Recherches sur l'intégration des équations différentielles aux différences finies, et sur leur usage dans la théorie des hasards.

P.S. Laplace


Three players A, B, C, of whom the respective abilities are represented by the letters p, q, r play together in a manner that, on a number x of trials, there lacks m to A, n to B and x-m-n to C; we propose to determine the respective probability of these three players for winning.

Let C(m,n,x) denote the probability of C winning. Taking p+q+r = 1 , we must have

C(m,n,x) = p*C(m-1,n,x-1)+q*C(m,n-1,x-1)+r*C(m,n,x-...

subject to the conditions C(0,n,x) = 0 , C(m,0,x) = 0 and C(m,n,m+n) = 1 .

Laplace's explicit solution to this problem is quite cumbersome and it would be extremely difficult to extend his methods to 4 or more players. One "direct" technique to the solution is by a Markov chain. However, there will be (m+1)*(n+1)*(x-n-m+1) different states in this system. On the other hand, computers allow us to make use of the recursive formulation of the problem.

We will let A(m,n,x) and B(m,n,x) be the probability of A and B winning respectively. The recurrence relations for A and B are the same as for C . Only the boundary conditions change.

> C:=proc(m,n,x)

> if (m=0 or n=0) then 0 elif x=m+n then 1 else (C(m-1,n,x-1)+C(m,n-1,x-1)+C(m,n,x-1))/3 fi;

> end:

> A:=proc(m,n,x)

> if (n=0 or x=m+n) then 0 elif m=0 then 1 else (A(m-1,n,x-1)+A(m,n-1,x-1)+A(m,n,x-1))/3 fi;

> end:

> B:=proc(m,n,x)

> if (m=0 or x=m+n) then 0 elif n=0 then 1 else (B(m-1,n,x-1)+B(m,n-1,x-1)+B(m,n,x-1))/3 fi;

> end:

Laplace illustrates his solution with the case where m = 2, n = 3 and x = 9. Here we have

> A(2,3,9);


> B(2,3,9);


> C(2,3,9);



This agrees with his solution. The procedures generalize readily to 4 or more players.