Gambler's ruin - problem 7

Probability Level 3

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

Again, 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 to keep playing, meaning that the only thing that will cause you to stop is losing all your money. How many games do you expect to play before you stop?

Note: As in problem 6 of this series, you may assume Martin's budget is infinite.

O(2^k)

O(k^k)

\infty

O(k^2)

1 solution

Mark Hennings
Feb 14, 2019

Suppose that the expected time to the end of the game was finite. Suppose that $E_k$ was the expected number of games until bankruptcy, given that you start with $k$ . Then $E_k \; = \; 1 + \tfrac12(E_{k+1} + E_{k-1}) \hspace{2cm} k \ge 1$ and hence we must be able to find constants $A,B$ such that $E_k = A + Bk - k^2$ for all $k \ge 0$ . Since $E_0=0$ we see that $A=0$ . But this would imply that $E_k < 0$ for all $k > B$ , which is not possible. Thus we deduce that the expected number of games to bankruptcy is infinite.

This reflects the fact that the simple symmetric random walk in one dimension is null recurrent.

What's the difference between this problem and the previous one?

Yes, infinite, or go bust eventually?

Bert Seegmiller - 2 years, 3 months ago

Problem 6 tells us that the probability of reaching any point on the Markov chain is $1$ (it is recurrent). This problem tells us that the expected time to reach any such point is infinite (it is null).

Mark Hennings - 2 years, 3 months ago

Gambler's ruin - problem 7

1 solution

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