I have a problem with the sequential game with random event at the event tree. The model of the game as follows:
- Player = $\{A,B\}$
- Pure strategy of player $A: A1, A2, A3$
- For each strategy of player $A$, we have some chances, which is presented by the probability as in game tree.
- Pure strategy of player $B$: for type $1$, strategies of $B$ is $B_{11}$ and $B_{12}$. For type $1$, strategies of $B$ is $B_{21}$ and $B_{22}$. For type $3$, strategies of $B$ is $B_{41}$ and $B_{42}$. And for type $4$, strategies of $B$ is $B_{41}, B_{42}$ and $B_{43}$.
The game tree is as follow.
https://drive.google.com/file/d/0BzdN8FPtLFScZU0wY2ROZzgzd3c/view?usp=sharing]
Sorry that I am the newbie, so I do not have enough reputations to post a figure.
The question is that:
- Find the subgame perfect equilibirum of the game above?
- Find a pure and a mixed Nash equilibrium points.
Thank you for your help!
As Théophile pointed out, the strategy names overlap the player names, which invites confusion, so I'll call player $A$'s strategies $A_1$, $A_2$ and $A_3$ instead.
Assuming $B$'s decisions to be maximally favourable for $A$ in case $A$ plays $A_2$ yields a maximum expected payoff of
$$ \frac15\cdot22+\frac17\cdot10+\frac14\cdot17+\frac{57}{140}\cdot15=\frac{1133}{70}\approx16.2 $$
for $A$, whereas assuming $B$'s decisions to be maximally adverse to $A$ in case $A$ plays $A_3$ yields a minimum expected payoff of
$$ \frac18\cdot13+\frac14\cdot13+\frac1{16}\cdot9+\frac9{16}\cdot20=\frac{267}{16}\approx16.7 $$
for $A$. The payoff of $A_1$ for $A$ cannot exceed $16$. Thus both $A_1$ and $A_2$ are dominated by $A_3$. It follows that $A$ plays $A_3$, and $B$ plays the corresponding optimal strategies, $B_{11}$, $B_{22}$, $B_{32}$ and $B_{42}$.
The corresponding payoffs are
$$ \frac18\cdot19+\frac14\cdot13+\frac1{16}\cdot9+\frac9{16}\cdot23=19.125 $$
for $A$ and
$$ \frac18\cdot18+\frac14\cdot22+\frac1{16}\cdot18+\frac9{16}\cdot23=21.8125 $$
for $B$.