Tic-Tac-Toe-Take-Two

First, we're going to assume that the game goes on until full, even if someone has already won (this doesn't affect the probability we're after).

There are 9!/5!/4! = 126 final configurations (the 5! is the number of ways the Xs can be arranged, and the 4! is the number of ways the Os can be arranged).

There are 8 ways to win. If X gets three in a row, there are these 8 wins, times the number of ways the other 2 Xs can be placed, which is 6!/2!/4! = 15; 8*15 = 120. However, we count some configurations twice, since X could get two three-in-a-rows. There are 2*6 = 12 ways X can have a diagonal and either a vertical or horizontal; there are 3*3 = 9 ways X can have a vertical and a horizontal; and there is 1 way X can have both diagonals. In total, there are 98 configurations where X has a row.

There are 8*6 = 48 ways for O to get three in a row (the 6 is the number of places for the 4th O).

In order for *both* players to have a row, neither can have a diagonal. There are therefore 6 ways for X to have a row, 2 remaining for O to have a row, and 3 ways for the 4th O to fall. So, there are 36 ways for both to have a row.

So, so far, there are 12 ways for only O to have a row (all diagonals), 62 ways for only X to have a row, 36 for both to have a row, and 16 configurations that are draws.

Now, of the configurations where both have a row, there are 5!4! = 2880 ways to reach these configurations. Consider X's last move. 3 of the 5 Xs are part of X's row, so there are 3*4! ways for X to move such that he completes his row on his last turn. There are still 4! ways for O to move. Therefore, there are 3*4!*4! = 1728 ways for X to finish his row on the last move (and O wins). In order for X to finish on his 4th move, one of the first 3 Xs must be in one of the 2 non-row spaces; there are 3*2*3! ways to do this. There are 3! ways for O to finish his row on his 3rd turn (and O wins); so, the number of ways both happen is 3*2*3!*3! = 216. These are the only two ways O wins if both have a row, so O wins 1944/2880 of these games, and X wins 936/2880 of them.

So, in summary:

X wins: (62 + 36*(936/2880))/126 = 737/1260 (.5849206...) O wins: (12 + 36*(1944/2880))/126 = 121/420 (.2880952...) Draw: 8/63 (.1269841...)