A Nash equilibrium is a set of strategies from which no player has an incentive to deviate, holding others' strategies constant. Take any putative set of (pure) equilibrium strategies; if there is any individual who loses when this set of strategies is played, then they have an incentive to change their guess to 2/3 of the average, and this set of strategies is not a Nash equilibrium. This implies that you are not in Nash equilibrium unless everyone wins.*
Holding other players' strategies constant, you have a single optimal strategy, which is to play 2/3 of the average. If there is another player who has already guessed 2/3 of the (new) average then you tie with probability 1; if there is not, you win with probability 1.
* Note that everyone winning is necessary, but not sufficient for a Nash equilibrium. Everyone playing 67 lets everyone win, but it is not a Nash equilibrium. If if anyone prefers not to tie, they could deviate and win by themselves.
So games in which there cannot be a tie have no Nash equilibrium?
I must have misread the wikipedia page; I thought the requirement was that there's no way to do better with an alternative strategy.
I was also assuming that everyone guesses at the same time, as otherwise the person to play last can always win (and so everyone will play 0). But this means it's no longer a perfect-information game, and that there's not going to be a Nash equilibrium. Thanks for your patience :)
I'd like to play a game with you. Send me, privately, a real number between 0 and 100, inclusive. (No funny business. If you say "my age", I'm going to throw it out.) The winner of this game is the person who, after a week, guesses the number closest to 2/3 of the average guess. I will reveal the average guess, and will confirm the winner's claims to have won, but I will reveal no specific guesses.
Suppose that you're a rational person. You also know that everyone else who plays this game is rational, you know that they know that, you know that they know that, and so on. Therefore, you conclude that the best guess is P. Since P is the rational guess to make, everyone will guess P, and so the best guess to make is P*2/3. This gives an equation that we can solve to get P = 0.
I propose that this game be used as a sort of test to see how well Aumann's agreement theorem applies to a group of people. The key assumption the theorem makes--which, as taw points out, is often overlooked--is that the group members are all rational and honest and also have common knowledge of this. This same assumption implies that the average guess will be 0. The farther from the truth this assumption is, the farther the average guess is going to be from 0, and the farther Aumann's agreement theorem is from applying to the group.
Update (June 20): The game is finished; sorry for the delay in getting the results. The average guess was about 13.235418197890148 (a number which probably contains as much entropy as its length), meaning that the winning guess is the one closest to 8.823612131926765. This number appears to be significantly below the number typical for groups of ordinary people, but not dramatically so. 63% of guesses were too low, indicating that people were overall slightly optimistic about the outcome (if you interpret lower as better). Anyway, I will notify the winner ahora mismo.