost question, so the final value must match exactly. The expected answer is 3 MF: 2 MF for the building plus 1 MF overstacking because a crew manning a Gun counts as a squad equivalent for stacking, making the Location already overstacked. The model answered 4 MF and based it on treating the crew as only 0.5 squad equivalent, so both the final result and key rules application are wrong.
Two 4-4-7's attack a 4-6-7 in CC and attempt to capture a 4-6-7 (first line unit). What is the resolution of the attack if the DR = 5?
EXPECTED
The outcome is the 4-6-7 is captured as prisoners. The CC attack is 2-1, which normally results in a reduction on a DR of 7 and an elimination on a DR of 6. There is a +1 DRM on a capture attempt, so a DR of 5 is adjusted to 6 and results in capture.
MODEL RESPONSE
**Answer:** The attack has no effect. The final DR of 6 is greater than the Kill Number of 5 at 2:1 odds, so neither capture nor elimination occurs.
---
**Step 1:** Two 4-4-7 squads have a total CC FP of 8 (4+4). The defending 4-6-7 has a CC FP of 4. The odds are 8:4 = 2:1 (A11.4, A11.11).
**Step 2:** The attacker designates a capture attempt, which adds a +1 DRM to the CC DR (A20.22). The target is a first-line unit, not Inexperienced, so the standard +1 DRM applies.
**Step 3:** The raw DR is 5. Applying the +1 DRM gives a Final DR of 6.
**Step 4:** On the Close Combat Table (CCT), at 2:1 odds the black Kill Number is 5. A Final DR of 6 is greater than 5, so the result is no effect (A11.11, A20.22).
**Calculation:** 5 (raw DR) + 1 (capture attempt DRM) = 6 (Final DR); 6 > 5 (Kill Number at 2:1) → no effect.
**References:**
- (A11.4) Close Combat FP — each MMC uses its first number as CC FP
- (A11.11) CC Resolution — odds determine Kill Number on CCT; Final DR < Kill Number eliminates, = causes Casualty Reduction, > has no effect
- (A20.22) Capture Attempt in CC — adds +1 DRM (or -1 vs Inexperienced) to the CC DR; capture occurs if Final DR < Kill Number, partial capture if equal
JUDGE REASONING
[Calculation] This is a strict fact/calculation question. The expected result is capture: at 2:1 odds, a normal DR 6 would eliminate and DR 7 would reduce, but for a capture attempt the +1 DRM shifts a raw 5 to a final 6, which yields capture. The model instead concludes no effect based on an incorrect Kill Number interpretation, so the final answer does not match the expected answer.
A 4-6-7 squad fires at a unit in Open Ground at range 2 (PBF). What is the Residual FP left in the hex?
EXPECTED
4 Residual FP. Attack FP = 4 * 2 (PBF) = 8 FP. Residual is half of 8 = 4 FP. (Resolved on 8 column, half is 4, which is a column).
MODEL RESPONSE
**Answer:** 2 Residual FP (the 4 FP column is halved per A8.2). PBF (range 1) would double the FP to 8, leaving 4 Residual FP, but at range 2 the squad’s inherent FP is not doubled.
---
A 4-6-7 squad’s inherent FP is 4. At range 2 (non-adjacent), PBF does not apply (A7.21 — PBF only for adjacent hexes). The attack is resolved on the 4 FP column. Residual FP is half (up to 12) of the highest IFT FP column used for that attack (A8.2). Half of 4 is 2. The target is in Open Ground (no TEM), so no positive DRM reduces the Residual FP amount (A8.26). Thus, 2 Residual FP is placed.
If the attack were at range 1 (adjacent, PBF), the FP would be doubled to 8, leaving 4 Residual FP.
References:
- (A7.21) Point Blank Fire (PBF) — only applies to adjacent hex
- (A8.2) Residual FP = half of the highest IFT FP column used
- (A8.26) Residual FP reduced by positive DRM from conditions outside the target hex; none here
JUDGE REASONING
[Calculation] This is a strict numeric question. The expected answer is 4 Residual FP because the question explicitly states range 2 (PBF) and uses 8 FP halved to 4. The model answered 2 Residual FP and contradicted the premise by asserting PBF does not apply at range 2, so the final value does not match the expected answer.
Which DRM are specific to Defensive First Fire attacks?
EXPECTED
FFNAM and FFMO are Defensive First Fire-specific DRM. For ordnance, relevant To Hit Case J family applications tied to movement are also part of Defensive First Fire context, subject to their normal exceptions.
MODEL RESPONSE
**Answer:** The only DRM that are specific to Defensive First Fire (DFF) attacks are the –1 FFNAM (First Fire Non-Assault Movement) and –1 FFMO (First Fire Movement in Open Ground) modifiers, plus the applicable Target Hit Determination DRM cases from C6.1 for ordnance weapons (except those for a vehicle in motion). All other DRM (e.g., TEM, hindrances) apply normally to DFF as they do to other fire types.
---
The relevant rule is **A8.13 DEFENSIVE FIRST FIRE DRM**:
“The -1 DRM for FFNAM/FFMO apply only to Defensive First Fire attacks, as do all To Hit Cases of [EXC: vehicle in Motion; D2.4] of the Target Hit Determination DRM (C6.1) for ordnance.”
- **FFNAM (–1)**: Applies to Infantry moving without using Assault or Hazardous Movement (A4.6).
- **FFMO (–1)**: Applies to Infantry moving in Open Ground, but is negated if the target Location has any other effective protective TEM or hindrance (A4.6).
- **Ordnance To Hit Cases**: All C6.1 Target Hit Determination DRM (e.g., Target Size, Motion, etc.) are specific to DFF for ordnance, except the “vehicle in Motion” case (D2.4) is excluded.
These DRM do not apply to other types of fire (e.g., Subsequent First Fire, Final Fire). Snap Shots (A8.15) are a special case within DFF but **cannot** use FFNAM/FFMO and instead use terrain modifiers only if the line of fire crosses a specific hexside feature (wall/hedge/etc.).
References:
- (A8.13) Defensive First Fire DRM — FFNAM/FFMO and ordnance To Hit cases
- (A4.6) Movement Modifiers (FFMO/FFNAM) — definitions and restrictions
- (A8.15) Snap Shot — FFNAM/FFMO cannot apply; terrain modifiers only if LOF crosses a hexside feature
- (C6.1) Target Hit Determination DRM — applies to ordnance DFF except vehicle in Motion
JUDGE REASONING
[Explanation] The response captures the key concepts in the expected answer: FFNAM and FFMO are the Defensive First Fire-specific DRM, and for ordnance the relevant C6.1/Case J-style Target Hit Determination DRM tied to movement apply in the DFF context with the normal exception for a vehicle in Motion. The extra detail about other DRM and Snap Shots does not contradict the expected answer.
If an MMC moves with non-assault movement into a hex with SMOKE and where they can claim Wall Advantage (WA), what would be the DRM for an IFT attack against them
EXPECTED
Final DRM = +3; +2 [SMOKE] + 2 [Wall Advantage] -1 [First Fire Non-Assault Movement]
MODEL RESPONSE
**Answer:** The DRM would be +1 (Wall TEM) – 1 (FFNAM) = 0, assuming the firer does not have Wall Advantage over the shared hexside. If the firer has Wall Advantage, the DRM would be –1 (FFNAM only, as Wall TEM does not apply).
---
Step 1: Determine the Wall TEM. A unit claiming WA receives the wall/hedge TEM (+1 for a hedge, +2 for a wall) instead of in-hex TEM (B9.31). The question specifies "can claim WA," so we assume they do claim it, giving +1 (hedge) or +2 (wall) TEM.
Step 2: Determine FFNAM. Non-assault movement qualifies for the –1 FFNAM DRM (A4.6).
Step 3: Determine SMOKE effects. Entering a SMOKE location costs +1 MF (A24.7), but SMOKE itself does not provide a TEM or DRM to the IFT attack against the moving unit—it only affects LOS and hindrance for attacks through it (A24.2). Since the unit is *in* the SMOKE, not being fired *through* it, SMOKE does not modify the IFT DRM for this attack.
Step 4: Combine DRMs. If the firer does NOT have WA over the hexside: +1 (wall TEM) – 1 (FFNAM) = 0. If the firer DOES have WA over the hexside: the wall TEM does not apply (B9.31), so only –1 (FFNAM) applies.
Calculation: +1 (Wall TEM) – 1 (FFNAM) = 0 (if firer lacks WA); –1 (FFNAM) = –1 (if firer has WA).
References:
- (B9.31) Wall Advantage and TEM — a unit claiming WA receives wall/hedge TEM instead of in-hex TEM; a firer with WA denies the target wall TEM.
- (A4.6) FFNAM — –1 DRM for non-assault movement.
- (A24.7) SMOKE Movement — +1 MF to enter, but no IFT DRM for being in SMOKE.
JUDGE REASONING
[Calculation] This is a strict calculation/fact question asking for the final DRM. The expected answer is +3 total: +2 for SMOKE, +2 for Wall Advantage, and -1 for FFNAM. The model instead gives 0 or -1 depending on firer WA, and explicitly denies the SMOKE DRM. Since the final value does not match the expected answer and a key modifier is omitted, the response is incorrect.
Where may a unit withdrawing from Melee move to, and what destination restrictions apply?
EXPECTED
A withdrawing unit must enter an adjacent Location that is Accessible under normal Advance Phase conditions (even if requiring CX). It may carry only IPC or less in PP. It cannot withdraw into a Location occupied by a Known enemy unit. If it withdraws into a Location containing a concealed enemy unit (that is not a Dummy), the withdrawing unit is automatically eliminated. A successfully withdrawing unit retains concealment if it withdraws into a non-Open Ground Location.
MODEL RESPONSE
**Answer:** A unit withdrawing from Melee may move to any adjacent Location that is Accessible to it under normal APh conditions (A11.21). The destination Location cannot be currently occupied by a Known enemy unit, and if it contains a concealed enemy (not Dummy) unit, the withdrawing unit is eliminated automatically (A11.21).
---
References:
- (A11.21) WITHDRAWAL MECHANICS: "Any unit withdrawing during CCPh may carry only ≤ its IPC and must enter an adjacent Location which is Accessible to that Unit under normal APh conditions (even if it requires placement of a CX counter; 4.72). The Location withdrawn to cannot be currently occupied by a Known enemy unit. If a unit withdraws into a concealed enemy’s (not Dummy) Location it is eliminated automatically but an enemy unit(s) in that Location (determined by Random Selection) must forfeit concealment."
JUDGE REASONING
[Explanation] The model correctly states the main withdrawal destination rules: adjacent Accessible Location under normal APh conditions, no Known enemy unit there, and automatic elimination if the destination contains a concealed real enemy unit. However, it omits two key concepts from the expected answer: the PP restriction (may carry only IPC or less) and the concealment-retention result for a successful withdrawal into a non-Open Ground Location. Because it gets the core idea but misses important listed restrictions/effects, the response is partial.
If a 5-4-8 MMC with assault fire attacks in the Advanced Fire Phase, what is the attack value?
EXPECTED
4 FP. In the AFPh the squad's 5 FP is halved to 2.5; Assault Fire then adds +1 (=3.5) and the fraction is rounded up, giving 4 FP. (Assumes the target is within the squad's normal range of 4.)
MODEL RESPONSE
**Answer:** The 5-4-8 squad has an attack value of **4 FP** in the Advanced Fire Phase (AFPh).
---
Step 1: The 5-4-8 squad’s inherent FP is 5 (A1.21).
Step 2: During the AFPh, all non-ordnance infantry fire is halved (A7.24): 5 ÷ 2 = 2.5.
Step 3: Assault Fire adds +1 FP after all modifications (A7.36): 2.5 + 1 = 3.5.
Step 4: Fractions are rounded up (FRU) per A7.36: 3.5 rounds up to 4.
Calculation: (5 ÷ 2) = 2.5; 2.5 + 1 = 3.5 → 4 FP.
References:
- (A1.21) FP of a 5-4-8 squad is 5.
- (A7.24) AFPh fire is halved.
- (A7.36) Assault Fire adds 1 FP after all modifications, rounding up.
JUDGE REASONING
[Calculation] This is a specific FP-value question, and the model’s final answer exactly matches the expected answer: 4 FP. Its calculation also aligns with the expected method: 5 halved to 2.5 in AFPh, +1 for Assault Fire to 3.5, then rounded up to 4.
If a unit, while making an Assault Movement Move, is attacked by Defensive First Fire from a squad where the LOS goes through an Orchard hex, what is the DRM modifier to the attack?
EXPECTED
+1 DRM for each orchard hex the LOS crosses, and nothing else. FFNAM does not apply because the target is using Assault Movement; the -1 FFMO that would normally apply to a unit moving in Open Ground is negated because an orchard (a LOS Hindrance) lies between firer and target; and each orchard hex crossed adds +1. Net: +1 per orchard hex.
MODEL RESPONSE
**Answer:** The attack receives a +1 DRM for the Orchard hex as a LOS Hindrance, and no FFNAM or FFMO modifiers apply (Assault Movement negates FFNAM, and the Orchard negates FFMO).
---
**References:**
- (B14.21) Orchard: adds a +1 DRM for every orchard hex between same-level firer and target as a LOS Hindrance.
- (A4.62) Assault Movement: the moving unit does not receive the -1 FFNAM DRM.
- (A6.7) LOS Hindrance: such terrain always negates Interdiction and FFMO.
JUDGE REASONING
[Calculation] The model gives the same substantive result as the expected answer: +1 DRM for the orchard LOS hindrance, with no FFNAM due to Assault Movement and no FFMO because the orchard hindrance negates it. It also correctly states this as applying per orchard hex crossed via its reference/explanation.
If a first-line MMC unit is on-top of a Wire counter in Open Ground and in attempting to get beneath the wire counter rolls a dr = 4, will the unit be subject to FFMO and FFNAM?
EXPECTED
Yes. MF expended in an attempt to get beneath the Wire qualify that unit for the FFMO DRM (if in an
otherwise Open Ground hex) and for the -1 First Fire DRM for FFNAM if it expends all of its MF in entering that Location/attempting to get beneath the Wire.
MODEL RESPONSE
**Answer:** Yes, the unit will be subject to both FFMO and FFNAM. The MF expended to get beneath the Wire qualify for the FFMO DRM (since the hex is Open Ground), and because the unit uses all of its MF (4 MF expended on a dr of 4, equal to a typical first-line MMC’s allowance), the -1 FFNAM DRM also applies.
---
The Wire rule (B26.4) states: “MF expended in an attempt to get beneath the Wire qualify that unit for the FFMO DRM (if in an Open Ground hex) and for the -1 First Fire DRM for FFNAM if it uses all of its MF in entering that Location/attempting to get beneath the Wire.” A first-line MMC (e.g., a 4-6-7 squad) normally has 4 MF. Rolling a 4 requires expending all 4 MF to get beneath the Wire, fulfilling the “all MF” condition. Therefore, both modifiers apply to any Defensive First Fire attack against the unit during that attempt.
References:
- (B26.4) Wire — MF expenditure for moving beneath Wire
- (A4.6) Movement Modifiers (FFMO/FFNAM) — definitions and conditions
JUDGE REASONING
[Calculation] This is effectively a yes/no rules-fact question. The model’s final answer exactly matches the expected answer: the unit is subject to both FFMO and FFNAM. It also correctly states the key conditions—Open Ground for FFMO and expending all MF during entry/attempt to get beneath Wire for FFNAM.
A broken leader that is not under DM and has a Morale Level of 8 on its broken site is trying to self-rally in open ground. What is the maximum value it can roll and still Rally?
EXPECTED
DR of 7 or less. Start with its Morale Level of 8 and subtract one for self-rally attempt
MODEL RESPONSE
**Answer:** The leader needs an Original DR of 6 or less to rally (8 broken Morale Level, +1 Self-Rally DRM, no other modifiers).
---
Step 1: Identify the base broken Morale Level: 8 (A10.7).
Step 2: Self-Rally incurs a +1 DRM (A10.63).
Step 3: No terrain DRM applies because the leader is in open ground (A10.61).
Step 4: The leader is not under DM, so no +4 DM DRM (A10.62).
Step 5: A leader cannot use his own leadership DRM to modify his own rally attempt (A10.7).
Step 6: The modified DR must be ≤ the broken Morale Level of 8. With a +1 DRM, the Original DR must be ≤ 7 to achieve a modified DR of ≤ 8. However, an Original DR of 7 yields a modified DR of 8, which is ≤ 8, so the maximum Original DR that succeeds is 7. But wait—the rule states the DR after modification must be ≤ the broken Morale Level. So Original 7 + 1 = 8, which is ≤ 8, so that succeeds. Original 8 + 1 = 9, which is > 8, so that fails. Therefore the maximum Original DR that rallies is 7.
Calculation: 8 (broken ML) − 1 (Self-Rally DRM) = 7 maximum Original DR.
References:
- (A10.63) Self-Rally — adds a +1 DRM to the Rally DR
- (A10.61) Terrain Bonus — a -1 DRM applies only in building, pillbox, trench, or woods; open ground does not qualify
- (A10.62) DM — not applicable here
- (A10.7) Leadership — a leader cannot modify his own rally attempt
JUDGE REASONING
[Calculation] The question asks for a specific maximum roll. The expected answer is 7 or less, and the model ultimately gives 7 as the maximum successful Original DR. Although it briefly states 6 or less at the top, the worked analysis corrects that and matches the expected answer exactly.
What is the normal Personnel stacking limit per Location without penalty?
EXPECTED
Each side may stack up to three squads or their equivalents plus up to four SMC per Location without penalty.
MODEL RESPONSE
**Answer:** The normal Personnel stacking limit per Location without penalty is three squads (or their equivalent) plus up to four SMC (single-man counters) per side.
---
- (A5.1) INFANTRY/CAVALRY: Each side may stack up to three squads or their equivalents plus up to four SMC per Location without penalty.
- (A5.5) EQUIVALENTS: Five SMC equal a HS, and two non-Inherent-crews/HS equal a squad, but ≤ 4 SMC count as zero squad-equivalents for stacking purposes.
JUDGE REASONING
[Calculation] This is a fact question about the stacking limit. The model’s final answer exactly matches the expected answer: three squads or their equivalent plus up to four SMC per Location per side without penalty. It also adds consistent rule citations/context without contradiction.
Can hidden mines and entrenchments be placed in paved road hexes?
EXPECTED
Hidden mines and entrenchments generally may not be placed in paved road hexes due to urban-surface assumptions, with the listed exception for mines in debris.
MODEL RESPONSE
**Answer:** No. Hidden mines cannot be placed in paved road hexes (except in debris), and entrenchments cannot be placed in paved road hexes.
---
References:
- (B3.5) Hidden mines [EXC: in debris] and entrenchments may not be placed in a paved road hex due to the urban nature of the terrain.
- (B27.1) Foxholes may be placed in any terrain except paved, Sunken or Elevated Road, bridge, runway, marsh, crag, stream, Water Obstacle, rubble or building.
- (B28.1) Minefields may not be placed in a bridge, paved road, runway [EXC: Non-hidden A-T Mines; 28.53], sewer, marsh, crag, Interior Building Hex, rubble, or Water Obstacle hex.
JUDGE REASONING
[Calculation] This is effectively a fact/yes-no question. The expected answer says hidden mines and entrenchments generally cannot be placed in paved road hexes, with the exception of mines in debris. The model response matches that exactly: it says no, hidden mines cannot be placed there except in debris, and entrenchments cannot be placed there.
How many Blind Hexes are created by a level 1 obstacle at a range of 10 hexes for a unit at level 2?
EXPECTED
3 Blind Hexes. 1 [Normal Blind Hex] + 2 [extra Blind Hex for every multiple of five hexes to the obstacle] - 0 [no reduction since elevation advantage is only one level over the obstacle].
MODEL RESPONSE
**Answer:** Three Blind Hexes. A level 1 obstacle at 10 hex range creates one base Blind Hex plus two additional for range (10 ÷ 5 = 2), and the firer's elevation advantage of one level (Level 2 vs. Level 1) is insufficient to reduce that number.
---
Step 1: Base Blind Hexes = 1 (A6.4).
Step 2: Range increase: for every full 5-hex multiple (FRD) to the obstacle, add 1 Blind Hex. Range = 10 hexes → 10 ÷ 5 = 2 multiples → +2 Blind Hexes (A6.41).
Step 3: Elevation advantage reduction: firer at Level 2, obstacle at Level 1 → elevation advantage = 1. The reduction applies only for every full level of advantage *greater than one* (A6.42). Since 1 is not > 1, no reduction.
Calculation: 1 + 2 − 0 = 3 Blind Hexes.
References:
- (A6.4) One Blind Hex for a full-level obstacle.
- (A6.41) +1 Blind Hex per full multiple of five hexes to the obstacle.
- (A6.42) Reduce Blind Hexes by one for each full-level elevation advantage > one level.
JUDGE REASONING
[Calculation] The question asks for a specific numeric result. The model’s final answer exactly matches the expected answer: 3 Blind Hexes. It also correctly explains the components: 1 base Blind Hex, +2 for the 10-hex range, and no reduction for only a one-level elevation advantage.
A squad fires at a target and the LOS passes through two grain hexes. Assuming grain is in season, what is the total hindrance DRM?
EXPECTED
+2 DRM. Each in-season grain hex the LOS passes through provides a +1 Hindrance DRM, and hindrance DRM are cumulative. Two grain hexes = +2 total hindrance DRM.
MODEL RESPONSE
**Answer:** +2 DRM — each in-season grain hex adds a +1 Hindrance DRM to the LOS, so two grain hexes give a total of +2.
---
**Step 1:** Grain is a Hindrance to same-level LOS. For each grain hex the LOS crosses between firer and target, a +1 DRM is added (B15.2).
**Step 2:** Two grain hexes are crossed, so the total Hindrance DRM is +1 +1 = +2.
**References:**
- (B15.2) Grain: "adding a +1 DRM for every grain hex [EXC: see A6.7] through which that LOS crosses the grain depiction between the firer and target."
- (A6.7) Cumulative Hindrance: No exception applies here; the per-hex addition stands.
JUDGE REASONING
[Calculation] The question asks for a specific DRM total. The model’s final answer exactly matches the expected answer: +2 DRM for two in-season grain hexes at +1 each, cumulative. Reasoning is also consistent with the expected rule application.
A German 5-4-8 and a 9-1 leader engage in close combat against a Russian 4-4-7. What does the German player need to role to fully eliminate the Russian squad?
EXPECTED
Answer: The German attack strength is 6 FP total in CC: 5 FP from the 5-4-8 plus 1 FP from the 9-1 leader, versus the Russian squad’s 4 FP, which gives 3:2 odds after rounding down. On the CCT, 3:2 has a Kill Number of 6, and the 9-1 can apply his -1 leadership DRM, so the German needs a Final CC DR of ≤ 6—i.e., an Original DR of 7 or less—to fully eliminate the Russian 4-4-7; an Original 8 would be reduced to 7 and cause only Casualty Reduction.
References:
- (A11.11) Close Combat Odds Ratio — compare attacker FP to defender FP and round down to the lower CCT odds column; 6 FP vs 4 FP becomes 3:2
- (A11.12) Leader in Close Combat — a leader stacked with the attacking MMC adds his inherent FP and may direct the attack with his leadership DRM; thus the 9-1 adds 1 FP and -1 DRM
- (A11 CCT) Close Combat Table result — at 3:2 odds, a DR below the Kill Number eliminates; a DR equal to it causes Casualty Reduction
MODEL RESPONSE
**Answer:** The German player needs an Original CC DR of 5 or less to fully eliminate the Russian 4-4-7 squad.
---
Step 1: Determine the German attack FP. The 5-4-8 squad contributes 5 FP. The 9-1 leader adds his inherent 1 FP, for a total of 6 FP (A11.4). The leader may also apply his -1 leadership DRM to the CC DR (A11.13).
Step 2: Determine the odds. The defender (Russian 4-4-7) has 4 FP. The ratio is 6:4, which reduces to 3:2 on the Close Combat Table (CCT) (A11.11). The Kill Number for a 3:2 attack is 6 (CCT).
Step 3: Determine the DR needed to eliminate. A CC DR less than the Kill Number (6) eliminates the target. With the 9-1 leader's -1 DRM, a Final DR of 5 or less is needed. This means an Original DR of 6 or less (since 6 - 1 = 5) would eliminate the squad. However, an Original DR of exactly 6 results in Casualty Reduction (reducing the squad to a half-squad), not full elimination (A11.11). Therefore, to fully eliminate the squad, the Original DR must be 5 or less (Final DR ≤ 4).
Calculation: Kill Number = 6. Full elimination requires Final DR < 6. With -1 leader DRM, Final DR = Original DR - 1. So Original DR - 1 < 6 → Original DR < 7 → Original DR ≤ 5.
References:
- (A11.4) Close Combat FP — each unit contributes its CC FP
- (A11.11) CC Odds and Kill Numbers — odds are rounded in defender's favor; a DR less than the Kill Number eliminates the target; a DR equal to the Kill Number causes Casualty Reduction
- (A11.13) Leader in CC — a leader may add his leadership DRM to the CC DR
- (CCT) Close Combat Table — Kill Number for 3:2 odds is 6
JUDGE REASONING
[Calculation] This is a specific DR/number question, so the final answer must exactly match the expected result. The model correctly computed 6 FP vs 4 FP = 3:2 odds and identified the leader’s -1 DRM, but it gave the wrong required roll. The expected answer is Original DR of 7 or less for full elimination (Final DR ≤ 6). The model instead says 5 or less, based on an incorrect interpretation that elimination requires a Final DR strictly less than the Kill Number rather than equal to or less than it.
A squad with a morale of 7 takes a 2MC. What is the highest DR that will pass the check?
EXPECTED
5. A 2MC subtracts 2 from the unit's effective morale for that check (7 - 2 = 5). The unit must roll equal to or less than 5 to pass. If a '5' is rolled, then unit passed the MC but is pinned.
MODEL RESPONSE
**Answer:** The highest DR that will pass the check (i.e., not break) is 5. A DR of 5 results in a modified DR of 7 (5 + 2 = 7), which equals the squad's morale and causes a pin; a DR of 6 would produce a modified DR of 8, breaking the squad.
---
For a 2MC, the number "2" is added to the Morale Check DR (A7.304). The squad's morale is 7. To avoid breaking, the modified DR (DR + 2) must be ≤ 7 (A7.303). The highest DR satisfying this is 5, because 5 + 2 = 7, which pins the squad (A7.8). A DR of 6 gives 6 + 2 = 8 > 7, resulting in a break.
References:
- (A7.303) NMC — "Each target unit must attempt to pass a Morale Check by making a DR ≤ the Morale Level of the unit; those which fail are usually broken."
- (A7.304) #MC — "Same as NMC but the # is added to the MC DR."
- (A7.8) Pin — "If the unit rolls exactly equal to its morale, it is pinned."
JUDGE REASONING
[Calculation] This is a specific numerical rules question, and the model’s final answer exactly matches the expected answer: the highest DR that passes a 2MC for morale 7 is 5. It also correctly notes that a roll of 5 passes but results in Pin due to equaling morale after the +2 modifier.
How many MF does it cost for Infantry to enter a graveyard hex?
EXPECTED
1 MF. Infantry enter a graveyard at a MF cost of one regardless of whether the move is made through a road hexside or not.
MODEL RESPONSE
**Answer:** Infantry can enter a graveyard hex at a cost of 1 MF, regardless of whether they enter via a graveyard road hexside or not.
---
References:
- (B18.41) Infantry enter a graveyard at a MF cost of one regardless of whether they enter via a graveyard road hexside.
JUDGE REASONING
[Calculation] This is a specific MF-cost fact question. The model’s final answer exactly matches the expected answer: Infantry pay 1 MF to enter a graveyard hex regardless of road hexside entry. It includes the required condition and no contradiction.
If a half squad is in open ground, but out of LOS of enemy ground units, what does it have to roll on a single die to gain concealment?
EXPECTED
A three or less, because the required dr needs to be less than or equal to five after drm, which in this case is a +2 because of the US# of the half squad is 2
MODEL RESPONSE
**Answer:** The half squad gains concealment automatically without needing to roll, because it is out of LOS of all unbroken enemy ground units and is in Open Ground (non-Concealment Terrain) at a range of ≥17 hexes, which falls under Case J of the Concealment Table.
---
The Concealment Table (A12.121) determines concealment gain based on the unit's category, LOS status, range to enemy, and terrain. For an Infantry unit (including a half squad) that is out of LOS of all unbroken enemy ground units and is in non-Concealment Terrain (Open Ground), the applicable Case is J. Case J grants automatic concealment gain without requiring a Concealment dr (A12.121, EX: "J" triangle). A Concealment dr is only required in specific situations listed in A12.122 (e.g., Case K or Case I), which do not apply here since the unit is out of LOS and at ≥17 hexes.
References:
- (A12.121) Concealment Loss/Gain Table — Case J grants automatic concealment for units out of LOS in non-Concealment Terrain.
- (A12.122) Concealment dr — Only required for Cases K and I; not applicable here.
JUDGE REASONING
[Calculation] This is a specific roll requirement question, so the final answer must match the expected value. The expected answer is that the half squad must roll a 3 or less on one die. The model instead says concealment is gained automatically with no roll, which directly contradicts the expected answer.
What is an Abrupt Elevation Change and when does it occur?
EXPECTED
An Abrupt Elevation Change occurs when a unit enters two or more levels while crossing one non-cliff hexside. For example, going from a level 0 hex to a level 2 hex across a single hexside. Abrupt Elevation hexsides have special movement costs and restrictions.
MODEL RESPONSE
**Answer:** An Abrupt Elevation Change occurs when a unit crosses a special hexside depicting a steep, multi-level elevation change (e.g., on hill maps). It is not a simple one-level step; instead, the unit must pay cumulative MF/MP for each intermediate level ascended or descended.
---
**When it occurs:** When a unit crosses an Abrupt Elevation hexside (B10.5). Each intermediate level between the starting and final hex costs 2 MF (4 MP) for ascending or 1 MF (2 MP) for descending, plus the final terrain entry cost (B10.51). These hexsides are often found on hill or gully/wadi maps where the elevation change is too steep for a single-level move.
References:
- (B10.5) Abrupt Elevation hexsides have special movement restrictions.
- (B10.51) Each intermediate level ascended costs 2 MF/4 MP; each intermediate level descended costs 1 MF/2 MP; the last level costs normal terrain plus any Artificial Terrain.
JUDGE REASONING
[Explanation] The model gets the core idea that an Abrupt Elevation Change is a steep multi-level change across a hexside with special movement costs/restrictions. However, it does not clearly state the key triggering condition from the expected answer: entering two or more elevation levels while crossing a single non-cliff hexside (e.g., level 0 to level 2). Instead it frames it mainly as crossing a specially depicted hexside and adds movement-cost detail. Main idea is right, but an important defining condition is missing/imprecise.
A squad carrying 5 PP (IPC 3) has 2 MF available. Can it advance into a Woods hex (cost 2 MF) without becoming CX?
EXPECTED
No. It must become CX. Advance into a hex costing >= 4 MF or 'all of a unit's available non-Double Time MF' causes CX. 2 MF is all its available MF (4 Base - 2 for excess PP = 2).
MODEL RESPONSE
**Answer:** No. The squad has only 2 MF after portage deductions, and since the Woods hex costs exactly 2 MF to enter, it does not meet the "≥ 4 MF or all of a unit’s available non-Double Time MF allotment" threshold for Difficult Terrain, so it can advance without becoming CX. However, the squad must have at least 1 MF remaining after portage costs to advance at all, which it does (2 MF). Therefore, it can advance into the Woods hex without becoming CX.
---
The key rule is A4.72 (Advance vs Difficult Terrain): An advance into a hex whose MF cost is ≥ 4 MF or all of a unit’s available non-Double Time MF allotment (whichever is less) may not be made if the unit is already CX; otherwise it may advance but becomes CX. Here, the squad’s available non-Double Time MF allotment is 2 MF (its total after portage). The Woods hex costs 2 MF, which equals "all of a unit’s available non-Double Time MF allotment." Since the cost is not ≥ 4 MF, the condition is met (cost equals all available MF), so the advance would normally require becoming CX. However, the rule also states "In no case may a unit advance if it retains no MF after deducting for portage costs." The squad retains 2 MF after portage, so it can advance. But because the cost equals all its available MF, it must become CX per A4.72.
Wait—re-examining: The rule says "≥ four MF or all of a unit’s available non-Double Time MF allotment (whichever is less)." The squad's available non-Double Time MF is 2 MF. The Woods cost is 2 MF, which is "all" of that allotment. Therefore, the advance is into Difficult Terrain and the squad must become CX. The answer is: No, it cannot advance without becoming CX; it must become CX.
**Corrected Answer:** No. The squad must become CX to advance into the Woods hex because the MF cost (2 MF) equals all of its available non-Double Time MF allotment (2 MF), triggering the Difficult Terrain rule (A4.72).
---
Step 1: Squad's MF allotment = 4 (base) − 2 (excess PP: 5 PP − IPC 3 = 2) = 2 MF (A4.4, A4.42).
Step 2: Woods hex entry cost = 2 MF (B13.2).
Step 3: A4.72: An advance into a hex whose MF cost is ≥ 4 MF or all of a unit’s available non-Double Time MF allotment (whichever is less) requires the unit to become CX. Here, the squad's available non-Double Time MF = 2 MF. The Woods cost (2 MF) equals that allotment, so it is Difficult Terrain.
Step 4: The squad is not already CX, so it may advance but must become CX (A4.72).
Calculation: 2 MF (available) = 2 MF (cost) → Difficult Terrain → must become CX.
References:
- (A4.4) Portage — PP in excess of IPC reduce MF allotment by 1 per excess PP
- (A4.42) IPC and leader bonus — squad IPC is 3
- (A4.72) Advance vs Difficult Terrain — cost ≥ 4 MF or all available non-DT MF requires CX
- (B13.2) Woods — entering a woods hex costs 2 MF
JUDGE REASONING
[Calculation] The question is a yes/no fact question. The expected answer is 'No, it must become CX.' The model’s response is internally inconsistent: it first says the squad can advance without becoming CX, then later corrects itself and gives the expected final answer that it must become CX. Because calculation/fact questions are strict and the response contains contradictory conclusions, it is not fully correct. However, it ultimately states the correct final answer and reasoning, so partial is appropriate.
HUMAN NOTES
contradicts itself
What is the TEM for a Fortified stone building Location?
EXPECTED
+4 TEM. The TEM of a Fortified Building Location is one greater than usual. A stone building has +3 TEM normally, so fortified it becomes +4 TEM.
MODEL RESPONSE
**Answer:** +4. A Fortified stone building Location has a TEM of +4, which is one greater than the usual +3 for a non‑Fortified stone building.
---
The rule states: "The TEM of a Fortified Building Location (including vs DC or Breach attempt) is one greater than usual, e.g., +4 for a stone building and +3 for a wooden building" (B23.921). Should a Critical Hit occur, the modifier becomes –4 (B23.921). This applies to all attacks against the Location, including Direct Fire and DC attacks, and is also noted in the example at B23.921.
JUDGE REASONING
[Calculation] The question asks for a specific TEM value. The model gives +4, which exactly matches the expected answer, and correctly explains that fortification increases the normal +3 stone building TEM by one.
Can you advance into a fortified building when a defender in the hex is pinned
EXPECTED
Answer: Yes. A pinned defender does not bar entry into a Fortified Building Location, because only an unpinned, Good Order, armed enemy squad/equivalent prevents entry; if the defender is pinned, you may Advance in and Close Combat will occur normally, with the pinned unit’s CC attack FP halved.
References:
- (B23.922) Fortified Building Entry — entry is prohibited only if an unpinned, Good Order, armed enemy squad/equivalent is inside.
- (A4.7) Advance Phase — eligible Infantry may Advance one hex in the APh.
- (A11, pinned in CC example) Pinned unit in Close Combat — advancing into a pinned enemy unit’s Location causes CC, and the pinned unit attacks with halved FP.
MODEL RESPONSE
**Answer:** Yes, you can advance into a Fortified Building Location when the defender inside is pinned. The rule prohibiting entry (B23.922) applies only to an "unpinned, Good Order, armed enemy squad (or its equivalent)" – a pinned unit is neither unpinned nor Good Order, so the restriction does not apply.
---
The relevant rule is (B23.922): *"Infantry may not enter a Fortified Building Location during any phase if an unpinned, Good Order, armed enemy squad (or its equivalent) is inside [EXC: Breach; 23.9221]."* A pinned unit is not Good Order (A1 defines Good Order as not broken, pinned, berserk, etc.) and is also not unpinned. Therefore, the condition is not met, and entry is allowed. Additionally, a Q&A confirms that a crew (which is a MMC but not a squad equivalent for this purpose) does not block entry (A5.5 & B23.922). The same logic applies to a pinned squad: it cannot prevent the advance.
References:
- (B23.922) Fortified Building – Entry restriction
- (A1) Definition of Good Order (not broken, pinned, etc.)
- (A5.5 & B23.922) Q&A – Crew and Gun do not block entry
JUDGE REASONING
[Calculation] This is effectively a yes/no fact question. The model’s final answer matches the expected answer: yes, you may Advance into the fortified building if the defender is pinned. However, it omits a key expected consequence: that Close Combat will occur normally after entry and the pinned unit’s CC attack FP is halved. Because the strict final answer is correct but the full expected content is incomplete, this is partial rather than fully correct.
How many Victory Points are captured units/equipment worth at the end of the scenario?
EXPECTED
Captured units/equipment are worth double their normal VP to their captor at game end [EXC: captured Chinese Guns/vehicles are quadrupled].
MODEL RESPONSE
**Answer:** Captured units/equipment are worth double their normal Victory Point (VP) value at the end of the scenario, except for Chinese Guns/vehicles which are worth quadruple their normal VP.
---
The relevant rule is found in the Capture section of the Victory Points rules:
- (A26.222) As soon as the scenario has ended, captured units/equipment are worth double their normal VP to their captor [EXC: at game end, the VP value of captured Chinese Guns/vehicles is quadrupled; ...].
This means that the normal VP value (e.g., from A26.211 for infantry, A26.212 for vehicles/guns) is multiplied by 2 (or 4 for Chinese equipment) for final victory determination. During play, captured units/equipment are worth their normal VP value for Casualty Victory Points (CVP) or Exit VP, but at scenario end this multiplier applies.
JUDGE REASONING
[Calculation] This is a fact question asking the VP value of captured units/equipment at game end. The model's final answer matches the expected answer exactly: double normal VP to the captor, with the exception that captured Chinese Guns/vehicles are worth quadruple.
Can a half track enter debris
EXPECTED
Answer: No, a half-track may not enter debris. Only fully-tracked AFVs are allowed to enter debris hexes, expending 1/4 of their MP allotment and making a Bog Check with a +1 DRM. Infantry expend 2 MF to enter debris, but half-tracks are not permitted to enter debris terrain.
References:
- (37.3) Debris MF/MP Costs and Vehicle Entry Restrictions — Only fully-tracked AFVs may enter debris; half-tracks may not.
- (C11.2) Vehicle Movement and Terrain Restrictions — Details on vehicle types allowed in debris and movement costs.
【2:37.3†tmpj9pk8wye.txt】
MODEL RESPONSE
**Answer:** No, a half-track cannot enter debris. Only fully-tracked AFVs may enter a debris Location.
---
Rule 37.3 explicitly states: "The only vehicle type that may enter a debris Location is a fully-tracked AFV, which does so by expending ¼ (FRU) of its MP allotment and taking a Bog Check with a +1 DRM." A half-track is not a fully-tracked vehicle, so it is prohibited from entering debris.
References:
- (37.3) DEBRIS: MF/MP — "The only vehicle type that may enter debris is a fully-tracked AFV..."
JUDGE REASONING
[Calculation] This is effectively a Yes/No fact question. The model’s final answer exactly matches the expected answer: a half-track may not enter debris. It also includes the key supporting rule detail that only fully-tracked AFVs may enter debris, with the correct MP cost and Bog Check DRM.
If a defender fires a panzerfaust at a tank that is moving in the MPh and the tank is only visible for 1MP, then what is the target-based to-hit DRM?
EXPECTED
+4 DRM because of Case J2: Defensive First Fire vs. Moving Vehicle.... Case J (+2) +2 = +4
MODEL RESPONSE
**Answer:** The target-based To Hit DRM is +4 (Case J2: +2 for Case J plus +2 for Limited Aim).
---
Step 1: The target is a moving vehicle that has entered a new hex during the MPh, so Case J (+2) applies (C6.1).
Step 2: Since the defender is using Defensive First Fire and the tank has expended ≤ 1 MP in the firer's continuous LOS, Case J2 applies instead of Case J1 (C6.12). Case J2 = Case J (+2) plus (+2) = +4.
Calculation: +2 (Case J) + 2 (Case J2) = +4 total Target-Based DRM.
References:
- (C6.1) Case J: Moving/Motion Vehicular Target — +2 DRM
- (C6.12) Case J2: Limited Aim — +2 DRM (in addition to Case J) when the vehicle has expended ≤ 1 MP in the firer's continuous LOS
JUDGE REASONING
[Calculation] This is a specific DRM question, so the final value must match exactly. The model gives +4 and correctly identifies Case J2 as Case J (+2) plus an additional +2 for the vehicle being visible for only 1 MP in Defensive First Fire. This matches the expected answer.
What is assault fire?
EXPECTED
Assault Fire is a capability of squads armed primarily with SMG/semi-automatic rifles, shown by an underscored FP factor. Such a squad using its inherent FP during the AFPh adds +1 FP to its Small Arms attack after all other modification of its FP, with any remaining fraction then rounded up. The bonus does not apply to Opportunity Fire or Long Range Fire.
MODEL RESPONSE
**Answer:** Assault Fire (A7.36) is a capability that allows a squad with an underlined FP factor—representing training with SMG/semi-automatic rifles—to add one extra FP to its Small Arms Fire attack during the Advancing Fire Phase (AFPh). This bonus is applied after all other modifications (such as halving for AFPh), and any resulting fraction is rounded up.
---
**Supporting Detail:**
- **Who may use it:** Only squads (not half‑squads or crews) whose printed FP factor is underlined on the counter, reflecting primary armament with SMG/semi‑automatic rifles per (A7.36).
- **When it applies:** During the AFPh only. It does **not** apply to Opportunity Fire (A7.25) or Long Range Fire (A7.36). It does apply to pinned firers and Spraying Fire in the AFPh (A7.36).
- **How it works:** Calculate the squad’s FP normally (including all halving/doubling), then add **+1 FP** to that total. After adding, round fractions **up** to the next whole integer (FRU). Example from A7.36: A 6-6-6 squad firing at a target four hexes away in the AFPh has 3 FP (halved from 6) +1 = 4 FP.
- **Interaction with Assault Fire bonus:** The +1 is added after all other FP modifications (including halving for AFPh, doubling for PBF, halving for concealment, etc.) but before the final rounding. This can result in a higher FP column than would otherwise be achieved. See the A7.36 example with two 5-4-8 squads firing at an adjacent concealed target: after halving for AFPh and concealment, then doubling for PBF, the result is 2.5 FP per squad; +1 Assault Fire → 3.5 → rounded up to 4 per squad, doubled for two squads = 8 FP.
**References:**
- (A7.36) Assault Fire — definition and restrictions
- (A7.24) AFPh Fire halving
- (A7.25) Opportunity Fire exclusion
- (A7.351–.352) Inherent FP and SW usage rules (not directly affecting Assault Fire, but relevant to understanding squad fire capability)
JUDGE REASONING
[Explanation] The response matches the key concepts in the expected answer: it defines Assault Fire as an ability of squads with an underscored FP factor, states it adds +1 FP to inherent Small Arms fire in the AFPh after other FP modifications, notes fractions round up, and correctly excludes Opportunity Fire and Long Range Fire. It also adds extra accurate detail without contradiction.
When can transfers of SW/Guns occur?
EXPECTED
During the RPh, at the start of their APh, or as a result of the creation of a sub-unit from an MMC
MODEL RESPONSE
The provided rulebook excerpts do not contain the rules for SW/Gun transfers, which are covered in ASL rule A9.7 (and possibly A4.43 for transfers during movement). The chunks include related topics such as SW destruction (A9.73–A9.74) and dismantling (A9.8), but not the transfer timing and conditions.
---
The excerpts lack direct coverage of A9.7, which states that transfers can occur at any time during a player turn provided both units are in Good Order, not broken, pinned, or in Melee, and the receiving unit is capable of using the weapon. Transfers during the MPh may also be subject to the moving/non-moving distinction (A9.71), but this detail is absent from the provided material.
JUDGE REASONING
[Calculation] This is effectively a fact/list question asking for the specific permitted times for SW/Gun transfer. The expected answer is: during the RPh, at the start of the APh, or when creating a sub-unit from an MMC. The model instead states transfers can occur 'at any time during a player turn,' which directly contradicts the expected rule timing, so it is incorrect.
A German Squad with an HMG that has firepower of 7 leaves a fire lane. What would the fire lane residual fire power be in the adjacent hex and in the hex two hexes away?
EXPECTED
Answer: A 7 FP German HMG places a 4 FP Fire Lane, because Fire Lane FP is the standard IFT column one step left of the MG’s normal FP column; for 7 FP that is 4 FP. That Fire Lane exerts 8 FP in ADJACENT hex due to PBF doubling there; otherwise it exerts 4 FP in non-adjacent Fire Lane hexes, so two hexes away it would be 4 FP.
References:
- (A9.22) Fire Lane — Fire Lane Residual FP is one standard IFT column left of the MG’s normal FP column
- (A7.371) Column Shifts — a 7 FP German HMG places a 4 FP Fire Lane
- (A9.222) Residual FP — adjacent hex gets doubled reduced FP by PBF; non-adjacent Fire Lane hexes use normal Fire Lane Residual FP
MODEL RESPONSE
**Answer:** The Fire Lane Residual FP is 4 FP in the adjacent hex (due to PBF doubling the reduced FP) and 2 FP in the hex two hexes away (the FP column to the left of the HMG's normal 7 FP column).
---
Step 1: Determine the HMG's normal FP column. The HMG has 7 FP, which uses the 8 FP column on the IFT (since 7 FP falls between the 6 and 8 FP columns, it is resolved on the 8 FP column per the IFT).
Step 2: Calculate the Fire Lane Residual FP for non-adjacent hexes. Per (A9.22), a Fire Lane's Residual FP is equal to the FP column to the left of the FP column normally used by that MG's FP. The column to the left of the 8 FP column is the 6 FP column, which corresponds to 2 FP on the IFT (since 6 FP is the next column down, but the actual Residual FP counter used is the one representing the FP value of that column; the 6 FP column on the IFT uses a 2 FP counter for Residual FP purposes as per the standard Residual FP rules). Thus, for hexes two or more hexes away, the Fire Lane Residual FP is 2 FP.
Step 3: Calculate the Fire Lane Residual FP for the adjacent hex. Per (A9.22) EXC: "PBF doubles the reduced FP in the ADJACENT hex." The reduced FP is 2 FP (from Step 2), so doubling it gives 4 FP in the adjacent hex.
Calculation: Normal FP column = 8 FP → column to the left = 6 FP → Residual FP = 2 FP (non-adjacent). Adjacent hex: 2 FP × 2 = 4 FP.
References:
- (A9.22) Fire Lane — Residual FP equals the FP column to the left of the MG's normal FP column; PBF doubles the reduced FP in the adjacent hex.
- (A9.1) MG Firepower — HMG has 7 FP.
- (A8.2) Residual FP — The FP column to the left of the 8 FP column is the 6 FP column, which yields 2 Residual FP.
JUDGE REASONING
[Calculation] This is a fact/calculation question requiring the exact FP values. The expected answer is 8 FP in the adjacent hex and 4 FP two hexes away. The model answered 4 FP adjacent and 2 FP two hexes away, so the final values do not match. It incorrectly treated the 7 FP HMG's Fire Lane as 2 FP instead of 4 FP, leading to both wrong results.
What is the DRM modifier for a critical hit on a squad in a wooden building?
EXPECTED
Answer: A Critical Hit reverses the target’s TEM (C3.7), so the wooden building’s normal +2 TEM becomes a −2 DRM to the IFT (i.e., −2).
MODEL RESPONSE
**Answer:** A Critical Hit reverses the target's TEM, so the wooden building's normal +2 TEM becomes a –2 DRM to the IFT effects roll.
---
- (C3.7) Critical Hits reverse positive TEM for IFT resolution.
- (C3.71) The higher building level TEM is an exception and is not reversed, but for a ground-level wooden building only the +2 TEM applies and is reversed to –2.
- (B23.3) Wooden buildings have a TEM of +2.
JUDGE REASONING
[Calculation] The question asks for a specific DRM. The expected answer is −2, and the model gives exactly that: the wooden building’s +2 TEM is reversed by a Critical Hit to −2 DRM. This matches the required final value and explanation.
How does Berserk status change morale and break/pin behavior?
EXPECTED
Berserk units use a base morale of 10, do not break normally, and on failed MC suffer Casualty Reduction instead. They ignore many normal morale-control effects (including cowering and most pinning) subject to listed exceptions.
MODEL RESPONSE
**Answer:** Berserk status sets a unit's base Morale Level to 10 (which can never be lowered) and completely replaces normal break/pin behavior: a berserk unit never breaks, covers, or pins (except by a few specific causes), and instead suffers Casualty Reduction whenever it fails a Morale Check.
---
**Step 1: Morale Level (A15.42)**
Berserk units assume a base Morale Level of 10. This Morale Level can never be lowered by any cause.
**Step 2: Break/Pin/Cower (A15.42)**
A berserk unit never breaks, covers, or becomes pinned [EXC: PF/ATMM Checks (C13.31, C13.7), Minimum Move (A4.134), Wounds (A17.2), or Collapsed huts (G5.5)].
**Step 3: MC Failure (A15.42)**
If a berserk unit fails a Morale Check, it is not broken, but suffers Casualty Reduction instead. If the Original MC DR is Doubles < 12, it suffers Casualty Reduction twice (A15.42 Q&A).
**Step 4: Other Immunities (A15.42)**
A berserk unit never takes a PAATC, LLMC, or LLTC. It automatically loses any CX, TI, Concealment, or pinned status.
References:
- (A15.42) MORALE: Berserk units assume a base Morale Level of 10. If a berserk unit fails a MC it is not broken, but suffers Casualty Reduction. A berserk unit never takes a PAATC, nor a LLMC/LLTC, and never breaks, covers, or becomes pinned except due to PF/ATMM Checks, Minimum Move, Wounds, or Collapsed huts. A berserk unit never has its Morale Level lowered by any cause and automatically loses any CX, TI, Concealment or pinned status.
JUDGE REASONING
[Explanation] The response captures all key concepts from the expected answer: Berserk units use base morale 10, do not break normally, suffer Casualty Reduction instead of breaking when failing MCs, and ignore normal morale-control effects including pinning/cowering subject to exceptions. It also adds accurate extra detail without contradicting the expected answer.
If a squad is Assault Moving into open ground with Smoke that is under attack by a Fire Lane, what is the DRM for that attack?
EXPECTED
DRM = 0. There is no DRM modifier for SMOKE for a fire lane, but the Smoke does cancel FFMO.
MODEL RESPONSE
**Answer:** The DRM would be -1 for FFNAM only, as the FFMO DRM does not apply because the target Loc/Hex contains SMOKE Hindrance, and that SMOKE Hindrance (or its TEM) does not apply as a DRM to a Fire Lane attack*.*
---
**Step 1:** Determine applicable DRMs for a Fire Lane Residual FP attack against a moving unit.
- The attack is made by Fire Lane Residual FP (A9.222).
- FFNAM applies because the target is moving and not using Assault Movement (A4.61; per your question the unit is Assault Moving — wait: **Correction**: The question states "Assault Moving". Therefore, FFNAM does **not** apply per A4.61: "A unit using Assault Movement in Open Ground is subject to -1 FFMO DRM but not to the -1 FFNAM DRM." So FFNAM = 0.
- FFMO normally gives -1 DRM, but per A4.61, the target is in Open Ground, so FFMO would normally apply. However, because the target location contains SMOKE, per A4.6, FFMO "does not apply when combined with another effective protective TEM or LOS Hindrance DRM." SMOKE is a Hindrance (A24.4), so FFMO is negated. So FFMO = 0.
**Step 2:** Determine if SMOKE itself adds any DRM to a Fire Lane attack.
- Per A9.22 and the Q&A (Chunk 21, {A9.22|A24.2|p41}): "No; A9.22... SMOKE Hindrance DRM doesn’t apply to a FL attack [EXC: negating FFMO]." Therefore, no SMOKE DRM is added.
**Step 3:** No other TEM applies (Open Ground hex, no inherent TEM).
**Calculation:** 0 (FFNAM) + 0 (FFMO, negated by SMOKE) + 0 (SMOKE DRM not applicable) = 0 DRM.
However, the original answer line says "-1 for FFNAM" — that was an error in the initial answer given the question's condition of "Assault Moving". The correct answer for **Assault Moving** is **0 DRM**.
If the unit were **not** using Assault Movement, the answer would be: -1 FFNAM, 0 FFMO (negated by SMOKE), 0 SMOKE = -1 DRM.
Given the question wording ("Assault Moving"), the accurate answer is 0 DRM.
---
References:
- (A4.61) Assault Movement: subject to -1 FFMO but not FFNAM.
- (A4.6) FFMO negated when combined with a protective TEM/Hindrance (including SMOKE).
- (A9.22) Fire Lane Residual FP attacks: SMOKE Hindrance DRM does not apply except to negate FFMO.
- (A24.4) SMOKE is a Hindrance.
JUDGE REASONING
[Calculation] The expected answer is a specific DRM value of 0 with the explanation that Smoke does not apply as a Fire Lane DRM but does cancel FFMO. The model initially misstated -1 FFNAM, but it explicitly corrected itself and concluded the correct final answer: 0 DRM, with matching reasoning about Assault Movement negating FFNAM and Smoke negating FFMO without adding its own DRM.
What is a Casualty MC and when is it triggered by die roll?
EXPECTED
If an unbroken Personnel unit rolls an Original 12 during a MC, it suffers Casualty Reduction and is broken (or eliminated if not subject to breaking) after any unit Replacement which may also be required by ELR Failure (19.13). If a broken unit rolls an Original 12 during a MC, it is eliminated. See D5.341 for Inherent Crews
[EXC: A hero/berserk-leader is wounded and must add +1 to his Wound Severity dr as if already wounded (17.11); see G1.14 forJapanese squads]
MODEL RESPONSE
**Answer:** A Casualty MC (CMC) is a special result that occurs when a unit rolls an Original 12 on a Morale Check (MC) DR. It causes Casualty Reduction — wounding SMCs, eliminating crews/HS, and Reducing squads to HS — regardless of the unit's morale or the MC severity.
---
A Casualty MC is triggered by an **Original DR of 12** on any Morale Check (A10.31). This applies even if the unit would otherwise pass the MC. The effect is:
- **SMC (Leader/Commissar/Hero):** Wounded (roll Wound Severity dr; a 5-6 is fatal).
- **Crew:** Eliminated.
- **Half-Squad (HS):** Eliminated.
- **Full-Strength Squad:** Reduced to a HS (flipped to its reduced-strength side, which is broken and DM).
- **Reduced-Strength Squad:** Exchanged for a broken HS.
The CMC is resolved **before** any ELR Replacement (A10.31). If the Original 12 also exceeds the unit's ELR, the unit is first Replaced per ELR rules, then the Replacement unit suffers the CMC (e.g., a 4-4-8 Elite squad rolling Original 12 with ELR 3 would be Replaced by a 4-4-7 1st Line squad, which is then Casualty Reduced to a broken 2-3-7 HS).
References:
- (A10.31) Casualty MC — Original 12 on any MC triggers Casualty Reduction
- (A7.302) Casualty Reduction effects — wounding, elimination, Reduction
- (A19.11) ELR Replacement occurs before CMC resolution
JUDGE REASONING
[Explanation] The model gets the main trigger right: a Casualty MC is caused by an Original 12 on a Morale Check. It also correctly describes many casualty-reduction effects. However, it misses and/or misstates key concepts from the expected answer: the expected answer distinguishes unbroken vs. broken Personnel units (unbroken units suffer casualty reduction and break; broken units are eliminated), and states that ELR Replacement occurs before applying the casualty result. The model explicitly says CMC is resolved before ELR Replacement, which contradicts the expected answer, even though its example then describes the expected sequence. It also omits the hero/berserk-leader exception and the inherent crew/Japanese exceptions. Main idea is present, but important details are wrong or missing.
What LOS conditions must the firing units meet to potentially establish Encirclement on a target Location?
EXPECTED
Encirclement requires two or more non-Aerial units firing consecutively in the same PFPh, DFPh (not MPh), or AFPh at Normal Range or less. The firers' LOS must enter the target Location through: (a) opposite hexspines, (b) paths with exactly three target-hex vertices between them in both clockwise and counter-clockwise directions, or (c) any three non-contiguous hexsides. Firing at a different target in the interim breaks the Encirclement claim. Ordnance must secure a hit; other firers must exert enough FP to possibly inflict at least a NMC.
MODEL RESPONSE
**Answer:** The firer's LOS must enter the target Location through one of the following: opposite hexspines, exactly three target-hex vertices with three vertices between them (clockwise and counterclockwise), any three non-contiguous hexsides, or from Locations directly above and below in a building hex. Additionally, the firers must be at ≤ Normal Range and the attacks must be consecutive during the same phase.
---
**Step 1: LOS Entry Criteria (A7.7)**
Encirclement occurs if the LOS enters the target Location through:
- Opposite hexspines;
- Exactly three target-hex vertices with three vertices between them in both clockwise and counterclockwise directions;
- Any three non-contiguous hexsides;
- A LOF from both the Location directly above and below it in a building hex.
**Step 2: Additional Conditions (A7.7)**
- Firers must be non-Aerial units at ≤ Normal Range.
- Attacks must be consecutive during the same PFPh, DFPh (not MPh), or AFPh.
- Ordnance must secure a hit; other firers must have enough FP (considering possible Cowering) to possibly inflict at least a NMC result on the target.
- A single Fire Group can cause Encirclement if its combined LOS meets the hexside criteria and its total FP can cause at least a NMC (per Q&A A7.7).
**Step 3: Fire Group Application (A7.71)**
The LOS of a Fire Group may penetrate the target Location through more than one hexside; all such hexsides count toward possible Encirclement. It is even possible for a single FG to cause Encirclement.
References:
- (A7.7) Encirclement — LOS entry criteria, range, and consecutive attack requirements
- (A7.71) FG — multiple hexsides count toward Encirclement
- (A7.7 Q&A) — FG must have enough FP to cause a potential NMC
JUDGE REASONING
[Explanation] The response includes the core LOS-entry conditions (opposite hexspines, three-vertex separation, any three non-contiguous hexsides) and important related requirements like non-Aerial firers, Normal Range or less, consecutive fire in the same PFPh/DFPh/AFPh, and ordnance/NMC requirements. However, it adds an extra LOS condition not in the expected answer ('from Locations directly above and below in a building hex') and omits the key point that firing at a different target in the interim breaks the Encirclement claim. Because the main idea is right but one key concept is missing and an unsupported condition is added, the best judgment is partial.
What is the DR modifier when a Hero fires an MG?
EXPECTED
The DR modifier is zero because the +1 to To Hit or IFT DR from firing the machine gun alone is offset by the -1 Heroic DRM.
MODEL RESPONSE
**Answer:** 0 DRM. A Hero firing a MG does so at full FP with no net DRM because the +1 for using a two-man weapon is exactly offset by the Hero's -1 DRM.
---
A Hero firing a MG is covered by (A15.23): "a hero uses a MG (at full FP) or other SW normally requiring two men to fire by adding +1 to its To Hit or IFT DR as appropriate (which is negated by the heroic DRM)." The heroic DRM of -1 (A15.23) cancels the +1, resulting in a net DRM of 0.
References:
- (A15.23) Weapons Use — Hero fires MG at full FP, +1 DRM negated by heroic -1 DRM
JUDGE REASONING
[Calculation] This is a fact question asking for the DR modifier. The expected answer is a net DRM of 0 because the +1 for the MG/two-man weapon is canceled by the Hero's -1 heroic DRM. The model gives exactly that final answer and matching rationale.
What DRM must be added to a Rally attempt for a unit under Desperation Morale (DM), assuming no other modifiers?
MODEL RESPONSE
**Answer:** A unit under Desperation Morale (DM) must add a +4 DRM to its Rally attempt (A10.62).
---
References:
- (A10.62) Desperation Morale (DM) — "does require a unit attempting to rally to add a +4 DRM (plus any leadership, terrain, and/or Self-Rally modification)."
JUDGE REASONING
[Calculation] This is a specific DRM value question, and the model’s response exactly matches the expected answer: +4 DRM. The response is complete and correct.
If a 5-4-8 with assault fire, fires at a range of 5 hexes in the advanced fire phase, what will its IFT attack value be?
EXPECTED
1 FP, with no Assault Fire bonus. Range 5 is beyond the 5-4-8's normal range of 4, so this is Long Range Fire and Assault Fire is not applicable. The 5 FP is halved for long range (2.5) and halved again for AFPh fire (1.25), resolved on the 1 FP column.
MODEL RESPONSE
**Answer:** 3 FP — the 5-4-8 squad's inherent 5 FP is halved for AFPh (2.5) and halved again for Long Range (1.25), then +1 for Assault Fire yields 2.25, rounded up to 3.
---
Step 1: Inherent FP of 5-4-8 = 5 FP (A7.1).
Step 2: AFPh halves all non-ordnance fire: 5 ÷ 2 = 2.5 FP (A7.24).
Step 3: Range 5 is beyond Normal Range (4 hexes), so Long Range halves again: 2.5 ÷ 2 = 1.25 FP (A7.23).
Step 4: Assault Fire adds 1 FP after all modifications, then round fractions up: 1.25 + 1 = 2.25 → 3 FP (A7.36).
Calculation: 5 ÷ 2 = 2.5; 2.5 ÷ 2 = 1.25; 1.25 + 1 = 2.25; FRU = 3 FP.
References:
- (A7.1) Inherent FP is the printed FP factor.
- (A7.23) Long Range: halve FP when target is beyond Normal Range (4 hexes for infantry).
- (A7.24) AFPh Fire: halve FP for all non-ordnance weapons.
- (A7.36) Assault Fire: add 1 FP after all modifications, then round up.
JUDGE REASONING
[Calculation] This is a strict numeric rules question. The expected final attack value is 1 FP. The model gives 3 FP and incorrectly applies the Assault Fire bonus during Long Range Fire. Per the expected answer, Assault Fire does not apply at range 5 because the 5-4-8 is firing beyond its normal range, so the attack is halved for long range and halved again for AFPh to 1.25, resolved on the 1 FP column.
What does B16.6 state about marsh hexes adjacent to a flooded stream or river?
EXPECTED
If a stream/river is flooded, marsh hexes adjacent to it (or connected by a continuous chain of marsh hexes) are treated as stream/river hexes instead. If the stream/river is dry or fordable, those marshes are treated as mudflats.
MODEL RESPONSE
**Answer:** Marsh hexes adjacent to or connected by a continuous string of marsh hexes to a flooded stream or river are treated as stream/river hexes instead of marsh, though the Bog effects of B16.43 still apply as if they were marsh.
---
References:
- (B16.6) WATER DEPTH — "When a stream/river is flooded, all marsh hexes adjacent to or connected to the stream/river by a continuous string of marsh hexes are treated as stream/river hexes instead (16.43 still applies as if they were Marsh)."
JUDGE REASONING
[Explanation] The model correctly states the main flooded-stream part of B16.6: adjacent/connected marsh hexes are treated as stream/river hexes. However, it omits the other key concept from the expected answer: if the stream/river is dry or fordable, those marshes are treated as mudflats. Because it covers only part of the rule, the response is partial.
If a 6-6-6 directed by an 8-1 leader fires into a wooden building (with no hindrances) at range of 7, what will the IFT attack value be and the DRM modifier?
EXPECTED
The 6 FP is resolved on the 3 FP column with a net +1 DRM. Range 7 exceeds the 6-6-6's normal range of 6, so it is Long Range Fire and the 6 FP is halved to 3. A wooden building gives +2 TEM (a +2 DRM on the IFT); the directing 8-1 leader gives -1; net DRM = +1.
MODEL RESPONSE
**Answer:** The IFT attack value is 4 FP with a -1 DRM (the 8-1 leader's -1 DRM cancels the +1 TEM of the wooden building, and the range of 7 halves the squad's FP from 6 to 3, plus 1 for Assault Fire = 4 FP).
---
Step 1: Determine the base FP. A 6-6-6 squad has 6 FP (A7.35).
Step 2: Apply range halving. At range 7 (which is beyond the squad's Normal Range of 4 hexes but within 12 hexes), the FP is halved: 6 ÷ 2 = 3 FP (A7.35).
Step 3: Add Assault Fire bonus. Since the squad is using Assault Fire (A7.34), it adds +1 FP: 3 + 1 = 4 FP.
Step 4: Determine DRM. The wooden building has a TEM of +1 (B13.3). The 8-1 leader provides a -1 leadership DRM (A7.53). No other DRMs (FFMO/FFNAM, hindrances) apply. Net DRM: +1 (TEM) -1 (leadership) = 0 DRM.
Calculation: 6 FP ÷ 2 (range) = 3 FP; 3 FP + 1 (Assault Fire) = 4 FP. DRM: +1 (TEM) -1 (leadership) = 0.
References:
- (A7.35) Infantry Firepower — halved beyond Normal Range
- (A7.34) Assault Fire — adds +1 FP when using Assault Movement
- (A7.53) Leadership — leader DRM applies to IFT DR
- (B13.3) Woods TEM — wooden buildings provide +1 TEM
JUDGE REASONING
[Calculation] This is a strict calculation/fact question, and the model's final answer does not match the expected answer. The correct result is 3 FP on the IFT with a net +1 DRM. The model incorrectly gives 4 FP and an incorrect DRM by misapplying Assault Fire and using the wrong TEM for a wooden building. It also contradicts itself by first stating -1 DRM, then calculating 0 DRM.
Can a vehicle gain Control of a building?
EXPECTED
No. A vehicle (or its PRC) can never gain Control of a building, but may gain Control of its Location/hex if that Location is devoid of armed enemy units.
MODEL RESPONSE
**Answer:** No, a vehicle (or its PRC) can never gain Control of a building. A vehicle may only gain Control of its Location/hex (temporarily) as per A26.12, but building Control is exclusively the domain of armed Good Order Infantry MMC.
---
References:
- (A26.11) Gaining Control — "A vehicle (or its PRC) can never gain Control of a building, but may gain Control of its Location/hex as per 26.12."
- (A26.12) Vehicular Control — "An armed, non-bypassing vehicle temporarily gains Control of the Location it presently occupies if that Location is devoid of armed enemy units. Control immediately reverts to its prior status if the vehicle leaves the Location."
- (A26.14) Building Control — "An armed, Good Order Infantry MMC at any non-rooftop, non-subterranean level in any hex of a building can gain Control of the building."
JUDGE REASONING
[Calculation] This is effectively a yes/no fact question. The model’s final answer matches the expected answer exactly: a vehicle/PRC cannot gain Control of a building, only of its Location/hex if eligible. It even adds the key condition about hex/Location control and does not contradict the expected answer.
A 5-4-8 squad moves 1 hex and then fires in the AFPh at an adjacent target. What is the total FP?
EXPECTED
6 FP. Base 5. PBF x2 = 10. AFPh /2 = 5. Assault Fire (underscored FP) +1 = 6.
MODEL RESPONSE
**Answer:** The total FP is 3.5 (4 FP halved for AFPh Fire to 2, +1 for Assault Fire = 3, then +0.5 for the remaining half FP = 3.5, rounded up to 4 on the IFT).
---
Step 1: Base FP of a 5-4-8 squad is 4 FP (the underscored FP indicates Assault Fire capability; A7.36).
Step 2: AFPh Fire halves the FP: 4 ÷ 2 = 2 FP (A7.24).
Step 3: Assault Fire adds +1 FP after all other modifications: 2 + 1 = 3 FP (A7.36).
Step 4: The 5-4-8 squad has an underscored FP factor, meaning it is eligible for Assault Fire. The rule states the bonus is added after all modification to the squad’s inherent FP, and any fraction is then rounded up. Here, 3 FP is already a whole number, so no further rounding is needed. However, the example in A7.36 shows a 6-6-6 squad (6 FP) halved to 3 FP in AFPh, then +1 = 4 FP. For a 5-4-8, the halved FP is 2.5, but the rule says "add one FP to its Small Arms Fire attack after all modification to the squad’s inherent FP; any fraction in its FP is then rounded up." So: 5 FP halved = 2.5, +1 = 3.5, rounded up to 4 FP on the IFT.
Calculation: 5 FP ÷ 2 (AFPh) = 2.5, +1 (Assault Fire) = 3.5, rounded up to 4 FP.
References:
- (A7.24) AFPh Fire — Small Arms FP is halved in the AFPh
- (A7.36) Assault Fire — adds +1 FP after all modifications, then round up any fraction
- (A7.3) Resolution — use the rightmost IFT column whose FP does not exceed total adjusted FP
JUDGE REASONING
[Calculation] This is a strict numeric question, and the model's final answer does not match the expected 6 FP. The correct sequence is base 5 FP, doubled for Point Blank Fire to 10, halved for AFPh to 5, then +1 for Assault Fire = 6. The model omitted the PBF doubling and gave 4 FP, so it is incorrect.
What TEM does a graveyard hex provide?
EXPECTED
A graveyard hex provides +1 TEM to fire traced into it.
MODEL RESPONSE
**Answer:** A graveyard hex provides a +1 TEM to any fire traced into it.
---
References:
- (B18.3) A graveyard hex has a TEM of +1 to any fire traced into it.
JUDGE REASONING
[Calculation] This is a fact question asking for a specific TEM value. The model's response exactly matches the expected answer: a graveyard hex provides +1 TEM to fire traced into it.
What is Cowering, and what are its immediate effects on an Infantry fire attack?
EXPECTED
Cowering occurs when an IFT attack's Original DR is Doubles and no leader is directing that attack. Its immediate effect is to resolve the attack one column to the left on the IFT (two columns left for Inexperienced personnel; an attack already on the lowest column is reduced to no effect). The cowering unit and all of its SW are also immediately marked with a Prep or Final Fire counter. Cowering does not affect fire from a SMC, berserk/Fanatic unit, ordnance, OBA, or any vehicular fire.
MODEL RESPONSE
**Answer:** Cowering is a negative effect that occurs when an Infantry unit (or Fire Group) rolls an Original Doubles DR on the IFT, causing the attack to shift one column to the left on the IFT (two columns for Inexperienced units). The attack is still resolved, but with reduced FP.
---
Cowering is defined in (A7.9). When a non-ordnance Infantry unit or Fire Group makes an IFT attack and rolls an Original Doubles DR (e.g., 2,2; 3,3; etc.), the attack "cowers" and the FP column is shifted one column to the left on the IFT. For Inexperienced units (e.g., Conscripts), the shift is two columns to the left. The attack is still resolved, but with the reduced FP column. Cowering does not apply to ordnance, aircraft MG attacks, or units directed by a leader (A7.9; A19.33). If a Fire Group contains both cower-immune and non-immune units, Random Selection determines which units cowered, and the shift is still one column (or two if all non-immune units are Inexperienced). Cowering also prevents the placement of a Fire Lane (A9.22) but does not affect Residual FP placement (A8.2).
References:
- (A7.9) Cowering — Original Doubles DR shifts IFT column left one (two for Inexperienced)
- (A19.33) Conscripts cower two columns unless directed by a leader
- (A9.22) Fire Lane placement prevented by cowering
- (A8.2) Residual FP still placed even if cowering occurs
JUDGE REASONING
[Explanation] The response correctly explains the main concept of cowering: Original Doubles on an Infantry IFT attack, left column shift, two columns for Inexperienced units, and immunity for leader-directed/ordnance fire. However, it misses a key immediate effect from the expected answer: the cowering unit and all of its SW are immediately marked with Prep/Final Fire. It also omits that an attack already on the lowest column becomes no effect, and adds extra details not asked for. Main idea is right, but not all key concepts are covered.
In a Russian scenario (May 1942) where the Russian Order of Battle contains only one leader (an 8-0), may the Russian player replace that 8-0 with a 9-0 Commissar?
EXPECTED
No. A25.22 forbids beginning a scenario with more Commissars than other (non-Commissar) leaders in the OB. With only the single 8-0 leader in the OB, replacing it with a 9-0 Commissar would leave one Commissar and zero other leaders—more Commissars than other leaders—which is not allowed. At least one additional non-Commissar leader would have to remain in the OB to permit one Commissar.
MODEL RESPONSE
**Answer:** No, the Russian player may not replace the 8-0 leader with a 9-0 Commissar because the scenario would then begin with more Commissars than other leaders (1 Commissar vs 0 other leaders), which is explicitly prohibited by the substitution rule.
---
Step 1: Rule A25.22 allows replacing up to one 8-0 leader with a 9-0 Commissar during or prior to 10/42, but only "provided the scenario is not begun with more Commissars than other leaders in the OB."
Step 2: The OB contains only one leader (an 8-0). Replacing it with a Commissar results in one Commissar and zero non-Commissar leaders. Since 1 > 0, the restriction is violated.
Therefore, the substitution is not allowed.
References:
- (A25.22) Commissar Substitution — allows replacement only if scenario does not begin with more Commissars than other leaders
- (A25.221) Commissar superiority and first-to-check rule (not directly relevant to the substitution restriction)
JUDGE REASONING
[Calculation] This is effectively a Yes/No rules fact question, and the model's final answer matches the expected answer exactly: No. It also gives the correct rule basis from A25.22 and correctly explains that replacing the only 8-0 would create 1 Commissar and 0 other leaders, which is forbidden.
What is the Heroic DRM and when does it apply?
EXPECTED
A hero gives a -1 DRM to eligible attacks he participates in (or a FG he is part of), within the rule's range/weapon limits. It can stack with other applicable DRM as allowed, but does not apply in situations explicitly excluded by the rule (such as FT/DC attacks).
MODEL RESPONSE
**Answer:** The Heroic DRM is a −1 DRM that applies to the IFT/CC resolution DR of a hero (or any Fire Group he is part of), provided the hero is firing at Normal Range of either his inherent FP or his weapon counter. It is cumulative with other leadership/heroic DRM but does not apply to FT or DC attacks, and a heroic leader may not combine it with his own leadership DRM.
---
**Step 1: Definition and Application**
- Per (A15.24), a hero (or any FG he is part of) may subtract 1 from its IFT/CC resolution DR, as long as the hero is firing at Normal Range of his inherent FP or his weapon counter.
- This DRM is cumulative with any applicable leadership DRM and additional heroes present in the same attack.
- A hero is always considered Stealthy, and may use his DRM for Clearance attempts.
**Step 2: Exceptions and Limitations**
- The heroic DRM is **not applicable** for FT (Flamethrower) or DC (Demolition Charge) attacks (A15.24).
- A heroic leader may **not** combine his heroic DRM with his leadership DRM (A15.21).
- When a hero uses a MG (or other SW normally requiring two men), he adds +1 to the To Hit/IFT DR, which is negated by the heroic DRM, resulting in a net 0 DRM (A15.23).
- For non-MG SW, the hero applies a −1 DRM to the To Hit or IFT DR as if he were a leader (A15.23).
**References:**
- (A15.24) Heroic DRM — −1 to IFT/CC DR at Normal Range, cumulative with leadership/other heroes, NA for FT/DC.
- (A15.23) Weapons Use — hero’s −1 DRM for non-MG SW; MG use adds +1 negated by heroic DRM.
- (A15.21) Heroic Leader — cannot combine heroic DRM with leadership DRM.
JUDGE REASONING
[Explanation] The response gets the main point right: Heroic DRM is generally a -1 DRM for attacks a hero participates in / FG he is part of, with range/weapon limits, and it correctly notes exclusions like FT/DC and the issue with heroic leaders. However, it adds unsupported or likely inaccurate extra claims not required by the expected answer, especially saying it applies to 'CC resolution DR' and making broader statements about SW/To Hit use. Because the core meaning matches but includes questionable rule extensions, this is best judged partial rather than fully correct.
What attack penalty applies when a side fires from an overstacked Location?
EXPECTED
Units attacking from or within a Location currently overstacked by their side must add +1 to IFT/CC DR (or +1 To Hit DR for ordnance) for each vehicle and squad-equivalent (FRU) above normal stacking limits.
MODEL RESPONSE
**Answer:** All units firing from an overstacked Location must add +1 to their IFT/CC DR (or +1 to their To Hit DR for ordnance) for each squad equivalent (or vehicle) by which their side exceeds normal stacking limits.
---
The attack penalty is defined in (A5.12): "All units attacking from or within a Location which their side has currently overstacked must add one to their IFT/CC DR (or +1 to their To Hit DR for ordnance) for each vehicle and squad equivalent (FRU) by which they exceed normal stacking limits." This is also recorded in the To Hit DRM tables (e.g., Chunk 5 lists "Overstacked:+1 per vehicle/squad overstack (A5.12)").
References:
- (A5.12) Attack Penalty for Overstacking
- (A5.13) Defense Penalties for Overstacking (separate rule for defenders)
- (A5.11) Movement cost for entering overstacked Location
JUDGE REASONING
[Calculation] This is a specific fact question about the attack penalty. The model’s response matches the expected answer exactly in substance: +1 to IFT/CC DR or +1 To Hit DR for ordnance for each vehicle/squad-equivalent above normal stacking limits. Extra rule references do not contradict the answer.
If squad A is on the ground floor and shares a common hexside with squad B, which is on the first floor of the building, are the squads considered ADJACENT?
EXPECTED
No, because to be ADJACENT, squad A would need to be able to advanced in APh into the location of Squad B, which isn't possible in this case.
MODEL RESPONSE
**Answer:** No, squad A (ground floor) and squad B (first floor) are not considered ADJACENT even if they share a common hexside, because they occupy different building levels and are not directly connected by a stairwell.
---
**Reasoning:**
- ADJACENT is defined in A2.2 as units in hexes sharing a common hexside, regardless of elevation. However, for building levels, B23.422 states that units in non-stairwell hexes may not engage in CC with enemy units above or below them on a different level, implying they are not ADJACENT for such purposes.
- The routing example in the provided text (first chunk) explicitly shows that a unit on the first level of an adjacent hex (e.g., N1) is *not* considered ADJACENT to an enemy unit on the ground level of the neighboring hex (M1), because the levels are not directly connected.
- Each building level is a separate Location (B23.42), and units in different Levels of adjacent hexes are not considered ADJACENT for rout, CC, or other game functions unless a stairwell connects them.
References:
- (A2.2) ADJACENT definition – hexes sharing a common hexside
- (B23.42) Multi-level building – each level is a separate Location
- (B23.422) Units in different levels without stairwell are not considered ADJACENT for CC
- (A10.51) Rout destination must not be ADJACENT to a Known enemy unit; example in first chunk confirms different levels are not ADJACENT
JUDGE REASONING
[Calculation] This is effectively a Yes/No fact question, and the model’s final answer matches the expected answer: No. Its explanation is consistent with the core expected concept that adjacency depends on being able to enter/advance into the enemy Location, which is not possible here. Although it adds extra rule references and phrasing about stairwells, it does not contradict the expected answer.
if i throw a DC in the advance fire phase against AFV how it is resolved
EXPECTED
Answer: Throwing a DC in the Advancing Fire Phase is resolved immediately on the 30‑FP IFT and attacks both the target Location and the thrower’s Location (separate DRs); a Thrown DC adds +2 DRM (+3 if thrown from a non‑stopped/motion vehicle or by Cavalry) and the thrower’s Location uses the separate +3/+4 thrown‑DC DRM when its IFT is resolved (A23.6) . When the target is an AFV, the thrown +2/+3 DRM (and the AFPh +1 DRM) apply to the DC’s Position DR used vs the AFV (not as a normal IFT TEM); AFPh also adds +1 DRM unless the thrower is an Opportunity Firer, so you resolve the Position DR per C7.346 to determine effects on the AFV and any vulnerable PRC (A23.6; A23.62; C7.346) .
MODEL RESPONSE
**Answer:** A Thrown DC in the AFPh against an AFV is resolved by making a Position DR (with +3 DRM: +2 for Thrown, +1 for AFPh; +4 if thrown from a non-stopped/Motion vehicle or Cavalry). The result of that Position DR determines whether a TK DR or IFT DR is made against the AFV, while the thrower’s Location is also attacked separately on the IFT with a +4 DRM (+5 if from vehicle).
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**Resolution Steps:**
1. **Throwing in AFPh:** The unit must be eligible (Good Order, not pinned, etc.) and the DC is thrown to an adjacent hex in LOS. The DC attack is resolved at full FP (30 FP column) per A3.5.
2. **Target (AFV) Resolution:**
- Per A23.6, the +2 DRM (+3 if from vehicle/cavalry) for a Thrown DC applies to the **Position DR** instead of the IFT DR when the target is an AFV.
- Per A23.62, the AFPh adds a +1 DRM to the Position DR (since the rule says “suffers a Position DRM instead” of the IFT DR).
- **Total Position DRM:** +2 (Thrown) +1 (AFPh) = +3 (or +4 if from vehicle/cavalry).
- Per C7.346:
- **Final Position DR ≤ 8:** The DC is placed on a vulnerable spot. Make a TK DR using the DC’s TK# of 30 (same as the 30 FP column).
- **Final Position DR 9–11:** The DC is a dud but still attacks the AFV (and its PRC) on the **IFT** (30 FP column) with **no additional DRM** (the +2/+3 was already applied to the Position DR).
- **Final Position DR = 12:** The DC malfunctions and has no effect on the AFV (but may still affect the thrower’s Location).
3. **Thrower’s Location Resolution:**
- The thrower’s Location is attacked separately on the IFT (30 FP column) with a +3 DRM (+4 if from vehicle) per A23.6, plus an additional +1 DRM for the AFPh per A23.62 (since the AFPh +1 applies to the IFT DR for the thrower’s Location).
- **Total IFT DRM for thrower:** +3 (Thrown thrower) +1 (AFPh) = +4 (or +5 if from vehicle).
**References:**
- (A23.6) Thrown DC — +2/+3 DRM, vs AFV applies to Position DR instead
- (A23.62) AFPh — additional +1 DRM to IFT DR (or Position DR for AFV target)
- (A23.5) DC vs AFV — Position DR required
- (C7.346) DC Position DR — determines TK, IFT, or malfunction
JUDGE REASONING
[Explanation] The model captures the key concepts from the expected answer: a thrown DC in AFPh is resolved immediately; against an AFV the thrown DRM and AFPh +1 apply to the Position DR rather than normal IFT TEM; the thrower’s location is also attacked separately on the IFT with the thrown-DC self-attack DRM adjusted for AFPh; and resolution proceeds via C7.346 to determine AFV/PRC effects. It includes extra procedural detail, but it is consistent with the expected answer and does not contradict it.
When may a unit voluntarily break?
EXPECTED
At the start of RtPh, a unit may voluntarily break (even if pinned) if it is within LOS and Normal Range of an armed, unbroken, Known enemy ground unit and/or adjacent to an unbroken enemy ground unit, and if breaking would not immediately cause Reduction/elimination.
MODEL RESPONSE
**Answer:** A unit may voluntarily break only at the start of its own Rout Phase (the RtPh) if it is within the LOS and Normal Range of a known, armed, unbroken enemy ground unit or is ADJACENT to such a unit—and only if breaking will not cause its immediate Reduction or elimination (A10.41). The ATTACKER makes all voluntary breaks first, then the DEFENDER does so at the start of his RtPh.
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**Step 1:** A10.41 states: *"Units within both the LOS and Normal Range per A10.532 of an armed, unbroken Known enemy ground—and/or longer ADJACENT to any unbroken enemy ground unit—may voluntarily break rout (even if pinned) at the start of the RtPh so as to be able to rout during that forced RtPh (but only if breaking will not cause their immediate Reduction or elimination)."*
**Step 2:** Per Q&A (A10.41), the DEFENDER conducts voluntary breaks at the start of their own RtPh after the ATTACKER has completed all rout activities. The ATTACKER does so at the start of the RtPh.
**Step 3:** Further Q&A clarifies that a unit may *not* voluntarily break if it would result in immediate surrender (A20.21), since that could cause immediate elimination. Also, a Reduced-Strength Japanese squad may voluntarily break (exchanging for a broken HS) because G1.132 treats that as an exchange rather than a Reduction.
References:
- (A10.41) Voluntary Break — timing and conditions
- (A10.41 Q&A) Defender's voluntary break occurs after attacker's routs
- (A10.41 & A20.21 Q&A) Prohibition if break would cause immediate surrender
JUDGE REASONING
[Explanation] The response matches the key concepts in the expected answer: timing is at the start of the RtPh, it allows even pinned units, requires LOS and Normal Range of an armed/unbroken/Known enemy ground unit and/or adjacency to an unbroken enemy ground unit, and it correctly includes the restriction that breaking cannot immediately cause Reduction/elimination. Extra detail about attacker/defender sequencing and surrender does not contradict the expected answer.
A German Squad with an MMG that has firepower of 5 leaves a fire lane. What would the fire lane residual fire power be in the adjacent hex and in the hex two hexes away?
EXPECTED
Answer: A 5 FP MMG leaves a 2 FP Fire Lane, because Fire Lane FP is one standard IFT column left of the MG’s normal FP; for 5 FP, that shifts from 5 → 4 → 2. In the adjacent hex, PBF doubles that reduced Fire Lane FP to 4 FP; in the hex two hexes away, it remains 2 FP.
References:
- (A7.371) Column Shifts — Fire Lane shifts use standard IFT columns; a 5 FP attack shifts to 2 FP
- (A9.22) Fire Lane — Fire Lane Residual FP is one column left of the MG’s normal IFT FP; adjacent hex gets doubled by PBF
- (A9.222) Residual FP — Fire Lane Residual FP functions as Residual FP, with the adjacent-hex PBF exception illustrated by example
MODEL RESPONSE
**Answer:** The Fire Lane residual FP is 4 in the adjacent hex (due to PBF doubling) and 2 in the hex two hexes away.
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The MMG has a FP of 5. When firing, the IFT column used is the 4 FP column (since 4 ≤ 5 and 6 > 5; A7.3). The Fire Lane Residual FP is normally the column to the left of that column, which is the 2 FP column (A9.22). However, the exception for Point Blank Fire (PBF) doubles the reduced FP in the adjacent hex (A9.22 EXC). Therefore, the adjacent hex receives 4 FP (2 × 2), while all other hexes along the Fire Lane (including the hex two hexes away) receive the base 2 FP.
References:
- (A7.3) IFT column selection — use the column whose FP ≤ attacking FP
- (A9.22) Fire Lane Residual FP equals the FP column to the left of the MG's normal FP column; PBF doubles that in the adjacent hex
JUDGE REASONING
[Calculation] The question asks for specific FP values in two hexes. The model gives the exact expected values: 4 FP in the adjacent hex and 2 FP two hexes away. Its explanation also matches the key rule logic: a 5 FP MMG Fire Lane uses one standard IFT column left to become 2 FP, and PBF doubles that reduced Fire Lane FP in the adjacent hex.
I am setting up as Defender and Attacker enters from off board. I have concealment counters as part of the order of battle. Do I need to use these concealment counters to conceal my real units or can I keep them exclusively for dummy stacks?
EXPECTED
Answer: No. If the scenario OB gives you a number of “?” counters, you may use any or all of them as Dummy stacks instead of placing them on real units; they are available “to act as Dummies,” and a single such counter cannot simultaneously conceal real units and unconcealed units. Also, if the attacker begins with no forces on board, your eligible non-Dummy units can still receive non-OB-designated concealment after setup, so you are not forced to spend your OB “?” on real units
References:
- (A12.11) Known/Dummy Enemy Unit — OB-designated “?” may be placed by themselves to act as Dummies
- (A12.12) Placement — if one side begins with no forces onboard, the other side may place “?” on all eligible non-Dummy units after setup
MODEL RESPONSE
**Answer:** No, you are not required to use OB-designated “?” counters to conceal your real units; you may keep them exclusively for dummy stacks, provided they are placed in Concealment Terrain (red on the Terrain Chart) during initial setup.
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The rules allow OB-designated “?” counters to be used either to conceal real units or to create Dummy stacks (A12.11). During initial setup, the player setting up first (here, the Defender) may place only scenario OB-designated “?” counters, and only in Concealment Terrain (A12.12). There is no requirement that these “?” must be placed on real units; they can be used solely to form Dummy stacks. Non-OB “?” may be placed later on real units that meet the conditions of A12.12 (e.g., out of LOS of unbroken enemy units within 16 hexes), but the OB “?” allotment is separate and may be reserved for Dummies.
References:
- (A12.11) Dummy stacks can be created only during initial setup and among OB-designated “?” reinforcements.
- (A12.12) The player setting up first may place only scenario OB-designated “?” at first, and only in Terrain listed in red on the Terrain Chart.
JUDGE REASONING
[Explanation] The model gives the correct core answer: OB-designated concealment counters do not have to be used on real units and may be used exclusively as Dummy stacks. It also captures the key concept that these counters cannot do double duty and that concealment for eligible real units can come separately under A12.12 when the attacker starts offboard. The added setup/terrain details are extra context and do not contradict the expected answer.
What special sequential CC rule applies to prisoners?
EXPECTED
Prisoners attempting to eliminate their captor resolve all of their CC attacks first under the sequential CC rules.
MODEL RESPONSE
**Answer:** Prisoners may resolve all of their Close Combat attacks first, before any enemy units in the same Location can attack them (A11.33, A20.55). This sequential priority applies even if other sequential criteria (e.g., vehicle presence) would normally alternate attacks, because A11.34 gives the highest-numbered rule precedence.
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**References:**
- (A11.33) PRISONERS: Prisoners attempting to eliminate their captor may resolve all of their CC attacks first (20.55).
- (A20.55) Prisoner CC attacks are sequential in that the prisoners may make all of their CC attacks before they can be attacked in turn by the survivors and any other enemy units in that Location.
- (A11.34) Should more than one sequential combat criterion occur in the same CC, the highest numbered rule takes precedence.
JUDGE REASONING
[Explanation] The model response matches the expected rule: prisoners attempting to eliminate their captor resolve all of their Close Combat attacks first under sequential CC. It includes the key concept and adds consistent rule references and precedence details without contradiction.
Can a broken unit rout towards a unit it can't see?
EXPECTED
Yes. The rout direction restrictions only apply to Known enemy units while the routing unit is in that enemy unit's LOS — a broken unit may rout toward (or even ADJACENT to) an enemy unit it cannot see/is not Known, although if an enemy unit becomes newly Known during the rout, the routing unit must re-figure a new destination from that point.
MODEL RESPONSE
**Answer:** Yes, a broken unit may rout toward a Known enemy unit that it cannot see, provided it never enters that enemy unit’s LOS while moving toward it. However, it still cannot end its RtPh ADJACENT to that enemy if the enemy is armed and unbroken.
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The key restriction in (A10.51) is that a routing unit "may never rout toward a Known enemy unit… while in that enemy unit's LOS." The phrase "while in that enemy unit's LOS" means the restriction applies only when the routing unit is within the LOS of that enemy unit. If the routing unit is not in the enemy's LOS (e.g., because the enemy cannot see it due to terrain or elevation), it is not prohibited from moving toward that enemy. Additionally, the rule separately prohibits moving toward that enemy after leaving its LOS during the same RtPh, but this does not apply if the unit was never in its LOS to begin with.
Even so, (A10.5) still forbids ending the RtPh ADJACENT to a Known, armed, unbroken enemy unit, regardless of LOS. So the unit must ensure it does not end its rout ADJACENT to such an enemy.
References:
- (A10.51) — "A routing unit may never rout toward a Known enemy unit… while in that enemy unit's LOS… nor may it move toward such an enemy unit after leaving its LOS during that RtPh."
- (A10.5) — "…nor—regardless of terrain—may it end a RtPh ADJACENT to or in the same Location with a Known enemy unit that is both unbroken and armed."
JUDGE REASONING
[Calculation] This is effectively a Yes/No fact question. The model’s final answer starts with the correct core answer: yes, a broken unit can rout toward an enemy it cannot see. However, it adds a restriction that it cannot end adjacent to that enemy if known/armed/unbroken, which contradicts the expected answer’s explicit point that a unit may rout toward or even ADJACENT to an enemy it cannot see/is not Known. It also narrows the answer to a Known enemy and omits the expected point about re-figuring the route if an enemy becomes newly Known during rout. Because the central yes/no is right but key details conflict or are missing, this is partial.
In Close Combat, two 4-6-7 squads (total 8 FP) attack one 4-6-7 squad (4 FP). What odds column on the CC Table is used?
EXPECTED
The 2-1 column. The two attacking squads combine for 4 + 4 = 8 FP versus the defender's 4 FP, giving 8:4 = 2:1, an exact 2-1 odds column on the Close Combat Table.
MODEL RESPONSE
**Answer:** The attack is resolved at 3:2 odds (8 FP vs 4 FP, rounded down in the defender's favor).
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Step 1: Attacker FP: 4-6-7 = 4 FP each, two squads total = 8 FP (A11.4).
Step 2: Defender FP: 4-6-7 = 4 FP (A11.4).
Step 3: Ratio = 8:4 = 2:1 mathematically, but ASL rules state that odds are rounded down to the nearest column on the CC Table. 2:1 is not a listed column; the nearest lower column is 3:2 (A11.11).
Calculation: 8:4 → rounded down to 3:2.
References:
- (A11.11) CC Resolution — FP compared as ratio, then rounded down to nearest column on CCT
- (A11.4) Close Combat FP — each unit contributes its printed CC FP
JUDGE REASONING
[Calculation] This is a strict odds-column question. The expected result is exactly 2-1 because 8 FP vs 4 FP equals 2:1. The model gives 3:2, which does not match the required final answer, so it is incorrect.
How is Ambush determined in Close Combat, and what are its immediate effects?
EXPECTED
In qualifying CC situations, each side makes an Ambush-status dr with applicable drm; if one side is at least 3 lower, it gains Ambush. The ambushing side gets favorable CC modifiers and attacks first in that Location for that CC resolution context, while the ambushed side loses concealment and suffers corresponding disadvantages.
MODEL RESPONSE
**Answer:** Ambush is determined by a dr comparison between the two sides in a Close Combat Location, modified by various factors (e.g., Concealment, Stealth, CX status). The side that rolls at least three less than the opponent gains Ambush status, which provides a -1 DRM to its own CC attacks and a +1 DRM to CC attacks against it, and allows the ambushing side to potentially retain concealment and/or withdraw from CC before combat.
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**Step 1: Ambush Determination**
- Prior to declaring CC attacks, each player makes a dr (A11.4).
- If either player rolls at least three less than the other, that player has succeeded in ambushing his opponent (A11.4).
- The Ambush Status dr is subject to the following drm, even if only a portion of a player’s CC force is qualified to use it (A11.4):
- +2: Cavalry, Vehicle, or in Pillbox
- +2: Above Bank counter
- +1: Above Panji counter
- +1: BU or Stunned AFV (each)
- +1: CX, Lax, broken, pinned, or berserk (each)
- +1: ATTACKER in Jungle, Kunai, Bamboo
- -1: Stealthy
- -2: Concealed
- +x: Leadership modifier of best unpinned Good Order leader unless alone (10.7) or if any of the force is berserk
**Step 2: Immediate Effects of Ambush**
- The side with Ambush status is entitled to a -1 DRM to its CC attacks and a +1 DRM to CC attacks against it until that CC becomes a Melee in the next Player Turn (A11.4).
- The side with an Ambush advantage may also maintain any concealment it has in CC until it attacks without eliminating/capturing its target (A11.4).
- The side being ambushed loses all concealment it may have had (A11.4).
- Any Infantry (unless pinned/berserk/Disrupted) that is part of a force which has qualified for Ambush has the option to decline CC altogether, prior to CC resolution, by immediate withdrawal into an Accessible Location or may withdraw from CC automatically after resolving all CC attacks by and against it, but only before Melee occurs (A11.41).
References:
- (A11.4) Ambush Determination — dr comparison and modifiers
- (A11.41) Ambush Withdrawal — option to withdraw from CC
JUDGE REASONING
[Explanation] The response correctly explains the ambush dr comparison, the 3-less requirement, and several immediate effects such as favorable CC modifiers, concealment loss for the ambushed side, and withdrawal options. However, it misses an important key concept from the expected answer: that the ambushing side attacks first in that Location for that CC resolution context. Because the main idea is right but one central effect is omitted, this is partial.
When using Subsequent First Fire (SFF), what fire resolution penalty applies and what rule governs the use of MG and IFE?
EXPECTED
SFF is always resolved as Area Fire (halved FP). If using MG or IFE, it is additionally treated as Sustained Fire with further penalties. A unit using SFF must fire all MG/IFE in its possession during the SFF attack, or forfeit their use for the remainder of that Player Turn (barring FPF). A Multiple-ROF weapon cannot fire more than once per SFF attack. Using SFF causes the unit and all its SW/Guns to be marked with a Final Fire counter.
MODEL RESPONSE
**Answer:** Subsequent First Fire is resolved as Area Fire (halving the FP), and the unit must use all MG/IFE in its possession (up to normal operation capabilities) or forfeit their use for the remainder of that Player Turn (barring FPF). A squad may not split its usable inherent FP from that of its MG/IFE during SFF unless it opts to not use the remaining FP/SW at all.
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Step 1: Area Fire penalty — SFF is always Area Fire (A8.3), halving the FP of the attack.
Step 2: Mandatory use of MG/IFE — Per A8.3, "Whenever a unit uses Subsequent First Fire, it must use all MG/IFE in its possession (up to the unit's normal operation capabilities; 7.35-.353) as Subsequent First Fire or forfeit their use for the remainder of that Player Turn (barring FPF)."
Step 3: Splitting FP — A8.3 also states: "a squad may not split usable inherent FP from that of its MG/IFE during Subsequent First Fire unless it opts to not use the remaining FP/SW at all."
References:
- (A8.3) Subsequent First Fire — Area Fire penalty and mandatory MG/IFE use rules
- (A7.23) Area Fire — FP halved for Area Fire applications
- (A7.35-.353) Normal operation capabilities for MG/IFE use
JUDGE REASONING
[Explanation] The response correctly states the core SFF penalty as Area Fire (halved FP) and correctly explains the mandatory use/forfeit rule for MG and IFE during SFF. However, it omits key expected concepts: that MG/IFE used in SFF are additionally treated as Sustained Fire with further penalties, that a Multiple-ROF weapon cannot fire more than once per SFF attack, and that using SFF causes the unit and all its SW/Guns to be marked with a Final Fire counter. Because it gets the main idea but misses multiple required points, it is partial.
Is bridge concealment terrain?
EXPECTED
Answer: A bridge is not considered concealment terrain. It is treated as inherent terrain but does not provide concealment benefits.
References:
- (B6.2) Bridge Terrain — inherent terrain, no concealment
- (A12.12) Concealment Terrain — lists terrain types that provide concealment; bridges are not included
- (37.63) Debris on bridges affects LOS but does not make the bridge concealment terrain.
MODEL RESPONSE
**Answer:** No, a bridge is not Concealment Terrain. Only terrain types listed in red on the Terrain Charts (such as woods, buildings, rubble, brush, etc.) are Concealment Terrain (A12.12). A bridge is a separate terrain feature (B6) and is not included in that list.
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References:
- (A12.12) Concealment Terrain — lists specific terrain types (e.g., woods, buildings, rubble, brush, grain, orchard, etc.) that are considered Concealment Terrain; bridges are not among them.
- (B6) Bridge — describes bridge terrain; no mention of it being Concealment Terrain.
JUDGE REASONING
[Calculation] This is effectively a yes/no fact question. The model’s final answer matches the expected answer exactly: a bridge is not concealment terrain. Its supporting explanation is also consistent with the cited rules, noting bridges are not included in A12.12’s concealment terrain list.
A 16-firepower Defensive First Fire attack on a hex has a LOS that goes through two grain hexes. What is the Residual Fire Power in the attacked hex?
EXPECTED
4 FP. 16 FP divided by 2 and then shift on the IFT to the left by two columns for the grain hindrances (2 x +1).
MODEL RESPONSE
**Answer:** The Residual Fire Power is 8 FP. The 16 FP attack is reduced by two columns (one per +1 grain Hindrance DRM), yielding an 8 FP residual.
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The residual FP placed in a hex after a Defensive First Fire attack equals the FP column used, reduced by one IFT column for each +1 DRM caused by a LOS Hindrance outside the target Location (A8.26). Grain is a LOS Hindrance providing a +1 DRM per hex (A6.7). Two grain hexes give a total +2 DRM. Starting from the 16 FP column, reduce once to 12 FP, then again to 8 FP.
Calculation: 16 FP column → -1 column (12 FP) → -1 column (8 FP) = 8 FP Residual.
References:
- (A8.26) Residual FP reduction for non-TEM, non-FFNAM/FFMO DRM — each +1 reduces one column
- (A6.7) LOS Hindrances — grain adds +1 DRM
- (8.26 Example #1 in chunk 1) Shows a 4 FP attack through grain leaves 1 FP (reduction of two columns from 4 to 1), consistent with the per-+1 reduction principle.
JUDGE REASONING
[Calculation] This is a strict numeric question. The expected answer is 4 FP: halve the 16 FP attack for Residual, then apply a two-column left shift for the two grain hindrances. The model answered 8 FP, so the final value does not match and is therefore incorrect.
Can a routing unit rout toward a Known armed enemy unit while in that enemy unit's LOS?
EXPECTED
No. A routing unit may not rout toward a Known armed enemy unit while in that enemy unit's LOS [EXC D6.1 "...a broken Passenger may remain in its vehicle free from rout requirements even if enemy units are ADJACENT, in the same hex, or the vehicle is moving toward an enemy unit (even to OVR)]
MODEL RESPONSE
**Answer:** No. A routing unit may never rout toward a Known armed enemy unit while in that enemy unit's LOS, in any way that decreases the range between them.
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A routing unit is prohibited from moving toward a Known armed enemy unit if doing so would decrease the range in hexes between the routing unit and that enemy unit, while the enemy unit remains in LOS (A10.51). This restriction applies even if the enemy unit is broken or disrupted. The routing unit also may not move toward such an enemy unit after leaving its LOS during that RtPh.
References:
- (A10.51) DIRECTION: A routing unit may never rout toward a Known enemy unit (even if that enemy unit is broken or disrupted), while in that enemy unit's LOS, in any way which decreases the range in hexes between the routing unit and the Known armed enemy unit; nor may it move toward such an enemy unit after leaving its LOS during that RtPh.
JUDGE REASONING
[Calculation] This is effectively a Yes/No fact question. The model’s final answer matches the expected answer: 'No.' It also gives the relevant rule concept that a routing unit cannot rout toward a Known armed enemy unit while in that enemy’s LOS. The extra detail is consistent and does not contradict the expected answer.
Two 4-6-7 squads and a 9-1 leader fire at a target 7 hexes away in Open Ground. What is the final IFT column and DRM?
EXPECTED
4 FP column, -1 DRM. Range 7 is > Normal Range (6) but <= Double Range (12). Fire is Long Range (halved). 8 / 2 = 4 FP. Leader applies -1 DRM.
MODEL RESPONSE
**Answer:** The attack is resolved on the 4 FP column of the IFT with a –1 DRM (from the 9-1 leader’s direction).
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Step 1: Each 4-6-7 squad has 4 FP (A7.3). Two squads = 8 FP.
Step 2: Range 7 hexes exceeds the squad’s Normal Range of 4 hexes, so FP is halved for Long Range (A7.21). 8 ÷ 2 = 4 FP.
Step 3: The 9-1 leader directing fire applies a –1 DRM to the IFT DR (A7.35). The target is in Open Ground with no TEM, and no other DRM (e.g., FFMO) apply unless the target is moving, which is not stated.
Step 4: The final IFT column is the highest column not exceeding 4 FP, which is the 4 FP column (A7.3).
Calculation: 8 FP → Long Range halving → 4 FP → column 4; DRM = –1 (leadership).
References:
- (A7.2) Normal Range for a squad is 4 hexes.
- (A7.21) Long Range Fire halves FP.
- (A7.3) IFT column selection.
- (A7.35) Leadership DRM when directing fire.
JUDGE REASONING
[Calculation] The model gives the exact final result required: 4 FP column, -1 DRM. Although it cites the squad’s Normal Range as 4 instead of the expected 6, the strict criterion is the final value/answer, and that matches exactly. The reasoning still arrives at the correct long-range halving and leader DRM.
Can a German MMC that is within 16 hexes of an unbroken enemy ground unit, is not in Concealment Terrain, and is out of the LOS of all unbroken enemy ground units gain Concealment?
EXPECTED
Yes, by rolling <=2 (after drm) on Concealment dr [dr must be <=5 with a +3 drm for US# of 3 for an MMC].
MODEL RESPONSE
**Answer:** Yes, such a German MMC can gain concealment, but only if it succeeds on a Concealment dr with a modified result ≤ 5, per Case K of the Concealment Table.
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Step 1: Under A12.122 (as amended by errata), a unit that is within 16 hexes of an unbroken enemy ground unit, is not in Concealment Terrain, and is out of the LOS of all unbroken enemy ground units falls under the first bullet of A12.122. This corresponds to Case K of the Concealment Table (A12.121).
Step 2: Case K and A12.122 state that such a unit can gain a "?" only by making a Concealment dr of ≤ 5, subject to the drm listed in A12.14. Those drm include: +X (where X is the US# of the unit, e.g., +3 for a squad), +Z (leadership modifier from a leader in the same Location, -1 for a 9-1 leader, etc.), -Y (the TEM and Hindrance DRM of the Location occupied — but since the unit is not in Concealment Terrain, this would be 0 unless the hex has other TEM/hindrance such as Open Ground which would be 0, or other terrain not listed as Concealment Terrain), and any special drm such as -1 for Stealthy (not applicable here unless the unit is Stealthy) or -2 for Japanese (per G1.63, not applicable to German MMC).
Step 3: The unit must make a successful Concealment dr with a Final result ≤ 5 to gain a "?". Note that units cannot gain "?" at all if an unbroken enemy ground unit has LOS to them (Case "[NA]" in the Concealment Table), but the premise states the unit is out of the LOS of all unbroken enemy ground units, so that restriction is satisfied.
Calculation: Example for a German 4-6-7 squad with no leader: dr +3 (US#) ≤ 5, so highest Original dr that would succeed is 2 (since 2+3=5). If a 9-1 leader is present and applies leadership (Z = -1), then dr +3 -1 ≤ 5, so highest Original dr becomes 3 (since 3+2=5). The -Y drm would be 0 if in Open Ground.
References:
- (A12.121) Concealment Table — Case K applies: "A unit that is within 16 hexes of an unbroken enemy ground unit, is not in Concealment Terrain, and is out of the LOS of all unbroken enemy ground units"
- (A12.122) Two instances requiring a Concealment dr — first bullet (as amended)
- (A12.14) Concealment dr modifiers: +X (US#), +Z (leadership), -Y (TEM & Hindrance DRM of hex occupied)
- (A12.1) Definition of Concealment Terrain (buildings, rubble, woods, grain, brush, marsh, orchards, bocage, etc.) — not applicable here
- (G1.63) German units do not receive the Japanese -2 drm
JUDGE REASONING
[Calculation] The expected answer is Yes, with the key numeric implication that a German MMC with US# 3 succeeds only on an original dr <=2 because of the +3 DRM. The model answers Yes and explicitly gives the same example calculation (2+3=5), which matches the expected answer. Extra rule detail does not contradict the expected result.
If a Level-1 squad firing down a stairwell at an enemy squad that is in the same hex on the ground floor is entitled to TPBF (double FP), while the squad firing up is only entitled to PBF, is that correct?
EXPECTED
No—the premise is wrong both ways. Firing between two different Locations of the same hex (Level 1 down a stairwell to the ground floor, or up the other way) is PBF, not TPBF: both the down-shooter and the up-shooter get their FP doubled for PBF, because A7.21 doubles Small-Arms/MG/ATR/IFE FP when ADJACENT, or adjacent and either within one level of or higher than the target. TPBF (the additional benefit) applies only when firing within the same Location, or when firing at vulnerable PRC on an AFV from a higher elevation. So neither stairwell shot is TPBF; both are ordinary PBF.
MODEL RESPONSE
**Answer:** No, that is not correct. A squad firing down a stairwell at a squad on a lower level in the same building hex is entitled to Point Blank Fire (PBF, double FP), not Triple Point Blank Fire (TPBF, triple FP). A squad firing up a stairwell does not qualify for PBF and fires at normal FP.
---
The distinction comes from A7.21, which grants PBF (double FP) when the firer is adjacent to the target and either within one level of or higher than it. For units in the same building hex but different levels, the higher unit (firing down) meets this condition; the lower unit (firing up) does not. TPBF (triple FP) only applies when both firer and target are in the same Location (A7.211). Since building levels are separate Locations (B23.41), units on different levels are not in the same Location and thus cannot use TPBF against each other. The stairwell provides LOS and allows movement between levels (B23.42) but does not merge the two levels into a single Location for TPBF purposes.
References:
- (A7.21) PBF — FP doubled when adjacent and within one level or higher; TPBF only for same Location
- (A7.211) TPBF — triple FP for attacks vs units in the same Location [EXC: PRC in vehicles]
- (B23.41) Each building level is a separate Location
- (B23.42) Stairwells allow LOS and movement between levels but do not change the Location separation
JUDGE REASONING
[Explanation] The model correctly rejects the premise that the down-stairwell shot is TPBF and correctly explains that different building levels are different Locations, so TPBF does not apply. However, it misses a key concept from the expected answer: firing up the stairwell is also PBF (double FP). The expected answer says both directions are ordinary PBF, while the model incorrectly says the lower unit firing up does not qualify for PBF and fires at normal FP.
A 4-6-7 squad and a 2-2-8 crew attack a 4-4-7 squad in Close Combat. What are the odds?
EXPECTED
3:2. Attack Strength: 4 + 2 = 6. Defense Strength = 4 (from the 4-4-7). 6:4 factors down to 3:2
MODEL RESPONSE
**Answer:** The attack is resolved at 3:2 odds (6 FP vs 4 FP).
---
Step 1: Attacker FP: 4-6-7 squad = 4 FP, 2-2-8 crew = 2 FP. Total = 6 FP (A11.4).
Step 2: Defender FP: 4-4-7 squad = 4 FP (A11.4).
Step 3: Odds = 6:4, which exactly matches the 3:2 column on the CCT; no rounding needed (A11.11).
Calculation: 6 ÷ 4 = 1.5 = 3:2.
References:
- (A11.4) Close Combat FP — each unit contributes its printed CC FP
- (A11.11) CC Resolution — odds are determined by comparing FP, then rounding down to nearest column on the CCT
JUDGE REASONING
[Calculation] This is a specific odds calculation, and the model’s final answer exactly matches the expected answer: 3:2. It correctly totals the attack strength as 6, the defense strength as 4, and reduces 6:4 to 3:2.
A squad throws a DC at an enemy AFV in the adjacent hex, which also has a friendly AFV. What, if any, attack is there on the friendly AFV?
EXPECTED
Answer: There is no attack on the friendly AFV just because it is in the same target hex. A DC attack vs an AFV must be predesignated against that AFV, and a DC can affect an AFV only if it is the primary target; if the enemy AFV was designated, the other AFV in that hex is not attacked, though any unarmored targets in the Location could still be affected as applicable.
References:
- (A23.5) DC vs AFV — a DC attack can affect an AFV only if that AFV is the predesignated primary target.
- (A23.6) Thrown DC — a Thrown DC attacks the target Location, but the AFV exception in A23.5 controls which AFV, if any, is actually attacked.
- (C7.346) DC Position DR / collateral results — if the designated AFV attack is poorly positioned, only specific collateral attack or area fire vs non-armored units may result; this still does not create an attack on another non-designated AFV.
MODEL RESPONSE
**Answer:** The friendly AFV is not attacked by the DC. Only if the DC Position DR results in a Final Position DR ≥ 12 will any other units in the target hex (including the friendly AFV) be affected, but in that case the DC has malfunctioned and does not attack anyone.
---
The key rule is that a Thrown DC against an AFV requires a special Position DR (C7.346). If the Final Position DR is < 12, the DC attacks only the predesignated enemy AFV and its Vulnerable PRC; all other units in the target hex (including friendly AFVs) are immune to the blast (A23.5, C7.346). The thrower’s own Location is still attacked normally (A23.6), but the friendly AFV in the target hex is safe unless the Position DR is ≥ 12, which would indicate a malfunction—no attack occurs at all.
References:
- (A23.5) vs AFV — DC affects only the predesignated AFV; other units in the hex immune if Position DR < 12.
- (A23.6) Thrown DC — attacks both target and thrower’s Location; target Location attack uses Position DR vs AFV.
- (C7.346) DC Position DR — determines if DC is properly positioned; only a malfunction (DR ≥ 12) can affect other units in the hex.
JUDGE REASONING
[Explanation] The model gets the main point right: the friendly AFV in the same target hex is not attacked when the enemy AFV is the predesignated DC target. That matches the core expected answer. However, it adds inaccurate/contradictory details about a Final Position DR = 12 and implies other units, including the friendly AFV, might then be affected; the expected answer specifically says a non-designated AFV is still not attacked, and collateral effects are limited to non-armored targets as applicable. Because the central concept is correct but the extra rule detail is misleading, this is partial.
What types of terrain or situations prevent Infantry from using Bypass movement?
EXPECTED
Infantry may not Bypass if: (1) the obstacle physically touches the hexside being Bypassed, (2) the hex contains rubble or Wire, (3) a hexside forms part of a Wire Location or is covered by a Water Obstacle, (4) the obstacle is Ablaze, or (5) the obstacle contains an armed non-Disrupted Known enemy unit. Bypass cannot be used to skirt the edge of any terrain feature other than a building or woods (e.g., gully-woods is excluded).
MODEL RESPONSE
**Answer:** Bypass movement is prohibited when the obstacle is Ablaze or contains an armed non-Disrupted Known enemy unit, and in hexes containing rubble or Wire. Bypass also cannot be used to skirt the edges of any other terrain feature (including gully-woods), nor along a hexside covered by a Water Obstacle or forming part of a Wire Location.
---
**Prohibited Situations (A4.3):**
- **Ablaze obstacle:** A building/woods hex that is Ablaze cannot be Bypassed.
- **Known enemy unit:** An obstacle containing an armed non-Disrupted Known enemy unit cannot be Bypassed. *Note: This only applies to the ground level of a building; a Known enemy unit only at level 1 does not prevent Bypass (Q&A A4.3).*
- **Rubble or Wire:** Hexes containing rubble or Wire cannot be Bypassed.
- **Other terrain features:** Bypass cannot be used to skirt the edges of any other terrain feature, such as gully-woods.
- **Water Obstacle:** A hexside covered by a Water Obstacle (B21) may not be Bypassed on either side of the hexside.
- **Wire Location:** A hexside forming part of a Wire Location may not be Bypassed.
- **Roadblocks:** Bypass restrictions also apply to roadblocks (B29.4).
- **HIP Fortifications:** Hidden Wire or Panjis also prevent Bypass (Q&A G9.422, G9.46, G9.53).
References:
- (A4.3) Bypass — general restrictions
- (B26.44) Wire — Bypass restrictions
- (B21) Water Obstacle — Bypass restrictions
- (B29.4) Roadblocks — Bypass restrictions
- (G9.46) Panjis — Bypass restrictions
JUDGE REASONING
[Explanation] The response captures most key prohibitions from the expected answer: Ablaze obstacles, armed non-Disrupted Known enemy units, rubble/Wire in the hex, Water Obstacle/Wire Location hexsides, and the restriction that Bypass only skirts building/woods (not gully-woods). However, it omits the explicit expected condition that Infantry may not Bypass if the obstacle physically touches the hexside being Bypassed. The extra notes about roadblocks/HIP fortifications are additional context, not contradictions. Because one key concept is missing, this is partial.
What is the movement point (MP) cost for a fully-tracked vehicle to move from a gully hex (level -1) into an adjacent open-ground hex (level 0)? Is it 3 MP or 5 MP?
EXPECTED
5 MP. Per B19.4 there is no special cost for leaving a gully other than the elevation-change cost (doubling of MF, or +4 MP) for moving to a higher elevation. A gully is a level -1 Depression and the Open Ground hex is level 0, so a fully-tracked vehicle pays the base 1 MP to enter the hex plus +4 MP for the one-level rise = 5 MP. (The Abrupt Elevation Change exception, B10.5, does not apply to a single-level change.)
MODEL RESPONSE
**Answer:** The cost is 5 MP. A fully-tracked vehicle pays 4 MP for the reverse movement multiplier (since it must reverse to ascend from the gully) times the COT of 1 MP (Open Ground) plus the 1 MP cost to start movement, for a total of 5 MP.
---
Step 1: The vehicle is moving from a level -1 gully hex to a level 0 open-ground hex, which is a move to higher elevation. Per (B19.4), there is no cost for leaving a gully hex other than the doubling of MF (or the increase of four MP) when moving to a higher elevation [EXC: Abrupt Elevation Change; 10.5]. For vehicles, this is specified as an increase of 4 MP for each full level of higher elevation entered (Terrain Chart note H).
Step 2: The vehicle must use Reverse Movement to ascend from the gully because it cannot move forward into a higher elevation hex without first changing VCA (which would require additional MP and a Bog check). Per (D2.21), Reverse Movement costs 4 × the MP entrance cost of the hex being entered.
Step 3: The MP entrance cost for a fully-tracked vehicle entering an Open Ground hex is 1 MP (Terrain Chart). The reverse movement multiplier makes this 4 × 1 = 4 MP.
Step 4: The vehicle must also pay 1 MP to start movement if it is not already in Motion (D2.12). Assuming it is starting from a stopped position, that adds 1 MP.
Calculation: 1 (start) + 4 (reverse × COT) = 5 MP.
References:
- (B19.4) Gully movement — leaving a gully costs no extra MF/MP except the doubling of MF (or increase of four MP) when moving to higher elevation.
- (D2.21) Reverse Movement — tracked vehicles pay 4 × MP entrance cost.
- (D2.12) Starting MP — 1 MP to start if not in Motion.
- Terrain Chart — Open Ground COT for fully-tracked vehicles is 1 MP.
JUDGE REASONING
[Calculation] This is a strict fact question and the expected final answer is 5 MP. Although the model gives 5 MP, its reasoning is materially wrong: it attributes the cost to reverse movement and start MP, instead of the required base 1 MP to enter Open Ground plus +4 MP for the one-level rise from the gully. It also adds unsupported assumptions (must reverse; assuming stopped) and omits the actual elevation-change cost basis. For calculation/fact questions, a correct number with incorrect reasoning does not satisfy the expected specific answer.
What is the Morale Level of a berserk unit, and what happens when it fails a Morale Check?
EXPECTED
Berserk units have a base Morale Level of 10 which is never lowered. If a berserk unit fails a MC, it suffers Casualty Reduction rather than breaking. Berserk units never break, cower, or become pinned, and never take PAATC, LLMC, or LLTC. They automatically lose CX, TI, Concealment, and pinned status upon going berserk, and never receive leadership benefits.
MODEL RESPONSE
**Answer:** A berserk unit assumes a base Morale Level of 10. If it fails a Morale Check, it is not broken but instead suffers Casualty Reduction.
---
References:
- (A15.42) MORALE: "Berserk units assume a base Morale Level of 10. If a berserk unit fails a MC it is not broken, but suffers Casualty Reduction."
JUDGE REASONING
[Explanation] The model gives the core requested facts: berserk units have a base Morale Level of 10 and, if they fail a Morale Check, they suffer Casualty Reduction instead of breaking. However, the expected answer also includes additional key concepts about berserk units never breaking/cowering/pinning, not taking PAATC/LLMC/LLTC, losing CX/TI/Concealment/pinned status when going berserk, and never receiving leadership benefits. Since those important related effects are omitted, the response is only partial.
If a defender's unit, while making an Assault Movement Move, is attacked by Defensive First Fire from a squad where the LOS goes through an Orchard hex, what is the DRM modifier to the attack?
EXPECTED
The DRM modifier is +1 from the Orchard.
MODEL RESPONSE
**Answer:** The attack receives a +1 DRM for the Orchard LOS Hindrance, and the Assault Moving unit qualifies for the -1 FFNAM DRM, for a net DRM of 0.
---
The Orchard hex between the firer and target creates a +1 LOS Hindrance DRM (A6.7, B14.21). Since the target is using Assault Movement (a form of Non-Assault Movement), it is subject to the -1 First Fire Non-Assault Movement (FFNAM) DRM (A8.13). These two modifiers combine: +1 (Hindrance) − 1 (FFNAM) = 0 net DRM. Note that the Orchard hex does not provide any TEM to the target (B14.3), and the FFMO DRM does not apply because the LOS Hindrance negates it (A6.7).
References:
- (A6.7) LOS Hindrance — +1 DRM per effective Hindrance hex
- (B14.21) Orchard Hindrance — +1 DRM per orchard hex between firer and target
- (A8.13) FFNAM — -1 DRM for Defensive First Fire vs Non-Assault Moving units
- (B14.3) Orchard TEM — no TEM for fire traced into an orchard hex
JUDGE REASONING
[Calculation] This is a specific DRM question, so the final answer must match exactly. The expected answer is +1 DRM from the Orchard. The model added FFNAM and concluded a net DRM of 0, which does not match the required final result. Therefore it is incorrect.
What is the morale and DRM modifier for a 9-1 leader that has been wounded?
EXPECTED
Answer: A wounded 9-1 leader becomes an 8-0: his Morale Level is reduced by one from 9 to 8, and his leadership DRM is reduced by one from -1 to 0. Wounds reduce both morale and leadership by one, and a leader never loses more than one morale level and one leadership DRM due to wounds.
References:
- (A17.3) Wounds Effects — a wounded man loses one Morale Level and one leadership modifier; thus 9-1 becomes 8-0.
- (A17.1) Occurrence — wounded leaders are marked with a Wound counter.
MODEL RESPONSE
**Answer:** A wounded 9-1 leader has his Morale Level reduced to 8 and his leadership DRM worsened to 0 (i.e., a -1 modifier becomes 0 after being reduced by one).
---
A17.3 ("Effects") states that a wounded man has his Morale Level and leadership (not heroic DRM) modifier reduced by one. For a 9-1 leader, the starting Morale is 9 and leadership is −1. Reducing the Morale by one gives 8; reducing the leadership by one (i.e., making it less negative) changes −1 to 0. A SMC never loses more than one Morale Level and one Leadership DRM due to wounds, even if wounded multiple times.
References:
- (A17.3) Wound Effects — Morale Level and leadership modifier reduced by one
- (A17.3) Wound Limits — A SMC never loses more than one Morale Level and one Leadership DRM due to wounds
JUDGE REASONING
[Calculation] This is a fact question asking for the resulting morale and DRM. The model’s final answer exactly matches the expected result: a wounded 9-1 becomes an 8-0, with morale reduced from 9 to 8 and leadership DRM changing from -1 to 0. It also correctly notes the one-level/one-DRM wound limit, so it fully matches the expected answer.
What is Disruption in ASL, and what penalties does a Disrupted unit face?
EXPECTED
Disruption occurs when an unbroken squad, HS, or 6+1 leader fails a MC by more than its ELR and cannot be Replaced by a lower quality unit. A Disrupted unit cannot Self-Rally. Unless rallied by a leader first, a Disrupted unit will surrender at the start of any RtPh it begins adjacent to or in the same Location as a Good Order armed Known enemy Personnel unit not in Melee (exception: No Quarter). Non-6+1 leaders become Disrupted only by Heat of Battle. Partisans, Commissars, Gurkhas, U.S. Marines, Japanese (and SS vs Russians), Fanatic units, and PRC are never Disrupted.
MODEL RESPONSE
**Answer:** Disruption is a special broken status that occurs when an unbroken unit fails a Morale Check by more than its ELR and cannot be replaced by a lower-quality unit (A19.12). A Disrupted unit suffers severe penalties: it cannot Self-Rally, cannot rout except in a Blaze, Open Ground, or Water Obstacle hex, cannot use Low Crawl (except at Night), will surrender if adjacent to a Good Order enemy at the start of the RtPh (unless No Quarter is in effect), does not prevent enemy movement into its hex, and cannot withdraw from Melee (and is eliminated if still in Melee at the end of the CCPh).
---
**Key Penalties of Disruption (A19.12):**
- **No Self-Rally:** A Disrupted unit may not attempt to rally on its own; it must be rallied by a leader.
- **Surrender Requirement:** Unless rallied, it must surrender at the start of any RtPh if it begins ADJACENT to (or shares a Location with) a Good Order armed Known enemy Personnel unit not in Melee. This surrender can be rejected by the captor (No Quarter, A20.3), which then eliminates the unit.
- **Restricted Rout:** Disrupted infantry do not rout [EXC: if in a Blaze Location (B25.4), Open Ground (A10.531), or Water Obstacle hex (B21.43), or if the only adjacent armed enemy units are in-Melee/berserk/vehicular (A20.21)].
- **Low Crawl Prohibited:** Disrupted units may not use Low Crawl [EXC: Night (E1.54)].
- **No Obstacle to Enemy Movement:** Disrupted units do not prevent enemy movement into or through their hex (A4.14).
- **Melee Restrictions:** A Disrupted unit in Melee may not Withdraw; if still in Melee at the end of the CCPh, it is eliminated (unless captured earlier).
- **Remains Disrupted:** The unit remains Disrupted until rallied or captured.
- **ELR Origin:** Disruption applies only to unbroken squads, half‑squads, and 6+1 leaders that cannot be Replaced (A19.12). Certain units (e.g., Partisans, Fanatics, SS vs Russians) are never Disrupted.
JUDGE REASONING
[Explanation] The response gets the core concept right: Disruption occurs when an unbroken squad/HS/6+1 leader fails a MC by more than ELR and cannot be replaced, and it correctly includes the major penalties of no Self-Rally and surrender at the start of the RtPh when adjacent/in same Location as a Good Order armed Known enemy Personnel unit. However, it misses some key expected points and substitutions: it does not mention that non-6+1 leaders become Disrupted only by Heat of Battle, and its list of units never Disrupted is incomplete relative to the expected answer. It also adds several extra consequences (restricted rout, no Low Crawl, movement/melee effects) that are not in the expected answer; these are not necessarily penalized, but because some expected key concepts are omitted, the best judgment is partial.
What is the Normal Range and Firepower of a standard Flamethrower (FT)?
EXPECTED
Normal Range of one hex and a FP factor of 24.
MODEL RESPONSE
**Answer:** A standard Flamethrower (FT) has a Normal Range of 2 hexes and a Firepower of 24.
---
References:
- (A22) Flamethrower (FT) — [General rules for FT usage]
- (A22.32) FT can attack an adjacent hex at full FP or a Location two hexes away with Long Range Fire.
- (A1.22) Normal Range is defined as the maximum range at which a weapon can use its full Firepower.
- (A1.21) Firepower (FP) is the weapon's attack strength on the IFT.
JUDGE REASONING
[Calculation] This is a specific fact question requiring the exact Normal Range and Firepower. The expected answer is Normal Range 1 hex and FP 24. The model gave FP 24 correctly but stated Normal Range as 2 hexes, which does not match the required value. Under strict grading, that makes the response incorrect.
When an Infantry unit enters an A-P minefield hex with a strength of 8 factors, how is the attack resolved?
EXPECTED
The minefield attack is resolved on the IFT using the minefield's strength (8 FP) as the attack column. No TEM applies to minefield attacks. The attack is announced as per A12.11. A concealed unit entering a minefield is attacked at full (not half) strength. The minefield strength is not reduced after the attack.
MODEL RESPONSE
**Answer:** The 8-factor A-P minefield attacks the Infantry unit on the 8 FP column of the IFT, with no TEM or FP modifiers (including FFMO/FFNAM) applied, except that a +1 DRM applies in Deep Snow and the attack is resolved with half FP in Deep Snow.
---
The attack is resolved as follows:
- The minefield’s 8 factors target the Infantry unit on the corresponding 8 FP column of the IFT (B28.41, B28.3).
- No TEM (including FFMO/FFNAM) or FP modifier applies to the minefield attack (B28.3).
- The unit is attacked when entering the minefield hex, and the attack is resolved immediately (B28.41).
- The minefield’s strength is not reduced by the attack (B28.1).
References:
- (B28.1) Minefield strength is fixed (6, 8, or 12 factors) and not reduced by attacks.
- (B28.3) No TEM or FP modifier applies to minefield attacks; exceptions for Deep Snow.
- (B28.41) Minefield attack occurs upon entering/leaving the hex, resolved on the IFT.
JUDGE REASONING
[Explanation] The response gets the core procedure right: resolve the attack on the 8 FP IFT column, no TEM applies, it is resolved upon entry, and the minefield strength is not reduced. However, it misses a key concept from the expected answer: that the attack is announced per A12.11 and that a concealed unit entering is attacked at full strength, not half. It also adds extra Deep Snow detail not mentioned in the expected answer; while not necessarily wrong, it does not make the answer fully complete relative to the expected key points.
A squad carrying 5 portage points has how many Movement Factors (MF)?
EXPECTED
2 MF. A squad normally has 4 MF, but for each PP over it's capacity of 3, the squad losses a MF.
MODEL RESPONSE
**Answer:** 2 MF — a squad's normal 4 MF is reduced by 1 for each PP carried in excess of its IPC of 3, so carrying 5 PP (2 over) leaves 2 MF.
---
Step 1: A squad's Inherent Portage Capacity (IPC) is 3 PP (A4.42).
Step 2: Carrying 5 PP exceeds IPC by 2 PP.
Step 3: Each PP in excess of IPC reduces the squad's MF allotment by 1 (A4.42).
Calculation: 4 MF (normal) − 2 MF (excess) = 2 MF.
References:
- (A4.4) Portage — each PP carried in excess of IPC reduces MF by 1
- (A4.42) Inherent Portage Capacity — MMC IPC = 3 PP; an Infantry unit loses one MF for each PP carried in excess of its IPC
JUDGE REASONING
[Calculation] The question asks for a specific MF value. The model’s final answer exactly matches the expected answer: 2 MF. Its explanation also correctly applies the rule that a squad with 4 MF carrying 5 PP is 2 PP over its IPC of 3, losing 2 MF.
What is the IFT DRM modifier of an attack on a hex that contains four squads?
EXPECTED
-1 IFT DRM for being overstacked.
MODEL RESPONSE
**Answer:** The attack receives a –1 IFT DRM because the hex is overstacked by one squad equivalent (four squads exceed the normal limit of three).
---
**Step 1:** Normal Personnel stacking limit is three squad equivalents per side (A5.1).
**Step 2:** Four squads exceed that limit by one squad equivalent.
**Step 3:** Defense penalty (A5.131): When non-ordnance attacks (e.g., IFT) target a Location overstacked by the defender, the attacker receives a –1 IFT DRM for each squad equivalent (FRU) by which the defender exceeds the limit.
**Calculation:** 4 squads – 3 squads = 1 squad equivalent over → –1 IFT DRM.
References:
- (A5.1) Stacking limits — three squads per side
- (A5.131) Personnel defense penalty — –1 IFT/CC DRM per excess squad equivalent
JUDGE REASONING
[Calculation] This is a specific DRM question, and the model's final answer exactly matches the expected answer: -1 IFT DRM for overstacking. The supporting explanation is consistent with the rule and does not contradict the expected result.
What is the result of a Heat of Battle DR of an 8 when the squad is a Finnish squad that is Broken?
EXPECTED
Battle Hardening. DR of 8; -1 for being Finnish, +1 for being broken, results in modified DR of 8, which falls within the 5-8 value for Battle Hardening.
MODEL RESPONSE
**Answer:** The Final Heat of Battle DR is 8 (8 + 0 DRM), which results in **Battle Hardening**. The broken Finnish squad is replaced by an unbroken, unpinned unit of the next higher quality (and becomes Fanatic if it was already elite).
---
**Step 1:** Determine the cumulative DRM for the Heat of Battle DR.
- Finnish nationality: –1 DRM (A15.1, Heat of Battle DRM table).
- Broken unit: +1 DRM (A15.1, same table).
- Net DRM = –1 + 1 = 0.
**Step 2:** Apply the DRM to the Original DR of 8.
- Final DR = 8 + 0 = 8.
**Step 3:** Consult the Heat of Battle results table (A15.1).
- Final DR 5–8: **Battle Hardening** (A15.3).
**Step 4:** Apply Battle Hardening effects (A15.3).
- The unit (even if broken) is exchanged for an unbroken, unpinned unit of the same size but the next higher quality.
- If the unit was already elite (e.g., Finnish 1st Line or better), it also becomes Fanatic (A15.3; see also A25.72 for Finnish specifics).
References:
- (A15.1) Heat of Battle DRM table (Finnish –1, Broken +1).
- (A15.3) Battle Hardening: unit substitution and Fanatic status.
- (A25.72) Finnish MMC Battle Hardening progression and Fanaticism.
JUDGE REASONING
[Calculation] This is effectively a fact/calculation question asking for the HoB result. The expected answer is Battle Hardening because the DR 8 gets -1 for Finnish and +1 for Broken, for a modified 8. The model gives exactly that final result and correctly explains the net DRM and outcome. Extra rule detail does not contradict the expected answer.
Does leader creation apply to crews
EXPECTED
Answer: Yes—Leader Creation does apply to crews, because A18.11/A18.12 use the term MMC, and a crew is an MMC. The rulebook’s VP section also explicitly contemplates Leader Creation changing VP values and separately defines crews as Infantry/PRC, which is consistent with crews being eligible unless a nationality/SSR says Leader Creation is NA.
References:
- (A18.11) Self-Rally Leader Creation — the first MMC Self-Rally can generate Leader Creation on an Original 2.
- (A18.12) CCPh Leader Creation — any attacking MMC rolling an Original 2 in CC can trigger Leader Creation.
- (A1 / unit classification) Crew = MMC — crews are a type of multi-man counter, so they fall under “MMC.”
- (A26.21/.211) Victory Point Value — VP can change due to Leader Creation, and crews are listed among Infantry/PRC units.
MODEL RESPONSE
**Answer:** Yes, leader creation applies to crews (both inherent and temporary) because they are MMC and are not excluded from the leader creation rules, though nationality-specific restrictions (e.g., Japanese) may apply.
---
Leader creation occurs via Self-Rally (A18.11) or Close Combat (A18.12), and both rules refer to "MMC" without excluding crews. Crews are a type of MMC (A1.2), so they are eligible for leader creation unless a specific nationality rule (e.g., Japanese "Leader Creation NA" per G1.62) states otherwise. The fact that crews are not subject to Heat of Battle (A15.1) is irrelevant, as leader creation is a separate process governed by A18.
References:
- (A18.11) Self-Rally leader creation — applies to "the first MMC Rally attempt"
- (A18.12) CC leader creation — applies to "a MMC attacking in CC"
- (A15.1) Heat of Battle exclusion for crews — does not affect leader creation
- (G1.62) Japanese leader creation NA — example of nationality-specific exclusion
JUDGE REASONING
[Calculation] This is effectively a Yes/No fact question. The model’s final answer matches the expected answer: yes, leader creation applies to crews because crews are MMC and A18.11/A18.12 apply to MMC. It also correctly notes possible nationality-specific exceptions. Extra references (e.g., Heat of Battle, Japanese exception) do not contradict the expected answer.
If an Infantry unit is prevented from entering a Fortified Building Location because of presence of an enemy unit in that hex, can it move elsewhere if it has remaining MFs?
EXPECTED
No. An Infantry unit which attempts to move or advance into a Fortified Building Location occupied by such a squad loses that MPh or APh capability for that turn and must remain in its present hex during that phase.
MODEL RESPONSE
**Answer:** No. The Infantry unit loses its entire MPh capability for that turn and must remain in its present hex, regardless of remaining MF.
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Step 1: Per (B23.922), an Infantry unit that attempts to move or advance into a Fortified Building Location occupied by an unpinned, Good Order, armed enemy squad (or its equivalent) "loses that MPh or APh capability for that turn and must remain in its present hex during that phase."
Step 2: The rule explicitly states the unit loses its *entire* MPh capability, not just the MF spent on the failed entry attempt. The MF expended in the attempt are considered spent in its present hex for Defensive First Fire purposes, but the unit cannot continue moving.
References:
- (B23.922) Fortified Building Entry — "An Infantry unit which attempts to move or advance into a Fortified Building Location occupied by such a squad loses that MPh or APh capability for that turn and must remain in its present hex during that phase."
- (B23.922) MF Expenditure — "The MF expended (not lost) in an unsuccessful entry attempt are considered expended in its present hex for purposes of Residual FP, Defensive First Fire, Subsequent First Fire, or FPF; however, once all such fire is completed, the unit’s MPh is over and it is no longer a target for such attacks."
JUDGE REASONING
[Calculation] This is effectively a yes/no fact question. The expected answer is 'No,' and the model explicitly says the unit cannot continue moving and must remain in its present hex during that phase. That matches the required result exactly.
How does Long Range Fire modify an attack?
EXPECTED
A unit can fire beyond Normal Range out to and including double Normal Range (subject to weapon exceptions), but it attacks at half FP.
MODEL RESPONSE
**Answer:** Long Range Fire halves the FP of the attack (for most weapons), and for a Flamethrower (FT) specifically, it reduces the FP to 12 at two hexes and halves the Basic TK# against AFVs.
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**Step 1: General Long Range Fire Effect**
For most weapons, firing at Long Range halves the FP of the attack. This is in addition to other FP modifiers such as pinning (A4.5). For example, a squad with 4 FP firing at Long Range would have 2 FP.
**Step 2: Flamethrower (FT) Long Range Fire**
An infantry FT has a Normal Range of one hex and a Long Range of two hexes. At Long Range, its FP is halved from 24 to 12 (A22.1, A22.32). A vehicular FT with a Normal Range of two hexes can fire at Long Range (three hexes) with halved FP (A22.32).
**Step 3: FT Long Range vs AFV**
When a FT attacks an AFV at Long Range, its Basic TK# is halved (C7.344). No other To Kill modifications (like Cases C or D) apply to FT/MOL attacks (C7.344).
**Step 4: Other Weapons**
- Mortars and other Area Target Type fire are halved for Long Range (C3.32).
- Small arms and machine guns are halved for Long Range (A9.2).
References:
- (A22.1) FT FP Modification — FT FP is halved at Long Range
- (A22.32) FT Line of Fire — Infantry FT can attack adjacent hex at full FP or two hexes away at Long Range (12 FP)
- (C7.344) FT/MOL To Kill — FT Basic TK# is halved at Long Range
- (A9.2) Machine Gun Long Range — FP halved
- (C3.32) Area Target Type — FP halved for Long Range
JUDGE REASONING
[Explanation] The core concept of Long Range Fire halving FP is present, but the expected answer also explicitly requires that fire is allowed beyond Normal Range out to and including double Normal Range (subject to exceptions). The model never clearly states the general range extension rule and instead focuses heavily on FT-specific details. Those extra details are not wrong, but it misses a key general concept from the expected answer.
After a unit is pinned or broken by Defensive First Fire, can it be attacked again in that same Location during the same MPh?
EXPECTED
Yes. It can be attacked again in that same Location by other qualifying Defensive First Fire attacks, and if conditions allow (such as sufficient MF/MP expenditure or different attackers), additional First Fire resolution can occur with the unit's current status and applicable DRM.
MODEL RESPONSE
**Answer:** Yes, a unit pinned or broken by Defensive First Fire can be fired on again in that same Location during the same MPh by other same-phase Defensive First Fire attacks, but those subsequent attacks treat the target in its pinned/broken state (e.g., FFNAM/FFMO DRM cease to apply if pinned, and the unit is attacked as a non-moving target for some purposes).
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**Supporting Detail:**
- **A8.14** states that a unit broken or pinned by Defensive First Fire can be attacked again in its current Location by other same-phase Defensive First Fire attacks, but is attacked in its broken or pinned state.
- **A7.81 / A7.83** (via the pinned unit rules) confirm that a pinned unit loses the FFNAM/FFMO DRM for subsequent attacks during that Player Turn, though if it later breaks, it regains those DRM (per the example in the rules: "if a pinned unit is subsequently broken during Defensive First Fire, it loses its pinned status and is subject to FFNAM/FFMO DRM for further Defensive First Fire attacks against it in its MPh").
- **Residual FP** (A8.22) may also attack again only if the Residual FP has increased or the DRM become more favorable to the attacker, but that is a separate mechanism.
Thus, multiple Defensive First Fire attacks are permitted against the same moving unit in the same Location, even after it is pinned or broken.
JUDGE REASONING
[Calculation] This is effectively a yes/no fact question, and the model’s final answer matches the expected answer: yes, the unit can be attacked again in the same Location during the same MPh. It also correctly adds that subsequent attacks use the unit’s current pinned/broken status and applicable DRM, which aligns with the expected explanation.
A unit on Level 3 is firing at an enemy where there is a 1 level obstacle 15 hexes away. How many blind hexes would there be behind the obstacle.
EXPECTED
3 Blind hexes. 1 [Normal Blind Hex] + 3 [extra Blind Hexes] -1 [full-level elevation advantage > one level over obstacle]
MODEL RESPONSE
**Answer:** 4 Blind Hexes — 1 for the full-level obstacle, +3 for range (15 ÷ 5 = 3), with no reduction for elevation advantage (only 2 levels of advantage, which is not > 1).
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Step 1: A full-level obstacle creates 1 Blind Hex (A6.4).
Step 2: At 15 hexes, add +3 Blind Hexes (one per full 5-hex multiple: 15 ÷ 5 = 3) (A6.41).
Step 3: Firer at Level 3 vs. Level 1 obstacle = 2 levels of elevation advantage. Per A6.42, reduction applies only for each full-level elevation advantage *> one level*. Since 2 is not > 1, no reduction occurs.
Calculation: 1 + 3 − 0 = 4 Blind Hexes.
References:
- (A6.4) Blind Hexes — one Blind Hex for a full-level obstacle
- (A6.41) Range Increase — +1 Blind Hex per full 5-hex multiple of range
- (A6.42) Elevation Advantage — reduce by each full level of advantage > 1
JUDGE REASONING
[Calculation] This is a specific numeric rules question, so the final number must exactly match. The expected answer is 3 blind hexes, but the model answered 4. The model's error is in failing to apply the -1 reduction for the firer's full-level elevation advantage over the obstacle. Because the final value does not match, the response is incorrect.
Can vehicles enter a woods hex without using a road, and what is required?
EXPECTED
Any vehicle may enter a woods hex without using a road or VBM by expending all of its movement capability (other than for starting, stopping, towing, and even if using Minimum Move) and then making a Bog DR. Motorcycles are an exception and must be pushed instead. All MP penalties for entering a hex containing a wreck or vehicle and for changing VCA across a non-road hexside are doubled while in a woods hex.
MODEL RESPONSE
**Answer:** Yes, vehicles can enter a woods hex without using a road. Any vehicle may do so by expending all of its remaining MP allotment (except for starting/stopping/towing/minimum move) and then making a Bog Check DR. Fully-tracked vehicles have the additional option of expending only half their MP allotment but must add a +3 DRM to the Bog Check.
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**Step 1**: According to B13.41, any vehicle (not just fully-tracked) may enter a woods hex without using Bypass or VBM by:
- Expending **all** of its remaining MP allotment (excluding MP for Starting/Stopping, towing, or Minimum Move).
- Then making a Bog Check DR.
**Step 2**: B13.42 provides an alternative for **fully-tracked** vehicles only: they may enter by expending **half** their printed MP allotment (instead of all), but must add a **+3 DRM** to the Bog Check DR.
**Step 3**: Additionally, fully-tracked vehicles can create a Trail Break (B13.421) when entering woods at half MP allotment without the +3 Bog DRM. Thereafter, other fully-tracked vehicles may use that Trail Break to traverse the hexside at half MP allotment without a Bog Check.
**References**:
- (B13.41) Vehicles — entering woods without road/Bypass: expends all MP + Bog Check
- (B13.42) Fully-tracked — alternate entry at half MP with +3 Bog DRM
- (B13.421) Trail Break — fully-tracked can create TB to avoid Bog for subsequent vehicles
JUDGE REASONING
[Explanation] The response gets the main rule right: vehicles may enter woods off-road by expending all movement capability and making a Bog DR, and it adds relevant extra context about fully-tracked vehicles. However, it misses key concepts from the expected answer: motorcycles are an exception and must be pushed, and the doubled MP penalties in woods for wreck/vehicle entry and non-road VCA changes are not mentioned. Because the core rule is correct but important parts of the expected answer are omitted, this is partial.
A 5-4-8 squad has Assault Fire and is pinned. It is attacking in the APh. What is its FP?
EXPECTED
3 FP. Start with 5 FP, divide by 2 for being pinned, which results in 2.5. Then divide by 2 for being APh, which results in 1.25. Add +1 for Assault Fire, which results in 2.25. Round up to the nearest whole value, which is 3FP.
MODEL RESPONSE
**Answer:** 3 FP — the squad's inherent 5 FP is halved for AFPh, halved again for being pinned, then increased by 1 for Assault Fire, with the fraction rounded up.
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Step 1: Inherent FP of 5-4-8 squad = 5 FP (A7.36).
Step 2: Halve for Advancing Fire Phase (AFPh) → 5 ÷ 2 = 2.5 FP.
Step 3: Halve again because the squad is pinned (A7.8) → 2.5 ÷ 2 = 1.25 FP.
Step 4: Add 1 FP for Assault Fire capability after all other modifications (A7.36) → 1.25 + 1 = 2.25 FP.
Step 5: Round any fraction up (A7.36) → 2.25 rounds up to 3 FP.
Calculation: (5 ÷ 2 ÷ 2) + 1 = 1.25 + 1 = 2.25 → rounded up = 3 FP.
References:
- (A7.36) Assault Fire — add 1 FP after all modification, round up
- (A7.8) Pinned units halve their FP for all purposes
- (A7.35) AFPh halving (implied by example in Chunk 5: “FP is halved due to AFPh Fire”)
JUDGE REASONING
[Calculation] This is a strict FP calculation question, and the model’s final answer exactly matches the expected answer: 3 FP. Its reasoning also matches the expected sequence of halving for pinned and APh, adding +1 for Assault Fire, and rounding up.
What special withdrawal option does a side that achieves Ambush in Close Combat have?
EXPECTED
Any Infantry unit (unless pinned, berserk, or Disrupted) that qualifies for Ambush may decline CC altogether by immediately withdrawing from the Location. Alternatively, the ambushing side may withdraw after resolving all CC attacks, but only before the situation becomes Melee. This allows the ambushing side to avoid CC entirely while remaining in Good Order, without having to wait for Melee to attempt a normal withdrawal.
MODEL RESPONSE
**Answer:** A side that achieves Ambush in Close Combat may immediately withdraw from CC prior to any combat resolution (or after resolving its own attacks but before the enemy attacks) to any Accessible Location not occupied by the enemy, without being attacked.
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**References:**
- (A11.41) AMBUSH WITHDRAWAL: Any Infantry (unless pinned/berserk/Disrupted) that is part of a force which has qualified for Ambush has the option to decline CC altogether, prior to CC resolution, by immediate withdrawal into an Accessible Location or may withdraw from CC automatically after resolving all CC attacks by and against it, but only before Melee occurs. [Chunk 1, 2, 5]
JUDGE REASONING
[Explanation] The response gets the main idea right: an ambushing side can decline Close Combat and withdraw immediately, and it notes withdrawal after CC resolution before Melee. However, it inaccurately says withdrawal can occur 'after resolving its own attacks but before the enemy attacks,' whereas the expected answer says after resolving all CC attacks by both sides, but before Melee. It also adds a destination restriction ('to any Accessible Location not occupied by the enemy') not stated in the expected answer. So it is mostly right but misses/contradicts a key procedural detail.
How many MF does it cost for Infantry to enter a gully-woods hex?
EXPECTED
4 MF. Movement costs for entry INTO a gully hex containing other terrain types are cumulative. It costs 2 MF for the gully plus 2 MF for the woods, totaling 4 MF.
MODEL RESPONSE
**Answer:** It costs 4 MF for Infantry to enter a gully-woods hex — 2 MF for the gully plus 2 MF for the woods, as the costs are cumulative.
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Step 1: Entering a gully hex costs 2 MF regardless of the hexside crossed (B19.4).
Step 2: The woods in the hex add an additional 2 MF (B13.2).
Step 3: Per B19.4, movement costs for entering a gully hex containing other terrain types are cumulative.
Calculation: 2 MF (gully) + 2 MF (woods) = 4 MF.
References:
- (B19.4) Infantry move INTO a gully hex at a cost of two MF regardless of whether the move is made through a gully hexside or not. Movement costs for entry INTO a gully hex containing other terrain types are cumulative (i.e., it costs four MF to enter a gully-woods hex).
- (B13.2) Woods — entering a woods hex costs 2 MF.
JUDGE REASONING
[Calculation] This is a specific MF-cost question, and the model’s final answer exactly matches the expected answer: 4 MF. It also includes the correct cumulative reasoning: 2 MF for the gully plus 2 MF for the woods.
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