Nash Equilibria in simultaneous game with four players

Question

Four parliamentary parties are working on a necessary but highly unpopular law. Each party decides whether to put forward the law on its own behalf. If $n$ parties will put forward the law on its own behalf, where $1 \leq n \leq 4$, then each party will lose reputation of size $\frac{12}{n}$. If no party puts forward the law then each party will lose reputation of size 15.

How can I find pure strategy Nash equilibria in this game?

Thanks!

[{JonMarkPerry} A variation is to say not putting forward the law results in +1 rep, except for if no party does, in which case -15.]

Answer 1

Since you are asking about pure-strategy Nash equilibria, we can make some simple observations:

• We can never have an equilibrium where more than one party propose the bill. Say $m>1$ parties are proposing the bill, and consider one of those parties $i$. Then $i$ gets utility $-\frac{12}{m}$. However, they can deviate to not proposing the bill, and get utility 0.
• If no party proposes the bill, each party gets utility $-15$. In that case, any single party $i$ can increase their utility to $-12$ by deviating to proposing the bill.

From these it should be easy to deduce the pure-strategy Nash equilibria.

Answer 2

Consider what happens if Party A proposes. They have a 1/8 chance of losing -12, a 3/8 chance of losing -6, a 3/8 chance of losing -4 and a 1/8 chance of losing -3.

If Party A doesn't propose, these become -15, 0, 0, 0.

In the first case, the sum over all 8 events is -45, so the mean is -6 3/8. In the second, the mean is -1 7/8.

However, if every party was to use this philosophy, then each party would be down -15. As no consultation is allowed, the logic ends.