recherch_xix.mws

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

P.S. Laplace

PROBLEM XIX.

I suppose two players A and B, with an equal number m of écus, playing to this condition, that the one who loses will give an écu to the other; let the probability of A winning a trial be p ; let that of B be q ; but let it be able to happen that any of them not win, and let the probability of this be r . This put, we ask the probability that the game will end before or at the number x of trials.

Consider the probability that player A has a stake k at time x . We may compute this through a Markov chain with two absorbing states, entered when the player either loses his stake or gains the entire stake of the other.

Example. Suppose . The transition matrix P is clearly the one having r on the main diagonal excepting the corners, q on the sub-diagonal and p on the super-diagonal. The initial state a is the vector having 1 in the st position.

> restart:

> with(linalg):

Warning, the protected names norm and trace have been redefined and unprotected

> P:=matrix(7,7,[[1,0,0,0,0,0,0],[q,r,p,0,0,0,0],[0,q,r,p,0,0,0],[0,0,q,r,p,0,0],[0,0,0,q,r,p,0],[0,0,0,0,q,r,p],[0,0,0,0,0,0,1]]);

> a:=matrix(1,7,[0,0,0,1,0,0,0]);

The probability that player A holds a stake k at time x is given by the value in the k th position of the expression .

> f:=x->evalm(a&*P&^x);

Thus, for example, we can examine the game after 4 matches.

> f(4);

To determine the probability that the game end at or before time x , it suffices to sum the first and last components. To this end we define the column vector b with 1 in its first and last position and zeros elsewhere.

> b:=matrix(7,1,[1,0,0,0,0,0,1]):

Now, is this desired probability.

The game ends at or before time 4 with probability

> evalm(f(4)&*b);

>

Suppose now that p , q and r each equal 1/3.

> p:=1/3:q:=1/3:r:=1/3:

> f:=x->evalm(a&*P&^x);

> evalm(f(4)&*b);

>