My question follows from this question: https://math.stackexchange.com/questions/2132846/game-theory-subgame-perfect-nash-equilibrium-in-a-sequential-game-with-identica from Maths stackexchange.

Based on the explanation given in the answer, I tried to find the subgame perfect Nash equilibrium (SPNE) of another game:

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Here, there are 2 players, and both have to choose between A and B. Player 1 moves first, followed by player 2.

Let us define the strategies of the players as $(p_1, p_2)$ where $p_1$ and $p_2$ are the probabilities that player 1 and player 2 choose A respectively.

In the subgame to the lower left, the analysis is simple: player 2 will choose A because it gives a higher payoff.

In the lower right, player 2 faces identical payoffs, so any $p_2 \in [0, 1]$ is optimal. We divide the possibilities into 3 cases:

1. $p_2 > 0.5$: Here the expected payoff of choosing B for player 1 is $2p_2 + (1-p_2)4 < 3$, the payoff of choosing A. Therefore player 1 will choose A. This gives us the SPNE $(p_1, p_2) = (1, 1)$.

2. $p_2 < 0.5$: Here the expected payoff of choosing B for player 1 is $2p_2 + (1-p_2)4 > 3$, the payoff of choosing A. Therefore player 1 will choose B. This gives us a family of equilibria: $(0, p_2), p_2 \in [0, 0.5)$.

3. $p_2 = 0.5$: In this case, the expected payoff of choosing B for player 1 is $2p_2 + (1-p_2)4 = 3$, the payoff of choosing A. Thus, any $p_1 \in [0, 1]$ is optimal. This gives us another family of equilibria: $(p_1, 0.5), p_1 \in [0, 1]$.

This brings me to my first question:

Question 1: Is my characterisation of SPNEs correct? That is, are all the equilibria mentioned above actually SPNEs? And also, are there other SPNEs that I have missed?

If the answer to "are they all actually SPNEs" is yes, then I think there is a problem with the family of equilibria in point 2. Conditional on player 1 choosing B, of course all $p_2 \in [0, 1]$ are optimal including all $p_2 \in [0, 0.5)$. But player 2 knows that choosing any such $p_2 < 0.5$ would mean that player 1 chooses B, and player 2 ends up with payoff 1. But if she instead chooses $p_2 > 0.5$ she can be sure that player 1 chooses A and then player 2 ends up with payoff 4, which is better than 1. Then why would player 2 ever choose $p_2 < 0.5$?

This leads me to my second question:

Question 2: If $(0, p_2), p_2 \in [0, 0.5)$ are not SPNEs, what went wrong in my analysis? Does backward induction fail in such cases or did I make a mistake? On the other hand, if these are actually SPNEs, then is there any refinement of SPNE, some other solution concept that is immune to my objection? Where can I read about them?


1 Answer 1


Your analysis is correct in principle, but your notation is not. A (behavioral) strategy for a player in a perfect information game has to specify a probability distribution over actions for each decision node of that player. Since player 2 has two such nodes, a Left and a Right one, with two possible actions at each node, her strategies can be written as pairs $(p_2^L,p_2^R)$ of probabilities of choosing $A$ at the left and the right node, respectively. Strategy profiles are then denoted by $(p_1,(p_2^L,p_2^R))$. The three sets of SPNEs are then given by

  1. $(1,(1,p_2^R))$ with $p_2^R>\frac12$,

  2. $(0,(1,p_2^R))$ with $p_2^R<\frac12$, and

  3. $(p_1,(1,\frac12))$ with $0\le p_1\le 1$.

Your question "why would player 2 ever choose $p_2^R<\frac12$?" has already been answered by yourself: Because conditional on player 1 choosing B it is optimal. Note that your claim that "if she instead chooses $p_2^R>\frac12$ she can be sure that player 1 chooses $A$" is not true. This would only hold if player 1 could observe player 2's choice before deciding.

  • $\begingroup$ Thanks! So if I understand it correctly, player 2 cant commit to prob > 0.5 because once player 1 chooses B, there is no point in actually following through, all prob values are optimal. Maybe what I was thinking would work if we keep repeating the game though, to punish player 1? Somewhat like tit-for-tat? $\endgroup$ Apr 16, 2022 at 13:29
  • 1
    $\begingroup$ @IshanKashyapHazarika Well, player 2 could try to stubbornly play (1,1) until player 1 "learns" to play 1. But the repeated game is a very different thing and I haven't yet thought through it... $\endgroup$
    – VARulle
    Apr 16, 2022 at 14:10

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