<p>There can't be an equilibrium in which no one tells the police. In that case, it is a profitable deviation to tell since <span class="math-container">$v-c&gt;0$</span>.</p>
<p>There can't be an equilibrium in which more than one player tells the police. In that case, it is a profitable deviation for one of them to remain silent since <span class="math-container">$v&gt;v-c$</span>.</p>
<p>There are five pure strategy Nash equilibria in which only one player talks. This one player would be worse off by remaining silent, <span class="math-container">$0&lt;v-c$</span>, and the silent players would be worse off by telling, <span class="math-container">$v-c&lt;v$</span>.</p>
<p>Next, you consider mixed strategies. For the same reason as above, there is no equilibrium in which one or more players alway talk and others mix between talking and silence as pure silence is better given someone talks to the police for sure. Similarly, there is no equilibrium in which only one player randomizes and all others remain silent, as this one player would not be indifferent between both actions, talking is always better given the others don't.</p>
<p>There are equilibria in which two players randomize and the others are silent. Wlog, suppose player 1 and 2 talk with probabilities <span class="math-container">$p$</span> and <span class="math-container">$q$</span>, respectively. For each player, talking always gives payoff <span class="math-container">$v-c$</span>. Both players must be indifferent (otherwise they would not want to randomize). Hence,
<span class="math-container">$v-c = p v + (1-p)0 = q v + (1-q)0$</span> which we can solve for <span class="math-container">$p=q=\frac{v-c}{v}$</span>. Given this indifference condition, it must also be a best response for player 3,4,5 to remain silent. This is true as <span class="math-container">$v-c\leq p^2 v + 2p(1-p)v$</span>, because <span class="math-container">$v-c=pv&lt;(p^2+2p(1-p))v$</span> and <span class="math-container">$p&lt;(p^2+2p(1-p))$</span> for any <span class="math-container">$p\in(0,1)$</span>. There are <span class="math-container">$\binom{5}{2}=10$</span> such equilibria.
Similarly, there should be <span class="math-container">$\binom{5}{3}=10$</span> and <span class="math-container">$\binom{5}{4}=5$</span> equilibria in which 3 and 4 players talk and the others are silent, and one equilibrium in which all five players randomize over both options with some probability <span class="math-container">$z$</span> solving <span class="math-container">$v-c=(1-(1-z)^4)v$</span>.</p>
<p>All in all, there should be <span class="math-container">$5+10+10+5+1=31$</span> Nash equilibia.</p>

CC BY-SA 4.0
2020-12-28T13:12:02.477Z
@mariazag what do you mean?
Bayesian
CC BY-SA 4.0
2020-12-28T15:16:10.737Z
The problem gave me 4 possible solutions.. 1 2 3 or 5 equilibrium.. I can expect that 5 is the answer as we get only the choosing of person k
mariazag
CC BY-SA 4.0
2020-12-28T15:25:42.440Z
Then perhaps the question is only after Nash equilibria in pure strategies? In that case, 5 would be the correct answer.
Bayesian
CC BY-SA 4.0
2020-12-28T15:28:12.633Z
Yes! Thank you very much for your help!! Its gem!
mariazag
CC BY-SA 4.0
2020-12-28T17:20:02.607Z
You're welcome. Consider accepting the answer so the question does not remain "unanswered."
Bayesian
CC BY-SA 4.0
2020-12-28T18:05:22.020Z
How can I do that ?
mariazag
CC BY-SA 4.0
2020-12-28T18:08:14.983Z
I think I make it
mariazag
CC BY-SA 4.0
2020-12-28T11:35:01.970Z
2020-12-28T11:35:01.970Z
Bayesian
<p>5 people are eyewitnesses to a criminal incident. Each one of them wants the police to get informed but they prefer a third person to give the information.
Suppose k is one of the 5 people.
If the police receive information from the person l≠k, the utility of k will be equal with v.
If the police receive information from the person k, its utility will be equal with v-c, where v&gt;c.
If the police don’t receive information, the utility of k will be 0.</p>
<p>How many Nash equilibria are there in the above game?</p>

<microeconomics><game-theory>
How many Nash equilibria are there in the above game?
mariazag
# How many Nash equilibria are there in the above game?

(Asked by: username_0 on 2020-12-28)

5 people are eyewitnesses to a criminal incident. Each one of them wants the police to get informed but they prefer a third person to give the information.
Suppose k is one of the 5 people.
If the police receive information from the person l≠k, the utility of k will be equal with v.
If the police receive information from the person k, its utility will be equal with v-c, where v>c.
If the police don’t receive information, the utility of k will be 0.

How many Nash equilibria are there in the above game?

**Tags:** microeconomics, game-theory

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## 1 answers

### Answer by username_1 on 2020-12-28. Score: 3 | Accepted answer

There can't be an equilibrium in which no one tells the police. In that case, it is a profitable deviation to tell since $v-c>0$.

There can't be an equilibrium in which more than one player tells the police. In that case, it is a profitable deviation for one of them to remain silent since $v>v-c$.

There are five pure strategy Nash equilibria in which only one player talks. This one player would be worse off by remaining silent, $0, and the silent players would be worse off by telling, $v-c.

Next, you consider mixed strategies. For the same reason as above, there is no equilibrium in which one or more players alway talk and others mix between talking and silence as pure silence is better given someone talks to the police for sure. Similarly, there is no equilibrium in which only one player randomizes and all others remain silent, as this one player would not be indifferent between both actions, talking is always better given the others don't.

There are equilibria in which two players randomize and the others are silent. Wlog, suppose player 1 and 2 talk with probabilities $p$ and $q$, respectively. For each player, talking always gives payoff $v-c$. Both players must be indifferent (otherwise they would not want to randomize). Hence,
$v-c = p v + (1-p)0 = q v + (1-q)0$ which we can solve for $p=q=\frac{v-c}{v}$. Given this indifference condition, it must also be a best response for player 3,4,5 to remain silent. This is true as $v-c\leq p^2 v + 2p(1-p)v$, because $v-c=pv<(p^2+2p(1-p))v$ and $p<(p^2+2p(1-p))$ for any $p\in(0,1)$. There are $\binom{5}{2}=10$ such equilibria.
Similarly, there should be $\binom{5}{3}=10$ and $\binom{5}{4}=5$ equilibria in which 3 and 4 players talk and the others are silent, and one equilibrium in which all five players randomize over both options with some probability $z$ solving $v-c=(1-(1-z)^4)v$.

All in all, there should be $5+10+10+5+1=31$ Nash equilibia.

> **Comments:**
> **username_1**: @mariazag what do you mean?  
> **username_0**: The problem gave me 4 possible solutions.. 1 2 3 or 5 equilibrium.. I can expect that 5 is the answer as we get only the choosing of person k  
> **username_1**: Then perhaps the question is only after Nash equilibria in pure strategies? In that case, 5 would be the correct answer.  
> **username_0**: Yes! Thank you very much for your help!! Its gem!  
> **username_1**: You're welcome. Consider accepting the answer so the question does not remain "unanswered."  
> **username_0**: How can I do that ?  
> **username_0**: I think I make it  

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