I\'m looking for a proof or algorithm for which player would win if there were n

Question

I'm looking for a proof or algorithm for which player would win if there were n stones. Using the same play style, either take one or two stones and the person to take the last stone loses.

Theorem 1.If the game of Poison starts with 10 stones, then the second player to go can always win.

Proof.

Player 2 can use the following strategy to win Poison with 10 stones:

1. Whenever player 1 takes 1 stone, player 2 takes 2 stones in the next move.

2. Whenever player 1 takes 2 stones, player 2 takes 1 stone in the next move.

Explanation / Answer

If N is 1 (mod 3)

then player 2 can always win.

this is because if

1. Whenever player 1 takes 1 stone, player 2 takes 2 stones in the next move.

2. Whenever player 1 takes 2 stones, player 2 takes 1 stone in the next move.

so following this strategy, after player 2 select stone, the number of stone will always decrease by 3.

so if N is 1 (mod 3)

then finally 1 stone will be left for player 1 to pick.

so , player1 looses as he/she picked last stone.

we know 10 is 1 (mod 3)

so

If the game of Poison starts with 10 stones, then the second player to go can always win.

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