Two expert and jaded tictactoe players, after drawing for the
nth time, decided to add some randomness to their favorite game.
First, they used a coin to decide who would start. Then, that player would pick his initial move randomly. Next, the other player would also pick his answer randomly. Finally, from then on the game went on as usual, with each player playing in the best possible way.
For each player, what are the odds of winning, losing, or drawing?
(In reply to
re(2): Solution by Penny)
BTW, you say "If I'm the second player ("0"), and the center square is empty, put a "0" there..."
Do you not also do this if you are the first player? Say the situation is such that the first two random moves put neither X nor O in the center. Would X not also take the center in that circumstance?
Can your program summarize, or list for all 72 cases, whether that category of first two moves, or that specific two first moves, leads to a win for the first player or a draw? As you came up with an odd number (43) of wins for the first player, there must be some place where, say, X in a corner followed by O in an adjacent side leads to a win, but a different set of corner followed by adjacent side (i.e., a symmetrically the same pair of moves) leads to a tie. So if you list out all 72 possibilities, matching up initial pairs with their symmetric equivalents will point out some disparities that can be looked into.

Posted by Charlie
on 20041005 13:46:37 