> 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.

SimpleStacker

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

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

south_korea_ln

econ

[Economics Puzzle] John Nash was wrong?

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.

