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 :)
So games in which there cannot be a tie have no Nash equilibrium?
No, that’s not a general rule. It’s just the case that in this particular game, if you’re losing you always have a better option that can be achieved just by changing your own strategy. If your prospects for improvement relied on others changing their strategies too, then you could lose and still be in a Nash equilibrium. (For an example of such a game, see battle of the sexes))
I thought the requirement was that there’s no way to do better with an alternative strategy.
Sort of. It’s that there’s no way to do better with an alternative strategy, given perfect knowledge of others’ strategies.
I was also assuming that everyone guesses at the same time
They do in the actual game; it’s just that that’s not relevant to evaluating what counts as a Nash equilibrium.
But this means it’s no longer a perfect-information game, and that there’s not going to be a Nash equilibrium.
I’m not entirely clear what you mean by the first half of this sentence, but the conclusion is false. Even if everyone guessed in turn, there would still be a Nash equilibrium with everyone playing zero.
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 :)
No, that’s not a general rule. It’s just the case that in this particular game, if you’re losing you always have a better option that can be achieved just by changing your own strategy. If your prospects for improvement relied on others changing their strategies too, then you could lose and still be in a Nash equilibrium. (For an example of such a game, see battle of the sexes))
Sort of. It’s that there’s no way to do better with an alternative strategy, given perfect knowledge of others’ strategies.
They do in the actual game; it’s just that that’s not relevant to evaluating what counts as a Nash equilibrium.
I’m not entirely clear what you mean by the first half of this sentence, but the conclusion is false. Even if everyone guessed in turn, there would still be a Nash equilibrium with everyone playing zero.
No problem. ;)
Sorry I didn’t/can’t continue the conversation; I’ve gotten rather busy.