Gambler's ruin - problem 5

Probability Level 2

You walk into Martin Gale's Betting Room with an initial budget of $k$

As usual, 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 until you have increased your money to $n$ , and then you will stop. Here, $n>k$ . Of course you will also have to stop if you lose all your money (i.e. you are ruined).

How many games do you expect to play before you stop?

n-2k

n

\frac{n+2}{k}

n^2-k^2

k(n-k)

1 solution

Mark Hennings
Feb 8, 2019

If $E_k$ is the expected number of games to be played given that you start with $k$ , then $E_0 = E_n = 0$ and (conditioning on the outcome of the first round) $E_k \; = \; 1 + \tfrac12(E_{k+1} + E_{k-1}) \hspace{2cm} 1 \le k \le n-1$ Solving this recurrence relation, we deduce that $E_k = \boxed{k(n-k)}$ for $0 \le k \le n$ .

Gambler's ruin - problem 5

1 solution

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