Gambler's ruin - problem 2

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 expected number of games you will play before you stop?

The answer is 6.

1 solution

Varsha Dani
Feb 7, 2019

For $0 \le i \le 5$ , let $g_{i}$ be the expected number of games till you stop, from an initial budget of $i$ . We want to compute $g_{2}$ . 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

$g_0 = g_{5} =1$ Because those are the stopping criteria .
$g_{i} = 1 + \frac{1}{2} (g_{i-1} + g_{i+1})$ for $1\le i\le4$ The +1 comes from the game you play to change your budget in one direction or the other.

This gives a system of linear equations in $g_1 \dots g_4$ . Solving it gives $g_{2} = \boxed{6}$ .

Gambler's ruin - problem 2

The answer is 6.

1 solution

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