Gambler's ruin - problem 1

Probability Level 3

You walk into Martin Gale's Betting Room with $2.

You can play the following game any number of times (if you have what it costs)

You pay Martin a dollar.
Martin tosses a fair coin
If the coin comes up heads Martin pays you two dollars. If it comes up tails, you get nothing.

You decide that you will play for a while, but if you ever get up to $5 you will cash out and go home. And, naturally, if you are left with no money then also you will go home.

What is the probability that you will go home "ruined", i.e with nothing?

The answer is 0.6.

1 solution

Varsha Dani
Feb 7, 2019

For $0 \le i \le 5$ , let $r_{i}$ be the probability that you are ruined, meaning that you go home with $0, from an initial budget of $i$ . We want to compute $r_{2}$ . If you play the game once, then with probability 1/2 you increase your budget by 1 and with probability 1/2 you decrease it by one. And then you get to play again, as if you were starting with your new budget. Thus

$r_0 =1$ -- If you have no money you are already ruined.
$r_{i} = \frac{1}{2} (r_{i-1} + r_{i+1})$ for $1\le i\le4$
$r_5 =0$ -- Because of your decision to take the money and run if you ever reach 5.

This gives a system of linear equations in $r_1 \dots r_4$ . Solving it gives $r_{2} = 3/5$ or $\boxed{0.6}$

Gambler's ruin - problem 1

The answer is 0.6.

1 solution

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