Previous:  Craps

In this problem we are concerned with the fate of two gamblers who are playing a game of chance.  This game of chance could be a game of cards, or something as simple as tossing a coin.  The basic premise on which we will work is that the game can be played repeatedly, and that after each play of the game, the gambler that wins receives £1 (or $1 if you prefer - since I'm English, I prefer to work in pounds) from the other gambler.

Let us refer to the gamblers as A and B.  Let us further assume that gambler A has a probability p of winning any particular round of the game, and let us assume that gambler B has a probability q of winning.  If the game can only be won or lost (i.e. there are no draws), then these probabilities are related as follows:

We will assume that gambler A begins with an amount m pounds, and that the total fortune of the two gamblers is n pounds.  That is, gambler B has an initial amount n – m pounds.  (We choose this way of specifying the amounts because it makes subsequent algebra slightly less cumbersome).  We assume that the gamblers play the game repeatedly, exchanging money depending on the result of each play of the game, until one of the gamblers has no money left.

The problem we want to solve is this.  What is the probability that gambler A wins the total fortune of n pounds before he loses his initial fortune of m pounds and is left with nothing.  This of course is how many such games proceed – the game continues until one of the players loses all his money.

This problem has a very elegant solution.  Let us begin by defining a_m as the probability that gambler A reaches a total fortune of n pounds before losing all his money, given that he starts with a fortune of m pounds.  Let us further define the following events:

Let A_m be the probability that gambler A wins £1 on the first play of the game;

Let B_m be the probability that gambler B wins £1 on the first play of the game;

Let W be the event that gambler A does indeed win the total fortune of n pounds.

Note that these events are defined in terms of the first play of the game.  Now, since the events A_m and B_m span the sample space of outcomes on the first play of the game, we can use one of the results of the article on Bayes’ theorem to write

From the information that we have already been given:

Now for the clever bit.  The conditional probability in the first term on the right hand side of the above equation is the probability that gambler A wins overall, given that he wins the first play of the game.  Similarly, the conditional probability in the second term on the right hand side of the above equation is the probability that gambler A wins overall, given that gambler B wins the first play of the game.  But these two conditional probabilities are just

The expression above for the probability of the event W can therefore be written as

This is a difference equation that provides a relationship between the probabilities of gambler A winning when starting with different initial fortunes.  In the case that m = 0, gambler A has lost.  In the case that m = n, gambler A has won.  So, the difference equation above has the following “boundary conditions”:

To solve the difference equation for the various probabilities a, we can rewrite it slightly by noting that p + q =1.  Using this we can write

This difference equation can then be written out explicitly as follows:

                                                        (A)

The first of these equations can be rewritten as

Using this equation, the second of the equations A can then be written as

Continuing in this way and adding the resulting equations, we obtain

The sum on the right-hand side of this expression is a geometric series that is easily evaluated.  We therefore end up with

                                                      (B)

This is the solution that we seek.

Let us first of all consider a “fair” game, such that p = q = 1/2.  A trivial game such as tossing a coin would satisfy this criterion.  Alternatively, both gamblers could execute the game with equal skill.  In this case, it can be shown that the probability a_m is given by

That is, the probability of gambler A winning the entire fortune is equal to the ratio of his initial starting amount to the total fortune.  Thus if both gamblers start with the same amount of money, the probability of gambler A winning in the end is 1/2.  If gambler A starts with much more money than gambler B, then the odds of an overall win become stacked in gambler A’s favour.

Now suppose that we have an unfair game, such that gambler A has a probability p = 0.6 of winning any particular round of the game.  This could arise because even though the game is a game of chance, some skill is required to conduct the game and gambler A happens to be better at it than gambler B.  (Alternatively, the game could be “fixed” in gambler A’s favour …).

Let us suppose that gambler B has £100 to start with.  His probability q of winning is 0.4.  The probabilities of gambler A winning for various initial starting amounts is obtained by using equation (B) and are set out below:

            m = £1              probability of a win for gambler A is 0.33

            m = £2              probability of a win for gambler A is 0.56

            m = £3              probability of a win for gambler A is 0.70

            m = £5              probability of a win for gambler A is 0.87

            m = £10            probability of a win for gambler A is 0.98

            m = £20            probability of a win for gambler A is 0.9996.

An amazing result.  Even when the odds only slightly favour gambler A, a starting amount of £10, compared with gambler B’s starting amount of £100, is sufficient to ensure a 98% probability of an overall win in which Gambler A takes £100 from gambler B.

If the odds are stacked even more in favour of gambler A, such that p = 0.8, then a starting amount of £4 for gambler A is enough to ensure a probability of 0.996 of winning.

THE END