Trick-Tac-Toe

For simplicity, let the robots play all nine moves. This means that there are $9!$ games that can be played, all equally likely. Then consider the first stage when each game is resolved (after anything between $5$ and $9$ moves).

For the game to be resolved at the $5$ th move, player X must fill a winning line in moves 1,3,5. There are thus $8 \times 3! \times 6! = 34560$ games in which this happens.

For the game to be resolved at the $6$ th move, player O must fill a winning line in moves 2,4,6, while player X cannot have filled a line in moves 1,3,5. There are $8 \times 3! \times 6! = 34560$ games in which a winning line is filled in at moves 2,4,6, and there are $6 \times 2 \times 3! \times 3! \times 3! = 2592$ games in which a winning line is filled in in both moves 1,3,5 and 2,4,6 (the lines must be either both horizontal or both vertical, so there are 6 choices for O's line and then 2 choices for X's line). Thus there are $34560-2592=31968$ games that are resolved at the $6$ th move.

For the game to be resolved at the $7$ th move, player X must complete a line with either moves 1,3,7 or 1,5,7 or 3,5,7, and player O cannot have completed a line at moves 2,4,6. Since it is not possible in these circumstances for player X to complete a line in moves 1,3,5, we deduce that there are $3$ times as many games that complete at the $7$ th move as complete at the $6$ th. Thus there are $3 \times 31968 = 95904$ games that complete at the $7$ th move.

For a game to complete at the $8$ th move, player O must complete a line at moves 2,4,8 or 2,6,8 or 4,6,8 (and therefore not at 2,4,6), while player X has not completed a line by then. There are $2 \times 3! \times 6! \times 3=25920$ games in which this is done by completing a diagonal line (there are 2 line choices and 3 move combinations in which the line can be completed, and X cannot certainly not complete a line) and a further $6 \times 9 \times 3! \times 4! \times 2! \times 3 = 46656$ games in which this is done by completing a horizontal or vertical line (there are 6 choices for the winning line, 9 configurations of X's four plays that do not complete a line, and 3 move combinations in which the winning line can be completed). This means that there are $25920+46656=72576$ games that are resolved at the $8$ th move.

All other games are resolved at the $9$ th move. There are $127872$ such games.

Thus the expected number of moves until resolution is $\begin{aligned} E & = \; \frac{5 \times 34560 + 6 \times 31968 + 7 \times 95904 + 8 \times 72576 + 9 \times 127872}{9!} \;= \; \frac{3203}{420} \end{aligned}$ and hence $1000E = 7626.19$ , making the answer $\boxed{7626}$ .