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In a game theory textbook there is something similar to the table below where there is one pure strategy nash equilibrium and multiple mixed strategy nash equilibria. It is a simultaneous game with the payoffs presented below.

If we assume that this game is played twice, How do I identify all subgame perfect equilibria for this game, as well as nash equilibrium that is not a subgame perfect equilibrium?

For a game with multiple pure strategy nash equilibrium I think I can find a solution by using backward induction, but for a game like this with just one pure strategy nash equilibrium and multiple mixed strategy nash equilibria, I have no idea how to identify the subgame perfect equilibria and possibly a nash equilibrium that is not subgame perfect equilibrium, especially when there are mixed strategy equilibria included.

Any help in this would be appreciated.

\begin{array}{|c|c|c|c|} \hline & A & B & C \\\hline A & (1,1) & (0,0) & (0,0)\\\hline B & (0,0) & (2,1) & (1,2)\\\hline C & (0,0) & (1,2) & (2,1)\\\hline \end{array}

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  1. Check for the Nash equilibria (pure or mixed) of the one-shot game.
  2. Repetition of the strategy profile of the Nash equilibria of the one-shot version yields one set of subgame perfect equilibria: For instance, play $(A,A)$ in the first stage and for any action profile played at the first stage, play $(A,A)$ in the second stage. The same holds true for the mixed (complete or otherwise) as well. For instance, the totally mixed SPNE is :Play $\left(\frac{3}{5},\frac{1}{5},\frac{1}{5}\right)$ in the first stage and for any action profile (assuming the randomisation is observable), play $\left(\frac{3}{5},\frac{1}{5},\frac{1}{5}\right)$ in the second stage.
  3. Combinations of the Nash equilibria of the one-shot game yield another set of SPNE: For instance, Play $(A,A)$ in the first stage and play $\left(0,\frac{1}{2},\frac{1}{2}\right)$ for any action profile in the second stage and so on.
  4. Use credible threats to play a non-Nash action profile in the first period: Play $(B,B)$ in the first stage. At the second stage, if $(B,B)$ was played, play $\left(0,\frac{1}{2},\frac{1}{2}\right)$. For any other action profile at the first stage, play the totally mixed Nash equilibrium of the one-shot game. Why does this work? There is no incentive for player 1 to deviate anyway because they're playing their best reply in the first stage. For player 2, if they were to deviate to $C$. they'd receive $2$ in the first stage and $1/3$ in the second stage. Assuming no discounting, they get $7/3$ while if they obeyed the strategy, they would get $(1+3/2)$ which is strictly greater. Similarly $(C,C)$, $(B,C)$ and $(C,B)$ can be played as first stage action profiles in a sub-game perfect equilibrium.
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  • $\begingroup$ Could you explain where the (0, 1/2 , 1/2) came from? And is there a nash equilibrium that's not spe? $\endgroup$ – Robin311 Oct 29 '20 at 20:13
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    $\begingroup$ That is a Nash equilibrium too: Suppose the probability of player 1 playing strategy $A$ is 0 and that of playing $B$ and $C$ equal, then player 2 gets 0 by playing $A$, 3/2 by playing $B$ and 3/2 by playing $C$. Hence, player 2 is indifferent between $B$ and $C$ and strictly prefers them to $A$. Hence any convex combination of $B$ and $C$ is a best reply. The same holds true for player 1. $\endgroup$ – soslow Oct 29 '20 at 20:15
  • $\begingroup$ aha, I think I understand it now. Then can this be a subgame perfect equilibrium too? Play (A,A) on the first stage and then play (3/5, 1/5, 1/5) for any action in the second stage. You didn't mention this so I was wondering if this could be a subgame perfect equilibrium too. $\endgroup$ – Robin311 Oct 30 '20 at 3:19
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Adding to @soslow's answer: once you have an SPE, it should be easy to construct a non-subgame-perfect NE by modifying the off-equilibrium actions in such a way that 1) the players have no incentive to deviate to those actions and 2) the action profile is not a NE in any subgame.

For example, one SPE of the game is

play $(A,A)$ in stage 1, and play $(A,A)$ in stage 2 regardless of the outcome in stage 1.

We can modify this to

play $(A,A)$ in stage 1, and play $(A,A)$ in stage 2 if the outcome in stage 1 is $(A,A)$, otherwise play $(A,C)$.

This modified strategy profile is a NE, since players are still best responding to each other by playing $(A,A)$ in both stages (the off-equilibrium path outcome $(A,C)$ is Pareto dominated by $(A,A)$). However, since $(A,C)$ is not a NE in any subgame, the modified strategy profile is not subgame perfect.

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