It has been a while again since I last posted a puzzle, so…

You have once again been captured by mathematically-inclined pirates and threatened with walking the plank, unless you can win the following game: some number of black balls and white balls will be placed in an urn. The captain will randomly select and discard a ball from the urn, noting its color, then repeatedly draw and discard additional balls *as long as they are the same color*. The first drawn ball of a *different* color will be returned to the urn, and the whole process will be repeated. And repeated, and repeated, until the urn is empty. If the last ball drawn from the urn is black, you must walk the plank; if it is white, you will be set free.

You can choose any positive numbers and of black and white balls, respectively, to be placed in the urn at the start. How many of each should you start with to maximize your probability of survival?

### Like this:

Like Loading...

*Related*

Unless I’ve made a mistake, a simple Monte Carlo simulation suggests it doesn’t matter what you pick for b and w. It’s always a 50% shot (result plot at the bottom):

So the *real* puzzle is proving why this is the case. Perhaps some inductive proof? When a ball is removed, it’s essentially the same as choosing a smaller b or w, and so any choice of (b, w) can be reduced to (1, 1). While writing up my Monte Carlo simulation, I thought about using dynamic programming, which lead to this idea.

Right– actually, the “base cases” are (b,0) yielding probability 0 of survival and (0,w) yielding probability 1, but this is the idea. A search for “urn solitaire” yields a paper or two on the game, in particular the more challenging problem of determining the expected “duration” of the game (i.e., how many rounds of “draw until a different-colored ball”).