Associated Topics || Dr. Math Home || Search Dr. Math

### 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/
```
Associated Topics:
High School Probability

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search