While munching on some French toast with my 3-year old at the local chain-restaurant, I noticed one of the games on the back of the kid's menu. There were 10 ears of corn and two players were supposed to take turns coloring in either 1, 2 or 3 ears (the players choice). The winner is whomever colors in the last ear.

This sounded to me like the basis of a good puzzle. First, the rules:

1) 10 ears of corn
2) Each player takes a turn and can color in either 1, 2, or 3 ears
3) The players are smart enough to always color in the number of ears which maximizes their chance of wining

The puzzle:
A) For 10 ears of corn, who will win --- the player who goes first, or the player who goes second.
B) If you played this repeatedly with a random number of ears of corn (for simplicity, let's say between 1 and 100 ears, inclusive), how often would the first player win and how often would the 2nd player win.

Just walking backward, and looking at some choices the players would have to make:

-If it is your turn, and there are one two or three left, you have won.
-If it is your turn and there are exactly four left, you have lost.
-If it is your turn and there are five, six, or seven left, you can win, by leaving your opponent with four.
-If it is your turn, and there are exactly eight left, you have lost (against a smart opponent).
-If it is your turn and there are nine, ten, or eleven left, you can win by leaving your opponent with exactly eight.

With a random starting number, the first player wins 75% of the time.

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What antoniseb said.

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