**Sribes:
Akash M Kushal
Tarun Agarwal
**

**18 Sept. 2002
Lecture 13**

Cats seem to go on the principle that it never does any harm to ask for what you want.

**Moral of the story : **
Knowledge of Game Theory and Nash Bargaining solutions can help you get a good share
of bread.

At each iteration one player proposes a payment value. If it is not acceptable to the
other player she rejects it and proposes a new value in the next iteration.
With each iteration, the contenders lose precious time and thus their gain from this
bargain is discounted by a factor.

Also, the bargain breaks down after N iterations and both get no utility, i.e, for .

Let us formally define the game. The first player proposes a value for X. Now, player 2 has two choices (either accept the offer or decline). If she accepts then the game ends there but, if she declines then she proposes another value for X. Now, player 1 either accepts or declines and the same thing repeats till the other player accepts. The strategy of a player is the values of X that she is ready to accept, and the values to offer in the next round in case she rejects.

In the case when it can be easily seen that the Nash Equilibrium is .
That is, player 1 proposes . Since only one iteration is allowed player 2 has
no choice because this is the last iteration. The utilities are , respectively.
Note that by changing their strategies , none of the two players can increase her utility.
Hence is a Nash Equilibrium.

If , i.e., two iterations are allowed, the game tree is shown below. The subgame perfect equilibrium offer amounts are also shown.

If player 2 is offered a share of less than she can easily reject it and herself propose a share of 1 for herself (which will give her a utility of . This will have to be accepted by Player 1 because this would be the last iteration. In this case the utilities are , respectively. On the other hand they can get utilites of , respectively if player 1 proposes to player 2 who accepts it (she has no reason to reject it since she cannot do better by rejecting it). This strategy pair actually forms a subgame perfect Nash Equilibrium because none can deviate from it to increase her utility.

Now, let us consider the case for a general N. We will see that just like the case for N=2 the players are better off by finishing the game in the first step itself rather than carrying on bargaining.

Since, the last player, say player
can always get a utility of
by rejecting the previous offer and in the th
turn offer the other player 0. Since, this would be the last iteration so the other
player cannot do better by not accepting. So, the second last player (call her player )
must give the last player atleast in the th turn. This would offer
player a utility of
which she would have no reason to reject.
Going 1 step further since player can ensure a utility of atleast
so player must offer her atleast this much. That is
player must offer
.

Let us denote by the amount that player can ensure for herself in the first turn itself if the game runs for turns. That is she can ensure a utility of atleast .

Let us try to find a recurrence relation for the function . Let us consider the case where . The amount that player can ensure is . Suppose that the bargain goes on for steps. Now only more steps are allowed. The situation now is similar to a new game with . Thus player can ensure a amount of . Thus for the original game, after steps player can ensure a utility of atleast . Thus player must offer player atleast in the second turn. This would give player a utility of . Hence in the first step, player must offer player a utility of atleast if she wants her to accept. To do this she must offer her . With this she will get in the first turn.

So,

If this gives

If then player 1 will get

Let
,
.

Now if and are very small then ;

So, we can see that the patient player (that is one who has a smaller ) is the one who gets the bigger piece of the total.

Tarun Agarwal 2002-11-19