Symmetric mixed-strategy equilibrium: Entering markets [Solved] 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.
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.