The Last Chip Pays

Question

A bowl holds 24 poker chips. You and a colleague alternate turns, each removing 1, 2, or 3 chips. Whoever removes the last chip buys dinner for the desk. You move first. How many chips do you take, and how do you play from there?

Accepted Answer

Take 3, leaving 21. Then, whatever number k your opponent takes, respond with 4 - k. Backward analysis. A bowl of 1 is a forced loss for the player to move (they must take the last chip). From 2, 3, or 4 you can dump your opponent onto 1, so those are winning. From 5, every move (take 1, 2, or 3) leaves 4, 3, or 2, all winning for the opponent, so 5 is a loss. The losing positions repeat with period 4: 1, 5, 9, 13, 17, 21, 25, dots quad text(sizes equiv 1 bmod 4text). From 24 the unique winning move is to 21. Your 4-k replies walk the bowl down through 17, 13, 9, 5 and finally hand over a single chip. Each of your turns restores the invariant "size equiv 1 bmod 4 on their move," and they can never escape it because no move changes the count by a multiple of 4. The trap: people who know the normal-play version ("last chip wins, leave multiples of 4") apply it blindly and take 1 here, hoping to reach 20. That hands the win away: the misère target shifts by exactly one chip, from 0 bmod 4 to 1 bmod 4. The desk-relevant habit: in any alternating game, characterize the losing set and the move that re-enters it, then verify the opponent can never jump over it. One modular observation rep

The Last Chip Pays

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