# Game Theory: finding nash equilibria of an extensive form game (game tree) [duplicate] How do you find the pure Nash equilibria from this commitment game?

The SPNE is (U, ps)

Do you find the NE by finding the best response (BR) of a player to a specific strategy of the other player?

e.g.

BR(p)={D} BR(q)={U} BR(r)={D} BR(r)={U}

BR(U)={p} BR(D)={s}

For the SPNE, you want to proceed by backward induction as explained by Lee Sin. For all other NE you want to construct the normal form representation (the usual table for simultaneous games) and solve for NE as if players were choosing their actions simultaneously. Note that the actions of player 1 are $$\{U,D\}$$, while for player 2 are $$\{pp, pq, pr, ps, qp, qq, qr, qs, rp, rq, rr, rs, sp, sq, sr, ss\}$$. This will indeed be a very large table.

Another way to do it is simply to look for fixed points directly. For example, assume that Player 2 plays $$pp$$, then the best response for player 1 is $$D$$ but if player one chooses $$D$$, the best response for player 2 is either of the following actions $$\{ps, qs, rs, ss\}$$ since $$pp$$ is not in this set, it is not a fixed point. iterate that process for all the actions of player 2. Try to convince yourself that there is no other NE, except for the SPNE. And think about how would you change the payoffs so that the SPNE is the same, but now there are other NE's. (Hint: would changing the payoff of the terminal node after U,p from (2,1) to (1,1) work?)

• Thank you. In player I plays D, why wouldn’t the set {ps, qs, rs, ss} be the set of best responses. (Ie at the right node, player II plays s for sure as s (6) has a higher payoff than p, q and r (1, 4, 5) – Maths May 1 '19 at 7:54
• Sorry, you are completely right, let me edit my answer. $pp$ is still not in that set. Thanks for pointing it out. – Regio May 1 '19 at 13:59

Let me try and simplify this for you.

You're trying to calculate every possible outcome, but as you rightly assert we need to be looking at the best response of each player. At a given node (a place where a player makes a decision) they're trying to make the decision that gives them the best possible outcome.

Let's look at the left node of player 2. Here he can choose P,Q,R or S. Well his payoff is 4,3,2,0 respectively. If we ever reach this node player 2 will choose P (4). We can effectively "delete" the other options since they will never get selected.

Likewise for the right node of player 2. Here he can choose P,Q,R or S. Well his payoff is 1,4,5,6 respectively. If we ever reach this node player 2 will choose S (6). We can effectively "delete" the other options since they will never get selected.

Player one knows all of this. He knows that if he chooses U, player 2 will choose P, giving him a payout of 1. If he chooses D however, he will get a payout of 0 (when player 2 chooses S).

So he has to choose between U(1) or D(0). Clearly he chooses U.

This means that as long as the players are rational (an assumption we make in games like these) that we will end up at P1: U P2: P. With strategy sets of P1(U), P2(P,S).

Hope that helps!

• Thank you very much. Please could you assist in finding the other (non-SPNE) NE of this game (directly from the extensive form). Thank you – Maths May 1 '19 at 0:25