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How many combinations are possible in the game tic-tac-toe (Noughts and crosses)?

So for example a game which looked like: (with positions 1-9)

A1   --   B1

A2   --   B2

A3   --   --

[1][3][4][6][7] would be one combination

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closed as off-topic by user21820, A. Pongrácz, Gibbs, mrtaurho, Eevee Trainer Jan 7 at 2:19

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    $\begingroup$ se16.info/hgb/tictactoe.htm $\endgroup$ – Daryl Jan 2 '13 at 10:00
  • $\begingroup$ @Daryl, if you feel up to it, you should repackage that as an answer citing that website so this question can have an answer. $\endgroup$ – Mark S. Nov 22 '13 at 3:42
  • $\begingroup$ @MarkS. Sure thing. $\endgroup$ – Daryl Nov 22 '13 at 5:14
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This information is taken from this website.

A naive estimate would be $9!=362\,880$, since there are $9$ possible first moves, $8$ for the second move, etc. This does not take into account games which finish in less than $9$ moves.

  • Ending on the $5^\text{th}$ move: $1\,440$ possibilities
  • Ending on the $6^\text{th}$ move: $5\,328$ possibilities
  • Ending on the $7^\text{th}$ move: $47\,952$ possibilities
  • Ending on the $8^\text{th}$ move: $72\,576$ possibilities
  • Ending on the $9^\text{th}$ move: $127\,872$ possibilities

This gives a total of $255168$ possible games. This calculation doesn't take into account symmetry in the game.

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  • $\begingroup$ Accounting for symmetry, this can quickly be reduced by a factor of 6, as there are only 12 possible two move openings, not 8*9=72. $\endgroup$ – Pieter Geerkens Nov 22 '13 at 5:32
  • $\begingroup$ As that linked page makes clear at the bottom, symmetry allows a reduction by a factor of $8$ (i.e. rotations and reflections of a square) to $31896$, but arguably by a greater amount to $26830$. $\endgroup$ – Henry Jan 2 '15 at 11:18
  • $\begingroup$ I am assuming I am asking a naive question here, but why isn't the answer $9! - \sum\limits_{i=5}^{8} m_i = 235584$, where, $m_i$ are the number of games ending on $i$ moves (the numbers above). (From the overestimation of the games that take all 9 moves we subtract the ones that require less) $\endgroup$ – kyriakosSt Dec 3 '17 at 23:47
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I will say that the board combinations are 3^9, which is 19683 possibilities, and 2032 winning positions. The answer of 9! is related to how many ways we have to fell all the positions, rather than the possible combinations.

I have answered this question already in another post, please see the next link: https://stackoverflow.com/a/54035004/5117217

Cheers!

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