Can the first player always win when two boys are playing a game called last counter if there are no passes and the player who takes the last counter loses? From a pile of 8 counters, 1 or 2...

Can the first player always win when two boys are playing a game called last counter if there are no passes and the player who takes the last counter loses? From a pile of 8 counters, 1 or 2 counters are removed by each player taking turns.

Asked on by sinkin

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

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Yes. Say you're player 1 and your opponent is player 2. Note that you'll definitely win if you leave him with only one counter. You can force him to have only one counter if you have either 2 or 3 counters the move before him.

Since you'll have 2 or 3 counters if he has 4 the move before that, you'll win if you can force him to have exactly 4 counters at some point. But the same reasoning shows that you can guarantee this if you have 5 or 6 counters at some point. How can you guarantee this? Easy, force him to have exactly 7 counters. Since you go first, you can do this.

Thus, player 1's winning strategy is to take 1 counter to get the opponent to 7, then take either one or two to get him to 4, then take one or two to get him to 1, when he loses. I believe this is a version of the famous game "Nim".

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