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Let's say, two players are playing extended version of tic-tac-toe on 10X10 board (who first has 5 tiles next to each other, wins). Should both players play Nash equilibrium strategy, would it mean, that there would never be a winner?

If this claim is true, could it be shown based on the argument that the situation, when none of the players is willing to change her strategy given her oponent's strategy, can only occur when none of the players is going to lose, but neither of them is able to win?

Sorry, if the question seems vague or makes no sence. I am a newbie to GT. Edit suggestions welcome.

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    $\begingroup$ No, it just means that one player will not want to change his or her strategy given what the other is playing--so if playing their current strategy means they lose, then they won't do any better playing a different strategy. One of the earliest theorems in game theory is Zermelo's Theorem, which says that in any finite, two-person game of perfect information and alternating moves, either a) one of the sides can force a win, or b) both can force a draw. $\endgroup$ – Trurl Sep 25 '17 at 21:07
  • $\begingroup$ Thank you! The Zermelo's theorem is very interesting. Can it be decided, if a) or b) happens for a particular game? $\endgroup$ – Jan Vainer Sep 26 '17 at 10:32
  • $\begingroup$ No, the theorem itself only says either a) or b) happens (for example, it applies to chess) but in any particular game it might be possible to decide which is the case. $\endgroup$ – Trurl Sep 27 '17 at 12:48

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