# Game Theory: Subgame perfect nash equilibrium in a sequential game with identical payoffs

I'm having a hard time reasoning out this sample game. The game has two players, they make their decisions sequentially, and the entire game happens only once.

Here is an outline with the payoffs where Player 1 chooses A or B, and after that Player 2 chooses C or D:

If Player 1 chooses B, the outcome of the game is simple. Player 2 chooses the outcome that gives him more, and would thus choose option D, resulting in a payoff of 2,2 to each of them.

However, if Player 1 chooses A, Player 2 now has two choices where both of them yield Player 2 3 points, but Player 1 gets either 4 points or just 1. What would Player 2 choose in this scenario? Either way he gets 2 points, but does he prefer one option over another, or is it entirely a random choice?

• Minor correction: if 1 chooses A, "either way he [Player 2] gets 3 points''.
– mlc
Feb 9, 2017 at 16:28
• The answer to your question needs more background: are you thinking of subgame perfection/backward induction, or of more sophisticated issues (i.e. refinements)?
– mlc
Feb 9, 2017 at 16:30

## 1 Answer

Make P1 a male and P2 a female, for simplicity. P2 does not prefer one option over another (both give the same payoffs). However, being indifferent, she can pick anything and even make a random choice.

Let $p$ the probability that P2 play C if the game reaches the bottom-left node. P2's best reply is any $p \in [0,1]$. So this game has plenty of subgame-perfect equilibria. They can be conveniently arranged in three classes. The common element is that P2 plays D at the bottom-right node, so P1 knows that in a subgame-perfect equilibrium he gets $2$ if he plays B. Let us consider the rest.

1) P2 plays C with probability $p<(1/3)$ in the bottom-left corner. (She is indifferent, so any $p$ is optimal.) Then 1 anticipates that playing A leads to an expected payoff $4p+1\cdot (1-p) < 2$ so he prefers playing B.

2) P2 plays C with probability $p>(1/3)$ in the bottom-left corner. Now 1 anticipates that playing A leads to an expected payoff $4p+1\cdot (1-p) > 2$ so he prefers playing A.

3) P2 plays C with probability $p=(1/3)$ in the bottom-left corner. Then 1 anticipates that playing A leads to an expected payoff $4p+1\cdot (1-p) = 2$ so he is indifferent and may randomise over his choice between A and B.

A note of caution: introductory examples dealing with backwards induction leads to subgame-perfect equilibria in pure strategies. But, when there are ties in payoffs, there may be equilibria in mixed strategies. This is one of those cases.