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What are the perfect Bayesian equilibria in the following game?

enter image description here

The part that is confusing me is the right-hand side, where player 2 is of type N.

Thanks!

EDIT: My workings out so far are given below:

We start off by noticing that if player 2 is type $N$, it can be rational for him to play both $Steal$ and $Split$, as he does not know what player 1 plays.

Let us therefore first look at the case where player 2 plays $Steal$ if he is type $N$. Then player 1's expected payoff from stealing is $100\beta$ and his expected payoff from splitting is $50\alpha + 50\beta$. In this case player 1 steals when $100\beta > 50\alpha + 50\beta \Leftrightarrow \alpha < \beta$, and therefore splits if $\alpha > \beta$ and is indifferent when $\alpha = \beta$. This means we get the following potential PBE's in this case: \begin{align*} (Steal, Steal; \alpha \leq \beta) \text{ and } (Split, Steal; \alpha \geq \beta). \end{align*}

Let us next look at the case where player 2 plays $Split$ if he is type $N$. Then player 1's expected payoff from stealing is $100\beta + 100(1-\alpha-\beta)=100(1-\alpha)$ and his expected payoff from splitting is $50$. In this case player 1 steals when $100(1-\alpha) > 50 \Leftrightarrow \alpha < 1/2$, and therefore splits if $\alpha > 1/2$ and is indifferent if $\alpha = 1/2$. This means that we get the following potential PBE's in this case: \begin{align*} (Steal, Split; \alpha \leq 1/2) \text{ and } (Split, Split; \alpha \geq 1/2). \end{align*}

As such, we have can write up the set of pure-strategy PBE: \begin{alignat*}{1} \text{PBE} = \{&(Steal, Steal; \alpha \leq \beta),(Split, Steal; \alpha \geq \beta), \\ &(Steal, Split; \alpha \leq 1/2),(Split, Split; \alpha \geq 1/2)\}. \end{alignat*}

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    $\begingroup$ What exactly is your confusion? $\endgroup$ – Herr K. Nov 19 '18 at 17:39
  • $\begingroup$ I have added my working outs to the post. I guess my confusion is regarding if I can just say that it is rational for player 2 to play both Steal and Split and then leave it at that. $\endgroup$ – Greased Dog Nov 20 '18 at 18:17
  • $\begingroup$ Please edit your tex code into the question rather than the image. $\endgroup$ – Giskard Nov 20 '18 at 19:08
  • $\begingroup$ Okay, I have added tex code now. $\endgroup$ – Greased Dog Nov 20 '18 at 19:58
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Hint:

You can convert the game into the following normal form:

\begin{equation} \begin{array}{|c|c|c|} \hline & Steal & Split \\\hline Steal & 100\beta, 0& 100(1-\alpha),0 \\\hline Split & 50(\alpha+\beta),50+50(1-\alpha-\beta) & 50,50 \\\hline \end{array} \end{equation}

Note that Steal is a weakly dominant strategy for player 2, so (Split, Split) can never be an equilibrium.

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    $\begingroup$ Thanks for the answer, it was a big help realizing I could just write the game up in a bimatrix! I have one question however: Consider $\alpha = \beta = 0.5$. Wouldn't all four possibilities be equilibria in that case, including (Split, Split)? $\endgroup$ – Greased Dog Jan 1 at 22:35
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    $\begingroup$ @GreasedDog: You're right. Note that $\alpha=\beta=0.5$ implies state $N$ never occurs, so player 2's choice doesn't matter. The probability also implies that player 1 is indifferent between split and steal. $\endgroup$ – Herr K. Jan 2 at 2:14

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