Corrupt Official

A corrupted Chinese official brings $10 mln to Macau... and lost $7 mln of it. He'll be executed anyway, so he must win back $7 mln. His strategy is to play a 50/50 game, and bet $1 mln each time (not optimal, but that's not the point here). When are the chances of him winning back $7 mln?

Follow-up: what's the expected number of times this official will be playing this game before he learns his fate? (i.e. either lose it all or win back $7 mln)

Solution:

The gambler's ruin problem. The are many elegant solutions to this, but I like this the most. Based on the martingale stopping time theorem.

The expected value of his money is always the same--$3 mln, no matter when he stops playing the game, i.e. either when he's lost all or now back at $10 mln.

So, E[Money= ] $3 mln = Prob(going bust) * $0 mln + Prob (Get to $10 mln) * $10 mln

provided such probabilities exist (they do!). Prob (Get to $10 mln) = 1 - Prob(going bust), so we have:

3 = p*0 + (1-p)* 10 = 10 - 10p, 10p = 7/10 , p = 0.7.

In general, it's just a/(a+b). In this case, the chances of him getting executed is 70%.


Solution to the follow-up:

Yet another application of martingales. Martingales are just such wonderful animals.

The trick is to find a suitable martingale. You need to know that B²(t)-t is a martingale, where B(t) is a Brownian motion. This martingale is application here because we're dealing with a random walk.

So, we have:

E[B²(t) - t] = 0
E[B²(T) - T] = 0 T is when game ends; this substitution okay thanks to stopping time theorem
E[B²(T)] - E[T] = 0 expectations are linear
E[T] = E[B²(T)] rearrange

t is time. In this context, it's the number of times our corrupt official plays the game.

So, down to evaluating the left hand side. Substituting the numbers in:

E[B²(T)]
= P(win back $7 mln) (7²) + P(lose $3 mln) (3²)
= 3/10 * 49 + 7/10 * 9 = 21

So the expected number of games this official plays would be 21. Notice that we're essentially calculating the variance in the last step.