would it simply be $w/(w+b)$?

Question

There are w white balls and b black balls in a box. Balls are drawn one after another without replacement. What is the probability that the white balls will be exhausted before the black balls?

Answer 1

You can simplify this question by asking for the probability that the last remaining ball is black. This is just the probability of drawing a black ball, $\frac{b}{b+w}$.

An even better way to think about the problem (fewer distractions) might be to re-frame it as an ordering problem: If we randomly order the balls, what is the probability that any given ball is black? This way, we're dealing with the same basic scenario, but we avoid being tricked by the previous balls that were drawn, which we can't condition on since we don't know anything about them.