05 Apr '08 23:04

I am sure by now most of you have seen some version of the game where you try to be the one to take the last pearl/chip/token, or else force the other person to do the same.

I haven't analyzed it fully, but I was pondering the same sort of game only with three people taking turns. I have a suspicion how that will turn out but let me present to you the game set-up.

21 tokens of some sort are placed on the table in the center. Each player in turn can take either 1, 2, or 3 tokens at a time. You are given the option to pick whether you want to go first, second, or third, and you'll always take your turn in that order. (A, B, C, A, B, C, A, B, C...)

If there is no collusion between the other two, is there a way to guarantee a win, and if so how? If not, is there any way to maximize your chances? What if the object is to avoid taking the last token?

In your analysis, assume a player will make a winning move if given the chance, and if no definitively winning move is available will randomly pick 1, 2, or 3 tokens.

I haven't analyzed it fully, but I was pondering the same sort of game only with three people taking turns. I have a suspicion how that will turn out but let me present to you the game set-up.

21 tokens of some sort are placed on the table in the center. Each player in turn can take either 1, 2, or 3 tokens at a time. You are given the option to pick whether you want to go first, second, or third, and you'll always take your turn in that order. (A, B, C, A, B, C, A, B, C...)

If there is no collusion between the other two, is there a way to guarantee a win, and if so how? If not, is there any way to maximize your chances? What if the object is to avoid taking the last token?

In your analysis, assume a player will make a winning move if given the chance, and if no definitively winning move is available will randomly pick 1, 2, or 3 tokens.