Two expert and jaded tic-tac-toe players, after drawing for the n-th 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?

Each player has 1/2 chance to be X (the first player), for whom the following program found 43 wins out of 72 games. 0 (the second player) won no games. There were 29 draws.

So if in 144 games (144 = 72*2) where each player is x as many times as he is 0, he will win 43 times, lose 43 times and draw 58 times. 

So each players, the chances are:

Victory: 0.3 Defeat 0.3 Draw 0.4

The VB program ran in zero elapsed time. (When the random feature was removed, the result of a game was a draw).

[Too lengthy to post]

Edited on October 5, 2004, 9:08 am

Posted by Penny on 2004-10-05 09:06:49