Subject: Re: Random chewing
Date: Fri, 26 Mar 2000 18:05:12 -0800
From: Jason Swanson
```The general solution to this problem when n:="the number of pieces of gum originally contained in each pack" and "r:=the number of pieces remaining.

at the moment the first pack is emptied" is

AP(r)=(2n-r-1)!/[(n-1)!(n-r)!(2^(2n-r-1))]

To illustrate the method by which this solution is derived, consider the
the example n=10 and r=6. Label the packs x and y and suppose,
at first, that pack x is the one that is emptied. Let sequences
of x's and y's be understood to represent the sequence of choices
of packs (i.e. xyx represents choosing pack x the first time,
pack y the second time, and pack x the third time). Under this
definition, pack y will contain 6 pieces when pack x is emptied
if and only if a sequence occurs which

(i) contains 10 x's,
(ii) contains 4 y's, and
(iii) ends in an x. Let us call the set of all such sequences S.
Since each such sequence will have 14 characters, the probability
of each such sequence occurring is 1/(2^14). Thus the probability
that pack y will contain 6 pieces when pack x is emptied is
|S|/(2^14).

By symmetry, the probability that pack x
will contain 6 pieces when pack y is emptied is also |S|/(2^14).

Since either event will satisfy the conditions of the original
question, the probability we seek is P(6)=2*|S|/(2^14)=|S|C/(2^13).

Now, |S| is the number of sequences containing 10 x's and 4 y's and
ending in an x. This is the same as the number of sequences
containing 9 x's and 4 y's, which is 13C4=13!/[(9!)(4!)].

Therefore, P(6)=13!/[(9!)(4!)(2^13)]=0.087280... This procedure
can be easily repeated using the variables n and r in place%0D%0A
of 10 and 6 to obtain the general formula given above.

```

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