Can Nash Equilibrium be more than two?
Yes! In the Nash equilibrium none of the players gains more by deviating his/her strategy from the equilibrium point. For an example in the following two-player reward table, there exist "many" equilibria:
$$\left[ \begin{array}{ccc} 1/1 & 0/0 & 0/0& 0/0& 0/0& 0/0 \\ 0/0& 0/0 & 0/0& 0/0& 0/0& 0/0 \\ 0/0 & 0/0 & 0/0& 0/0& 0/0& 0/0 \\ 0/0 & 0/0& 0/0&1/1 & 0/0& 0/0 \\ 0/0 & 0/0 & 0/0& 0/0& 0/0& 0/0 \\ 0/0 & 0/0& 0/0& 0/0& 0/0 &1/1 \end{array} \right]$$
Most games have an odd number of Nash equilibrium. For example in the coordination game below: $$ \begin{array}{|c|c|c|} \hline P1\backslash P2 & PC & MAC \\ \hline PC & 2,2 & 0,0 \\ \hline MAC & 0,0 & 3,3 \\ \hline \end{array} $$ You have 3 Nash equilibria: (PC,PC), (MAC,MAC) and also one in mixed strategies where each player chooses PC with probability 3/5 and MAC with prob. 2/5.
Consider 3 hunters, Bob, Charlie, and Doug
with the following choices of things to hunt:
A moose: worth 9 units of food A wolf: worth 4 units of food A rabbit: worth 1.5 units
Now there are plenty of rabbits so if a hunter opts to hunt a rabbit they will definitely get their one rabbit
A wolf requires at least 2 hunters, if 2 hunters hunt it each then each gets 2 units of food, but if all 3 hunters go for the wolf then the pay off is is only 1.3333 units of food so it is more advantangeous to go for an individual rabbit,
Lastly we Moose requires all 3 hunters to be hunted successfully, naturally if hunted each hunter gets 3 units of food resulting in the highest payoof but highest risk.
If we set up our 3 dimensional payoff matrix by presenting its slices
Assume Bob hunts a rabbit (Bob = first index, Charlie = second index, Doug = third index) $$ \left( \begin{array}{ccc} & R & W & M \\ R & 1.5,1.5,1.5 & 1.5,0,1.5 & 1.5,0,1.5 \\ W & 1.5,1.5,0 & 1.5,2,2 & 1.5,0,0 \\ M & 1.5,1.5,0 & 1.5,0,0 & 1.5,0,0 \end{array} \right)\ $$
Assume Bob hunts a wolf
$$ \left( \begin{array}{ccc} & R & W & M \\ R & 0,1.5,1.5 & 2,2,1.5 & 0,0,1.5 \\ W & 2,1.5,2 & 1.33,1.33,1.33 & 2,2,0 \\ M & 0,1.5,0 & 2,2,0 & 0,0,0 \end{array} \right)\ $$
Assume Bob hunts a moose
$$ \left( \begin{array}{ccc} & R & W & M \\ R & 0,1.5,1.5 & 0,0,1.5 & 0,0,1.5 \\ W & 0,1.5,0 & 0,2,2 & 0,0,0 \\ M & 0,0,0 & 0,0,0 & 3,3,3 \end{array} \right)\ $$
Now there are 3 equillibrium strategies for the players:
- Nobody trusts anybody so they all go rabbit
Someone expects at least one person but not the other to be trustworthy so they go the way of wolf
Total trust results in the moose
Now if Bob, Charlie, and Doug take random strategies the expected payout for Bob g(and expected payout for Charlie and Doug too by symmetry is)
$$ 1.5 \frac{9}{27} + 2 \frac{4}{27} + \frac{4}{3} \frac{1}{27} + 3 \frac{1}{27} = 0.95...$$