Subgame perfect Nash equilibrium when there is a tie in payoffs seems problematic

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:

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

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.

• 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? Apr 16, 2022 at 13:29
• @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... Apr 16, 2022 at 14:10