Discounting (Time Preference) & Rubinstein Game

Published by Mario Oettler on

In this topic, we scrutinize how time preferences and discounting influence the outcome of bargaining situations. Therefor, we look at a simple game where two players negotiate about sharing a cake.

In particular, they negotiate about the shares each player should receive. Offers are submitted in rounds. The peculiarity is, that in each round the size (value) of the cake shrinks with the rate d:

d = 1/(1+r)

r: interest rate

The offer from A is a, and the offer from B is b.

B accepts the offer of A, if the discounted value is equal or higher than his demand:

(1) 1-a = dB*b

A agrees if he receives at least the discounted value of his offer.

(2) 1-b = dA*a

Now, we want to find a and b. Therefore, we rearrange equation (1)

(1’) a = 1-dB*b

We insert (1’) into (2) and rearrange it for b:

(3) b = (1-dA)/(a-dA*dB)

To get a, we insert (3) into (1’) and receive:

a = (1-dB)/(1-dA*dB)

But there is a problem.

Both offers together are higher than the size of the cake.

a + b > 1

(1-dB)/(1-dA*dB) + (1-dA)/(a-dA*dB) > 1

(2-dA -dB)/(1-dA*dB) > 1

This inequation is true for 0<=d<1

(for d = 1, we divide by zero)

In this situation, the first mover has an advantage. A demands

a = (1-dA)/(a-dA*dB)

This leaves B with:

1-a = (1-dA*dB-1+dB/(1-dA*dB) = dB(1-dA)/(1-dA*dB)

Should B accept the offer from A?

Yes, he should because in the next round, the cake shrinks. So, he can’t get more than in this round. The discounted value of his current demand is:

dB*b = dB*(1-dA)/(1-dA*dB) = 1- a

Discount factors represent different levels of (im)patience. They can also be interpreted as different negotiation strengths.