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.