A bag contains r red balls and w white balls. Each time, you take on ball out of the bag at random and without replacement. You stop as soon as all the red balls have been taken out of the bag. What is the expected number of white balls remaining in the bag when you stop?
Hello all! I'm self-studying, and came across this question in Henk Tijms' book, Probability, a Lively Introduction. So this is an exercise meant for practicing indicator variables, and my question is about a certain choice of it.
The author's solution labels the white balls using 1 <= k <= w, and makes an indicator variable $I_k$ for each of them: 1 for remaining in the urn till the end, 0 for otherwise. Then the author claims that when considering a particular white ball, we can "discard other white balls", and so E($I_k$) = P($I_k$) = $\frac{1}{(r + 1)}$.
What wizardry is this!? This result is consistent with the probability of having $r$ red balls and only 1 white ball, then making $r$ extractions without touching the white one. But why is this equivalent to the original situation?
Much thanks in advance!
Best Answer
$I_k$ is the indicator of the event that the $k$th white ball remains in the bag after all the red balls have been extracted. The insight here is that this event is determined entirely by the relative position of ball $k$ compared to the $r$ red balls; to answer whether the event occurred, you don't need to know the placement of the other white balls, or even the number of other white balls; or even whether there exist balls of other colors in the bag. It matters only whether ball $k$ is seen at the rightmost position of the combined list of $r+1$ balls (which has probability $1/(r+1)$).
You can reach the same conclusion by considering all possible placements of these $r+1$ balls among the hypothetical "extraction queue" (of length $r+w$), answering the question for a fixed placement, and then rolling up the answer; you'll discover that the answer is the same for every placement. This is the sense in which you can discard the other white balls.
A similar insight helps you solve the problem "If you shuffle a deck of cards and deal the cards one at a time until you've seen all the aces, what is the probability that the ace of spaces is seen last?" If you realize that it matters not what other cards are in the deck, or when/how other cards were dealt, you jump immediately to the answer: $\frac14$.