Suppose that, instead of playing in your turn, you have to submit in advance, an ordered list of the 9 board positions. Then a computer plays for you by using, whenever it is your turn, the leftmost position that is still untaken. For example, suppose you are "X" and you play first. Then the computer puts "X" in the first position on your list. Then the opponent plays. Then the computer puts "X" in the second position on your list (if it is still free), otherwise it pust "X" in the third position. And so on.

Note that the other player can see your list and plan accordingly. Can you force a tie when you play first? When you play second?

Let's place a few constraints on what X's list could be.

First, the first three moves in X's list must make a line. Otherwise, O can just let X make the first three moves and then make a line for themselves. No three moves that do not create a line prevent another line from happening (if they are not all on different rows or columns, then there is an empty row or column; otherwise it looks like:

(or rotated), for which there is a diagonal 3-in-a-row.)

If X does not take the center as your first move, then O plays in the center. If X's 3-in-a-row passes through the center, then X's first two moves are locked into two diametrically opposite points. Then O can simply look at the X's fourth move, and decide not to make a 3-in-a-row there. (there are two 3-in-a-row's left).

If X's 3-in-a-row does not pass through the center, it must be one of the edges.
If X's 4th move does not block O's line then O can play in X's 3rd and then 4th positions.

Otherwise X's first four moves will look like one of the following configurations:

O can disrupt the 3-in-a-row and make their own 3-in-a-row, without having to play in the center.

So we have shown that X's first move must play in the center, and then the next two moves must be diametrically opposite points.
But now we have a problem. If X plays in a diagonal, then O disrupts this diagonal and looks at X's 4th move on the list. Then O can make a line not involving X's 4th move.

If X plays in a horizontal or vertical, then O can check which line involving X's 2nd or 3rd move does not contain X's 4th move. Then O disrupts on that line, forming a 3-in-a-row.

Thus we can conclude that X cannot even force a tie (and indeed will lose after the third turn) with a predetermined list.
(If the person with the predetermined list plays second, then you can follow the same strategy as above, except that you gain priority on same turn moves, which you don't even need.)

• I believe the result is correct, but the reasoning is slightly wrong. In the paragraph "If X's 3-in-a-row does not pass through the center ...", your picture shows actually one possibility that X does not lose after the third turn. After X plays 2, O must play 3 in order not to lose. But then X's third move would be the original fourth move, which blocks O's winning move. In that situation, O should play the first move somewhere else (which can be done, because X's moves are all known in advance) to win in the third turn. Feb 6 at 20:27
• I'm offering to improve the graphics on your post, or is the style intended? Feb 7 at 1:57