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?

(In reply to re(3): Solution by Charlie)

You're right. x should only go for a corner square if that really is the first move of a game. In this case there are random moves ahead of it.

My posted algorithm is much too simplistic. For instance, when the first random moves are x in (0,0) and 0 in (0,1), the game continues with: x at (0,2) [???], 0 at (1,1), x at  (2,1), 0 at (2,0), x at (2,2), 0 at (1,2), 0 at (1,0) DRAW 

If instead x at (0,0), 0 at (0,1), x at (2,0) [!!!], 0 at (1,0), x at (2,2), then x WINS [0 at (1,1), x at (2,1) or 0 at (2,1), x at (1,1)]     

I'll try to update that algorithm to an intelligent one.

Later.....

 

Edited on October 5, 2004, 4:17 pm

Posted by Penny on 2004-10-05 16:06:02