Editing Nash equilibrium (section)

== Oddness of equilibrium points ==
{| class="wikitable floatright" style="text-align:center; font-size:95%; margin-left:1em; margin-bottom:1ex"
|+ Free money game
!scope="col" rowspan="2"| Player A<br/> votes
!scope="row" colspan="2"| Player B votes
|-
!scope="col" {{Yes}}
!scope="col" {{No}}
|-
!scope="row" {{Yes}}
| 1, 1
| 0, 0
|-
!scope="row" {{No}}
| 0, 0
| 0, 0
|}

In 1971, Robert Wilson came up with the "oddness theorem",<ref>{{Cite journal|last=Wilson|first=Robert|date=1971-07-01|title=Computing Equilibria of N-Person Games|url=https://epubs.siam.org/doi/abs/10.1137/0121011|journal=SIAM Journal on Applied Mathematics|volume=21|issue=1|pages=80–87|doi=10.1137/0121011|issn=0036-1399}}</ref> which says that "almost all" finite games have a finite and odd number of Nash equilibria. In 1993, Harsanyi published an alternative proof of the result.<ref>{{Cite journal|last=Harsanyi|first=J. C.|date=1973-12-01|title=Oddness of the Number of Equilibrium Points: A New Proof|url=https://doi.org/10.1007/BF01737572|journal=International Journal of Game Theory|language=en|volume=2|issue=1|pages=235–250|doi=10.1007/BF01737572|s2cid=122603890|issn=1432-1270}}</ref> "Almost all" here means that any game with an infinite or even number of equilibria is very special in the sense that if its payoffs were even slightly randomly perturbed, with probability one it would have an odd number of equilibria instead.

The [[prisoner's dilemma]], for example, has one equilibrium, while the [[battle of the sexes (game theory)|battle of the sexes]] has three—two pure and one mixed, and this remains true even if the payoffs change slightly. The free money game is an example of a "special" game with an even number of equilibria. In it, two players have to both vote "yes" rather than "no" to get a reward and the votes are simultaneous. There are two pure-strategy Nash equilibria, (yes, yes) and (no, no), and no mixed strategy equilibria, because the strategy "yes" weakly dominates "no". "Yes" is as good as "no" regardless of the other player's action, but if there is any chance the other player chooses "yes" then "yes" is the best reply. Under a small random perturbation of the payoffs, however, the probability that any two payoffs would remain tied, whether at 0 or some other number, is vanishingly small, and the game would have either one or three equilibria instead.