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0 sats \ 2 replies \ @SimpleStacker OP 3 Dec 2024 \ parent \ on: [Economics Puzzle] John Nash was wrong? econ

This is correct enough. In the end, they will randomize. To be fully exhaustive, there are actually $C_{3}^{4} + C_{2}^{4} + C_{1}^{4} + 1$ Nash equilibria!

$C_{3}^{4}$ . Three of the guys go for the blonde's friends, one guy goes for the blonde.
$C_{2}^{4}$ . Two of the guys go for the blonde's friends. Two of the guys randomize over the blonde.
$C_{1}^{4}$ . One of the guys goes for the blonde's friends. Three of the guys randomize over the blonde.
$1$ . All of the guys randomize over the blonde.

The exact probabilities in each equilibrium depend on the relative utilities of the blonde, her friends, and getting no one.

30 sats \ 1 reply \ @south_korea_ln 3 Dec 2024

Thanks for formalizing it and hinting me in the right direction. And thanks for the sats :)

EDIT: so, (4+6+4+1=)15 equilibria?

119 sats \ 0 replies \ @SimpleStacker OP 3 Dec 2024

so, (4+6+4+1=)15 equilibria?

That should be correct, assuming all the males are actually identical. If they have different preferences, or if the blonde may select one over another when there's a conflict, then the solutions could change.

Fun fact: The number of Nash equilibria, including randomized ones, which we call "mixed strategies", is always odd.