Which matchbox empties first when they start unequal?

Question

A smoker's left box starts with 2 matches and his right box starts with 3. Each reach chooses a pocket with probability 	frac12, independently, and he draws until he reaches into a pocket and finds the box empty. What is the probability the left box is the one discovered empty, and how many matches tend to remain in the loser?

Accepted Answer

Left box discovered empty, k remaining in right (k = 0,1,2,3): all 2 left matches are drawn, 3-k right matches are drawn (leaving k), then a final left reach finds nothing. There are \binom{2 + (3-k)}{2} = \binom{5-k}{2} orderings of the pre-discovery reaches, and each full sequence of 6-k reaches has probability (	frac12)^{6-k}: P(	ext{left empty},\, k) = \binom{5-k}{2}\left(	frac12ight)^{6-k}. k \binom{5-k}{2} probability --- --- --- 3 1 1/8 = 4/32 2 3 3/16 = 6/32 1 6 3/16 = 6/32 0 10 5/32 P(	ext{left discovered empty}) = \frac{4 + 6 + 6 + 5}{32} = \frac{21}{32} \approx 0.656. Right b…

Which matchbox empties first when they start unequal?

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