The game is as follows- two people use the following procedure to split Rs. 100. Player 1 offers Player 2 an amount of money upto Rs. 100. If P2 accepts this offer then P1 receives the remainder of the Rs.100 and P2 receives the offered amount. If P2 rejects it, then neither person receives any payoff. In theory, each person cares only about the amount of money she receives and each player prefers to receive as much as possible. It is assumed that the offer can only be made in an integral number of rupees.
The game is formally defined thus:
Players: The Two People
Terminal Histories: A set of sequences (x,Z), where x is a number with 0 ≤ x ≤ 100 and Z is either Y (yes, I accept) or N (no, I reject).
Player Function: P(φ) = 1 and P(x) = 2 for all x.
Preferences: Each person’s preferences are represented by payoffs equal to the amounts of money she receives. For the terminal history (x,Y) P1 receives (100-x) and P2 receives x. Fr the terminal history (x,N) each person receives 0.
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Here x is an integer and the triangle has a finite number of values of x.
Thus the game has finitely many actions for both the payers; and it has a finite horizon so we can use backward induction to find its subgame perfect equiibria.
Let us first consider the subgame of length 1, in which P2 either accepts or rejects an offer of P1. For every possible offer of P1 there is such a subgame. In a subgame that follows an offer of x › 0 (x greater than 0) by P1, the optimal strategy of P2 is to accept the offer. In a subgame that follows the offer x = 0, P2 is indifferent between accepting and rejecting.
For both possible subgame perfect strategies of P2, there is an optimal strategy of P1. If P2 accepts all offers, then P1’s best strategy is to offer 0. If P2 accepts all offers except 0, then P1’s best strategy is to offer Re.1.
Thus the game has 2 subgame perfect equilibria - one in which P1 offers 0 and P2 accepts all offers;