Banach's matchbox problem
Oxford Dictionary of Statistics:
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Banach's matchbox problem
A classic problem in probability. A pipe smoker has two full boxes of n matches. When he wants a match, he is equally likely to select either box. The question concerns the number of matches left in the second box on the first occasion that he chooses a box and finds it empty. The probability that there are exactly k (k=0, 1, 2,..., n) matches left in the second box is
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Banach's matchbox problem
Banach's match problem is a classic problem in probability attributed to Stefan Banach. Feller [1] says that the problem was inspired by a humorous reference to Banach's smoking habit in a speech honouring him by H. Steinhaus, but that it was not Banach who set the problem or provided an answer.
Suppose a mathematician carries two matchboxes at all times: one in his left pocket and one in his right. Each time he needs a match, he is equally likely to take it from either pocket. Suppose he reaches into his pocket and discovers that the box picked is empty. If it is assumed that each of the matchboxes originally contained N matches, what is the probability that there are exactly k matches in the other box?
Solution
First consider the case where the matchbox in his right pocket has an unlimited number of matches and let M be the number of matches removed from this one before the left one is found to be empty. When the left pocket is found to be empty, the man has chosen that pocket (N+1) times. Then M is the number of successes before (N+1) failures in Bernoulli trials with p=1/2, which has the negative binomial distribution and thus
![P[M=m] = \binom{N+m}{m}\left(\frac{1}{2}\right)^{N+1+m}](http://wpcontent.answcdn.com/wikipedia/en/math/6/7/7/677e079627648e86e8ea1bf0b159a246.png)
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Returning to the original problem, we see that the probability that the left pocket is found to be empty first is P[M < N + 1] which equals 1/2 because both are equally likely. We see that the number K of matches remaining in the other pocket is
![P[K=k] = P[M=N-k|M<N+1] = 2P[M=N-k] = \binom{2N-k}{N}\left(\frac{1}{2}\right)^{2N-k}](http://wpcontent.answcdn.com/wikipedia/en/math/6/0/e/60ecd98710d2626b8f6693185f2b0e05.png)
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References
- ^ Feller, William, An Introduction to Probability Theory And Its Applications, Third Edition, Wiley, 1968, Chapter VI, section 8
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Oxford Dictionary of Statistics
A Dictionary of Statistics. Second edition revised. Copyright © Oxford University Press, 2008. All rights reserved.
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