Banach's matchbox with a favored pocket

Question

A smoker keeps a matchbox in each pocket, each starting with $n$ matches. This time the pockets are not equally likely: he reaches into the left pocket with probability $p$ and the right with probability $q = 1 - p$ each time, independently. Eventually he reaches into a pocket, pulls out the box, and finds it empty.

At that moment, what is the probability the other box contains exactly $k$ matches?

Accepted Answer

Left box discovered empty, right holds k. This requires all n left draws, exactly n-k right draws (leaving k), and finally one more left reach that finds nothing. Among the first n + (n-k) = 2n-k reaches, n are left and n-k are right in any order, and each specific sequence of those reaches plus the final left reach has probability p^{n} q^{n-k}\cdot p. There are \binom{2n-k}{n} orderings, so P(	ext{left empty},\, k 	ext{ in right}) = \binom{2n-k}{n} p^{\,n+1} q^{\,n-k}. Right box discovered empty, left holds k. By the mirror argument, swap p and q: P(	ext{right empty},\, k 	ext{ in left}…

Banach's matchbox with a favored pocket

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