# Probability and Stochastic processes – Gambler’s ruin problem

Let’s say there are 2 tennis players, A and B. Now, A is a better player than B and hence wins 2/3 of the matches that they play. However, they don’t just play tennis – they also gamble on the results. Initially, let’s say that A has $1 and B has$2. The rule is that whoever loses a tennis match must give $1 to the other. They keep on playing tennis matches until one of them becomes bankrupt, at which point they stop. What is the probability that B will become bankrupt? —— We note that whatever handing over of the money takes place between A and B, the total amount of money remains constant at$3.

There are 4 possible states —

State 0 — A has $0 State 1 — A has$1

State 2 — A has $2 State 3 — A has$3.

(Note that the amount of money that B has, in each of the above states, is completely determined by simply stating the amount of money with A since the total amount of money remains constant.)

We get the following transition diagram with probabilities depicted besides the arrows.

The probability that we wish to compute is the probability of reaching State 3 starting initially at State 1.

We can compute that quite easily. Let $p_{i}$ denote the probability of reaching State 3 from State $i$.

We directly get the following from the above transition probabilities —

$p_2 = \frac{2}{3} + \frac{1}{3} p_1$

$p_1 = \frac{1}{3} p_0 + \frac{2}{3} p_2$.

$p_0 = 0$.

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Therefore, our desired probability, $p_1 = 4/7$, which means that there is 4/7 probability that the series of tennis matches would end with B becoming bankrupt.

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