Two player game about maximizing earnings subject to an interesting condition

Question

Me and my friend had a bet. We each pick an integer between $1$ and $100$ inclusive and reveal it at the same time. Whoever picks the higher number has his number halved. Then the person whose number is lower (after this operation) has to pay the other person an amount equal to the difference.

For example, if my friend picks $80$ and I pick $48$ then he is forced to pay me $\$8$.

I have no idea what my friend will pick. What is the optimal number I should play to maximize my earnings?

Answer 1

Playing $1$ is a dominated strategy. So, suppose your friend is playing $y\in \{2,\dots,100\}$, then your best response is to choose $x$ in the following way.

• If you choose $x=y$, the your payoff is $0$.
• If you choose $x>y$, the your payoff is $\frac{x}{2}-y$. So if you are going to choose $x>y$ you should choose $x=100$, which gives you a payoff of $\frac{100}{2}-y$.
• If you choose $x<y$, the your payoff is $x-\frac{y}{2}$. So if you are going to choose $x<y$ you should choose $x=y-1$, which gives you a payoff of $\frac{y}{2}-1$.

You will be indifferent between the last two choices if $$\frac{100}{2}-y=\frac{y}{2}-1 \Rightarrow y=34 $$

Hence, your best response function can be summarized as

$$ x= \begin{cases} y-1 &\text{, if }y>34 \\[1.5ex] \{33,100\} &\text{, if }y=34 \\[1.5ex] 100 &\text{, if }y<34 \end{cases}$$

Your friend's best response is symmetric. Since the two best response functions do not cross, there is no (pure-strategy) Nash equilibrium to this game.