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 Here's a try by Rajal)

By symmetry, you just have to consider X going first. As you calculated, there are 72 cases:

X on corner, O on corner          -- Win (12)


X on corner, O on edge            -- Win (16)


X on corner, O on center          -- Draw (4)





X on edge,   O on center          -- Draw (4)


X on edge,   O on adjacent edge   -- Win  (8)


X on edge,   O on opposite edge   -- Draw (4)


X on edge,   O on adjacent corner -- Draw (8)


X on edge,   O on other corners   -- Win  (8)





X on center, O on corner          -- Draw (4)


X on center, O on edge            -- Win  (4)

Out of the 72 cases, I get 48 Wins (not 40) and 24 draws. So, the player who goes first has 2/3 odds of a win, and 1/3 tie -- adding the initial toss, you get 1/3 equal chances for win, draw, or loss.

Posted by Federico Kereki on 2004-10-04 12:41:13