<p>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 &quot;p&quot;. Assume that in a tie the reward is split equally among winners. I know that if p&lt;1 then the Nash equilibrium is 0. If p&gt;1 the Nash equilibrium is 100.</p>
<p>What about if p=1?</p>
<p>Is the Nash equilibrium a mixed strategy that assigns equal probability to all numbers between 0 and 100?</p>
<p>OR</p>
<p>Is the Nash equilibrium simply choosing 50 (the mean of all numbers between 0 and 100)?</p>
<p>OR is the Nash equilibrium something else entirely?</p>
<p>I am confused and so is my professor, please help :)</p>

CC BY-SA 4.0
2023-06-15T06:55:11.943Z
Is it possible there is not *a* Nash equilibrium but *several* Nash equilibria?
Giskard
CC BY-SA 4.0
2023-06-15T06:56:16.317Z
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.
Giskard
CC BY-SA 4.0
2023-06-15T08:13:34.527Z
This game is only really funny if played by 2 players.
VARulle
<game-theory><nash-equilibrium><behavioral-economics>
Nash equilibrium in p-beauty contest game where p=1
Arif
# Nash equilibrium in p-beauty contest game where p=1

(Asked by: username_0 on 2023-06-14)

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 :)

**Tags:** game-theory, nash-equilibrium, behavioral-economics
> **Comments:**
> **username_1**: Is it possible there is not *a* Nash equilibrium but *several* Nash equilibria?  
> **username_1**: 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.  
> **username_2**: This game is only really funny if played by 2 players.  

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## 0 answers

