Game theory -  A Tutorial

Jean Walrand

Cooperative: Rubinstein-Stahl

Assume Alice and Bob negotiate to split a unit pie. Each negotiation step is expensive and reduces the value of the pie by the factor a for Alice and by the factor b for Bob.  How should they split?

Consider that Alice starts by demanding a fraction x of the pie. If Bob refuses, he then demands a fraction y of the pie, and so on.

Bob refuses Alice’s demand if he thinks that he can get a fraction y such that yb ≥ 1 - x. Indeed, a fraction y is worth yb after one step for Bob.  Similarly, Alice refuses Bob’s demand if she thinks she can get a fraction x such that xa > 1 - y. Consequently, the maximum acceptable fractions x and y are such that yb = 1 - x and xa = 1 - y.  That is, xab = b - yb = b - (1 - x) = b - 1 + x, so that

1 - b = x(1 - ab). Hence, x = (1 - b)/(1 - ab).  Consequently, 1 - x = b(1 - a)/(1 - ab).

 

One can make this argument rigorous and show that

 

Theorem (Rubinstein-Stahl) [6]

Alice demands the fraction  x = (1 - b)/(1 - ab) of the pie and Bob accepts a share at least equal to 1 - x.