Hat tip to my son for telling me about the game below...
Suppose there are four players who can independently choose either 2 or 10 and will receive $2 for 2 and $10 for 10 EXCEPT that if all four choose 10 they will all get zero.
In that game, three players choosing 10 and one choosing 2 is Nash, in that no one has an incentive to deviate knowing the other's choices. Working out mixed Nash is trickier...the intuition is that if the others play a strategy mix that leaves you with a payoff of x from playing 10 and the same payoff of x from playing 2, that mix is Nash...so let's see...your payoffs are equal if there is a .8 chance the others will all choose 10 and a .2 chance they won't...the tech is then that the others "should" play 10 with a p = p to the third power = .8 or about .928 if mixed Nash is treated as normative.
Now, I personally believe in mixed Nash as a powerful guide to unlocking social reality. (Maybe that belief is quite wrong--but it makes sense given the academic project I'm engaged in, which tries to marry mixed Nash with social preferences like altruism and competitiveness.)
Give a belief in mixed Nash and egoistic preferences, one would expect that a lab experiment would show people playing 10 a bit over 90% of the time. My intuition is that that would in fact be borne out.
(A qualification to reflect my strong belief in analyses that consider social preferences: The prediction above is dependent on the framing of the experiment implying that self-interested preferences--as opposed to social preferences of one kind or another--are the relevant ones in the situation.)
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