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Setup: players must chose a number between 0 and 100. The winner of the game is the player whose chosen number is closest to the average of all chosen numbers multiplied by "p". Assume that in a tie the reward is split equally among winners. I know that if p<1 then the Nash equilibrium is 0. If p>1 the Nash equilibrium is 100.

What about if p=1?

Is the Nash equilibrium a mixed strategy that assigns equal probability to all numbers between 0 and 100?

OR

Is the Nash equilibrium simply choosing 50 (the mean of all numbers between 0 and 100)?

OR is the Nash equilibrium something else entirely?

I am confused and so is my professor, please help :)

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  • $\begingroup$ Is it possible there is not a Nash equilibrium but several Nash equilibria? $\endgroup$
    – Giskard
    Commented Jun 15, 2023 at 6:55
  • $\begingroup$ An example: If there are 10 players, each of them saying 79 is a Nash equilibrium, any single deviatior would lose out on the reward. $\endgroup$
    – Giskard
    Commented Jun 15, 2023 at 6:56
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    $\begingroup$ This game is only really funny if played by 2 players. $\endgroup$
    – VARulle
    Commented Jun 15, 2023 at 8:13

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