Banach Matchbox Problem
Date: 10/31/98 at 00:30:44 From: Patel Akshay Subject: Probability Mr. Stephan kept two matchboxes, one in each pocket. Each box contained exactly n matches. Whenever he wanted a match he reached at random into one of his pockets. When he found that the box he picked was empty, what is the probability that the other one has exactly k matches (k = 0, 1, 2,......n)? My approach to this example: There are two cases possible: (1) both are empty (2) one is empty and other is full For second case the probability of getting k matches out of n is k/n, but as we have two pockets there are two different possibilities. Either the left is empty or the right is empty. So to cover both cases I just multiplied the answer by 2 and got 2(k/n). But I think my answer is wrong, so I want your help for this example. Thank You.
Date: 10/31/98 at 10:43:14 From: Doctor Anthony Subject: Re: Probability This is a classic problem, sometimes called the Banach Matchbox problem. Note that the least number of 'trials' is n+1 and the maximum number is 2n+1. Suppose p = the probability that he uses the lefthand pocket and q = the probability that he uses the righthand pocket. Then: Prob(k=n) = C(n+1,n+1)p^(n+1) q^0 + C(n+1,n+1)p^0 q^(n+1) = p^(n+1) + q^(n+1) = 2(1/2)^(n+1) if p = q = 1/2 = (1/2)^n If he has used 1 match from the other box, then during the first n+1 occasions he must have chosen n matches from one box and 1 match from the second box. Then on the n+2 nd occasion he returns to the empty box. The probability of this is: Prob(k=n-1) = C(n+1,n)p^n q p + C(n+1,n)p q^n q = 2 C(n+1,n)(1/2)^(n+2) if p = q = 1/2 = C(n+1,n)(1/2)^(n+1) = (n+1)(1/2)^(n+1) If he has used 2 mmatches from the other box, then during the first n+2 occasions he must have chosen n matches from one box and 2 matches from the second box. Then on the n+3 rd occasion he returns to the empty box. The probability of this is: Prob(k=n-2) = C(n+2,n)p^n q^2 p + C(n+2,n)p^2 q^n q = 2 C(n+2,n)(1/2)^(n+3) if p = q = 1/2 = C(n+2,n)(1/2)^(n+2) The pattern is now clear: P(k=n-r) = C(n+r,n)(1/2)^(n+r) So if we want the answer in terms of k we replace r by n-k in this expression: P(k matches in other box) = C(n+n-k,n)(1/2)^(n+n-k) = C(2n-k,n)(1/2)^(2n-k) - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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