<p>I solved it, haha. Turns out I was being silly. I had completely neglected the fact that this was a 3-player game, and $\text{not entering}$ still had probability values attached to it. Let $A$, $B$, and $C$ be the three firms. So the game looks like this:
$$
\text{P(A enter;B,C enter)+P(A enter;only B enter)+P(A enter;only C enter)+}\\
\text{P(A enter;neither enters) = P(A dne;B,C enter)+P(A dne;only B enter)+}\\
\text{P(A dne;only C enter)+P(A dne;neither enters)}
$$
It's a giant mess of $p$'s and $(1-p)$'s, but at the end of the day, the final version $8=9(1-p^2)$ is derived.</p>

CC BY-SA 4.0
2020-12-10T23:14:12.267Z
if you solved the issue you can accept your answer to indicate that this problem is solved
1muflon1
CC BY-SA 3.0
2016-11-10T03:29:47.647Z
2016-11-10T03:29:47.647Z
peachyloaf
<p>Question: Suppose three identical, risk-neutral firms must decide simultaneously and irreversibly whether to enter a new market which can accommodate only two of them. If all three firms enter, all get payoff 0; otherwise, entrants get 9 and firms that stay out get 8. </p>

<p>My attempt: Arbitrarily choosing firm A, firm A is deciding whether to enter the market or not. Hence, the Nash equilibrium occurs when $\text{payoff of not entering = entering}$. Let $p$ be the probability of each firm entering the market. Then
$$\text{not entering = P(only firm who enters) + P(second firm to enter) + P(third firm to enter)}\\
8=9(p)(1-p)^2+9(p^2)(1-p)+0(p^3)\\
8=9(p-p^2)
$$</p>

<p>and I get a complex number as the answer. The correct equation should be $8=9(1-p^2)$, but I'm not sure how to get that. </p>

<p>Any help is greatly appreciated!</p>

<p>Edit: Solved it an hour later. I was being silly. I'm leaving my partial answer below for anyone who is curious. </p>

<game-theory><nash-equilibrium>
Symmetric mixed-strategy equilibrium: Entering markets [Solved]
peachyloaf
# Symmetric mixed-strategy equilibrium: Entering markets [Solved]

(Asked by: username_0 on 2016-11-10)

Question: Suppose three identical, risk-neutral firms must decide simultaneously and irreversibly whether to enter a new market which can accommodate only two of them. If all three firms enter, all get payoff 0; otherwise, entrants get 9 and firms that stay out get 8.

My attempt: Arbitrarily choosing firm A, firm A is deciding whether to enter the market or not. Hence, the Nash equilibrium occurs when $\text{payoff of not entering = entering}$. Let $p$ be the probability of each firm entering the market. Then
$$\text{not entering = P(only firm who enters) + P(second firm to enter) + P(third firm to enter)}\\
8=9(p)(1-p)^2+9(p^2)(1-p)+0(p^3)\\
8=9(p-p^2)
$$

and I get a complex number as the answer. The correct equation should be $8=9(1-p^2)$, but I'm not sure how to get that.

Any help is greatly appreciated!

Edit: Solved it an hour later. I was being silly. I'm leaving my partial answer below for anyone who is curious.

**Tags:** game-theory, nash-equilibrium

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## 1 answers

### Answer by username_0 on 2016-11-10. Score: 2

I solved it, haha. Turns out I was being silly. I had completely neglected the fact that this was a 3-player game, and $\text{not entering}$ still had probability values attached to it. Let $A$, $B$, and $C$ be the three firms. So the game looks like this:
$$
\text{P(A enter;B,C enter)+P(A enter;only B enter)+P(A enter;only C enter)+}\\
\text{P(A enter;neither enters) = P(A dne;B,C enter)+P(A dne;only B enter)+}\\
\text{P(A dne;only C enter)+P(A dne;neither enters)}
$$
It's a giant mess of $p$'s and $(1-p)$'s, but at the end of the day, the final version $8=9(1-p^2)$ is derived.

> **Comments:**
> **username_1**: if you solved the issue you can accept your answer to indicate that this problem is solved  

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