<p>Assume the players have to choose integers, otherwise a best response may not exist. Let the payoff of winning be $\alpha\cdot[\frac23\text{ of the average}]$, $\alpha&gt;0$. Consider the two player (A and B) case, and let's verify whether choosing above $0$ is optimal. </p>

<p>Suppose A chooses $x&gt;0$. Then B can guarantee a win by choosing $x-1$, since $\frac23(x-\frac12)$ is closer to $x-1$ than to $x$. However, if B chooses $0$, A choosing $x=1$ is a best response (A gets zero anyway). So we will have a NE where not both players choose $0$, and the one who chooses $0$ gets positive payoff.</p>

<p>This equilibrium can be generalized to the $n$-player case. Let $n-1$ players choose $0$ and the remaining one choose $1$. In this equilibrium, those who choose $0$ would share the positive payoff (each gets $\frac{\alpha}{n-1}\cdot\frac23\cdot\frac1{n}$). They are best responding because choosing any $x&gt;0$ would imply zero payoff. The one who chooses $1$ is also best responding since he gets zero anyway. (Of course this is not the only equilibrium. There is also one where everyone chooses $0$, which is obvious.)</p>

<p>Admittedly, the above equilibrium relies crucially on the assumption that players are only allowed to choose integers. But this is due to the fact that payoff is a function of the choices (compared to a fixed amount in the original version of the problem). Suppose players are allowed to choose any real number in $[0,100]$. Then if A chooses $x&gt;0$, B would best respond by choosing $x-\epsilon$, where $\epsilon=\min\{y:y&gt;0\}$ (B would win of course, but she also wants to maximize her earning by making her choice as close to A's as possible). 
However, such an $\epsilon$ does not exist. Therefore, the only equilibrium is for everyone to choose $0$. </p>

<p>The intuition is this: however large the winning is, you don't get it if you don't win. But winning requires that you choose small. So the incentive to win overwhelms the incentive to win big.</p>

CC BY-SA 3.0
2016-11-09T20:53:18.477Z
"those who choose 0 would share the positive payoff " Why do players who won by guessing zero get a positive payoff? I thought they win 5 times 0, which is still zero?
Giskard
CC BY-SA 3.0
2016-11-09T21:06:09.480Z
@denesp: OP says the payoff is equal to 2/3 of the average, which is positive, since 1 person chooses above $0$.
Herr K.
CC BY-SA 3.0
2016-11-10T07:25:30.053Z
Sorry, I misread that. You are correct.
Giskard
CC BY-SA 3.0
2016-11-09T19:52:43.417Z
2016-11-09T21:08:06.583Z
2016-11-09T21:08:06.583Z
Herr K.
<p>@denesp nailed the answer on the head.</p>

<p>This is not different from the original game in any meaningful way, therefore, the nash equilibrium remains the same. Size of payouts don't change the outcome of the game. This is because the strategy of choosing zero (assuming all parties are fully rational, a condition for Nash Equilibrium) dominates all other strategies.</p>

<p>Keep in mind that in reality not all parties are fully rational, so the results of the game you conducted with your students shouldn't be expected to reflect this.</p>

CC BY-SA 3.0
2016-11-09T19:21:10.573Z
2016-11-09T19:21:10.573Z
Lee Sin
<p>Recently I conducted a small game among students of our institute. The game was based on Keynes' beauty contest game. The participants had to guess a number between 0 to 100 and the participant whose guess was closest to 2/3rd of the average of all guesses would win the game. </p>

<p>The twist was this: We also awarded money to the winner in each round. The money was a multiple say, 5x of the final number (2/3rd of the average). </p>

<p>What will be the Nash Equilibrium of this modified game. The solution to the original problem is of course everybody choosing 0. Will the equilibrium shift in the above described game. Consider n to be greater than 30. 
Any help would be appreciated. </p>

CC BY-SA 3.0
2016-11-09T14:49:33.420Z
I am confused. What exactly is the difference between this and the original game? That if I win my winnings are proportional to my guess?
Giskard
CC BY-SA 3.0
2016-11-09T17:38:35.480Z
The money part. Won't people want the number to be high. 0 will not be their best choice. I want to be sure whether the game outcome will be the same or not.
Sub-Optimal
<game-theory><nash-equilibrium><repeated-games>
Nash Equilibrium of modified Keynes' beauty contest
Sub-Optimal
# Nash Equilibrium of modified Keynes' beauty contest

(Asked by: username_0 on 2016-11-09)

Recently I conducted a small game among students of our institute. The game was based on Keynes' beauty contest game. The participants had to guess a number between 0 to 100 and the participant whose guess was closest to 2/3rd of the average of all guesses would win the game.

The twist was this: We also awarded money to the winner in each round. The money was a multiple say, 5x of the final number (2/3rd of the average).

What will be the Nash Equilibrium of this modified game. The solution to the original problem is of course everybody choosing 0. Will the equilibrium shift in the above described game. Consider n to be greater than 30.
Any help would be appreciated.

**Tags:** game-theory, nash-equilibrium, repeated-games
> **Comments:**
> **username_1**: I am confused. What exactly is the difference between this and the original game? That if I win my winnings are proportional to my guess?  
> **username_0**: The money part. Won't people want the number to be high. 0 will not be their best choice. I want to be sure whether the game outcome will be the same or not.  

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## 2 answers

### Answer by username_2 on 2016-11-09. Score: 2 | Accepted answer

Assume the players have to choose integers, otherwise a best response may not exist. Let the payoff of winning be $\alpha\cdot[\frac23\text{ of the average}]$, $\alpha>0$. Consider the two player (A and B) case, and let's verify whether choosing above $0$ is optimal.

Suppose A chooses $x>0$. Then B can guarantee a win by choosing $x-1$, since $\frac23(x-\frac12)$ is closer to $x-1$ than to $x$. However, if B chooses $0$, A choosing $x=1$ is a best response (A gets zero anyway). So we will have a NE where not both players choose $0$, and the one who chooses $0$ gets positive payoff.

This equilibrium can be generalized to the $n$-player case. Let $n-1$ players choose $0$ and the remaining one choose $1$. In this equilibrium, those who choose $0$ would share the positive payoff (each gets $\frac{\alpha}{n-1}\cdot\frac23\cdot\frac1{n}$). They are best responding because choosing any $x>0$ would imply zero payoff. The one who chooses $1$ is also best responding since he gets zero anyway. (Of course this is not the only equilibrium. There is also one where everyone chooses $0$, which is obvious.)

Admittedly, the above equilibrium relies crucially on the assumption that players are only allowed to choose integers. But this is due to the fact that payoff is a function of the choices (compared to a fixed amount in the original version of the problem). Suppose players are allowed to choose any real number in $[0,100]$. Then if A chooses $x>0$, B would best respond by choosing $x-\epsilon$, where $\epsilon=\min\{y:y>0\}$ (B would win of course, but she also wants to maximize her earning by making her choice as close to A's as possible).
However, such an $\epsilon$ does not exist. Therefore, the only equilibrium is for everyone to choose $0$.

The intuition is this: however large the winning is, you don't get it if you don't win. But winning requires that you choose small. So the incentive to win overwhelms the incentive to win big.

> **Comments:**
> **username_1**: "those who choose 0 would share the positive payoff " Why do players who won by guessing zero get a positive payoff? I thought they win 5 times 0, which is still zero?  
> **username_2**: @denesp: OP says the payoff is equal to 2/3 of the average, which is positive, since 1 person chooses above $0$.  
> **username_1**: Sorry, I misread that. You are correct.  

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### Answer by username_3 on 2016-11-09. Score: 1

@denesp nailed the answer on the head.

This is not different from the original game in any meaningful way, therefore, the nash equilibrium remains the same. Size of payouts don't change the outcome of the game. This is because the strategy of choosing zero (assuming all parties are fully rational, a condition for Nash Equilibrium) dominates all other strategies.

Keep in mind that in reality not all parties are fully rational, so the results of the game you conducted with your students shouldn't be expected to reflect this.


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