If two people play tic tac toe choosing their moves randomly, what are the odds that they will tie? What is the probability that it is a tie game?
What is the probability that X wins?
What is the probability that O wins?
Assume X goes first and all moves are random.
 A: In 2001 I worked out the possible number of games
Win in 5 moves    1440
Win in 6 moves    5328
Win in 7 moves   47952
Win in 8 moves   72576
Win in 9 moves   81792
Draw             46080
Total           255168 

but these are not equally likely; you need to weight the wins in $5$ moves by $4!$, the wins in $6$ moves by $3!$, and so on.  If you do that then 


*

*the probability for the first player wining at random is $\frac{1440\times 4!+47952\times2!+81792\times0!}{9!}\approx 0.585$

*the probability for the second player wining at random is $\frac{5328\times3!+72576\times1!}{9!}\approx 0.288$

*the probability of a draw is $\frac{46080\times0!}{9!}\approx 0.127$

A: Tie: Since all final outcomes (with 5 Xs and 4 Os) are equally likely, we can count the number of outcomes which end in a tie.


*

*With an X in the center


*

*X must appear in 1 of each pair: (NW, SE), (NE, SW), (N, S), (E, W), giving $2^4 = 16$ options

*However, not all of these are  ties. There cannot be 3 Xs in a row, which can happen in one of $4$ different rows, and there will then be $2$ choices for placement of the last X, giving $8$ disallowed possibilities

*So there are $16 -8 = 8$ possibilities with an X in the center


*With an O in the center


*

*X in four corners and one edge: never a tie

*X in three corners and two edges: determined by the omitted corner, $4$ possibilities

*X in two corners and three edges: one choice given omitted edge, $4$ possibilities

*X in one corner and four edges: never a tie

*Total of $8$



So there are $16$ total final positions with ties, out of $\binom{9}{5}$ final positions
