Catalan Numbers and ProbabilityDate: 05/29/2003 at 07:18:43 From: Rangarajan Subject: Dollar Change Problem Twenty persons want to buy a $10 ticket each. Ten of them have a $10 note and others have a $20 note. The person at the ticket counter has no money to start with. What is the probability that the person at the ticket counter will not have a change problem? I started with finding the number of instances where it's unfavorable, i.e. where the ticket issuer WILL have a problem. These are when we have series like one of the following: a. 20,.... b. 10,20,20.... c. 10,20,10,20,20....etc. So, the number of ways in which 'a.' can be formed is 19!/[10!*9!] and the number of ways in which 'b.' can be formed is 17!/[9!*8!] and so on. Hence, I finally end up getting a series which on summation will give me the number of unfavorable situations (call it A). Then the total no. of arrangements that can be formed is 20!/[10!*10!] (call it B). Thus, no of favorable cases will be B-A and hence the probability of favorable situations will be (B-A)/B. I just want to know if there is some other route by which the process of summation and hence the time and effort required to solve the problem can be minimized. Date: 05/29/2003 at 08:31:24 From: Doctor Anthony Subject: Re: Dollar Change Problem This is an example where we use Catalan numbers. The problem is identical to that of considering the number of routes from one corner of a grid to the diagonally opposite corner such that you always remain above the diagonal. See the example below. Example: A city lawyer works n blocks north and n blocks east of his place of residence. Every day he walks 2n blocks to work. How many routes are possible if he never crosses (but may touch) the diagonal line from home to office? Each acceptable route is above the diagonal. Each route is a sequence of n northerly blocks and n easterly blocks. We identify north with +1 and east with -1. Thus each route corresponds to a sequence a(1), a(2) , a(3), ......, a(2n) of n +1's and n -1's and in order that the route not dip below the diagonal we must have a(1) + a(2) + a(3) + ........ + a(k) >=0 for k=0, 1, 2, ......, 2n The number of sequences a(1), a(2), a(3), ......, a(2n) of 2n terms that can be formed by using n +1's and n -1's whose partial sums satisfy a(1) + a(2) + ... + a(k) >=0 .....(1) (k=1, 2, 3, ....,2n) equals the nth Catalan number (see proof below) C(n) = 1/(n+1) C(2n,n) n >= 0 We call a sequence of n +1's and n -1's acceptable if it satisfies (1). and unacceptable otherwise. [This is equivalent to the number of routes consisting of n northerly blocks (+1) and n easterly blocks (-1) that never cross the diagonal.] Let A(n) denote the number of acceptable sequences of n +1's and n -1's and let U(n) denote the number of unacceptable ones. The TOTAL number of sequences of n +1's and n -1's is (2n)!/(n!n!) = C(2n,n) So we must have A(n) + U(n) = C(2n,n) and we evaluate A(n) by first evaluating U(n) and subtracting from C(2n,n). Consider an unacceptable sequence of n +1's and n -1's. Because the sequence is unacceptable there is a smallest k such that the partial sum: a(1) + a(2) + ... + a(k) < 0 Because k is smallest there is an equal number of +1's and -1's preceding a(k) and we have: a(1) + a(2) + .... + a(k-1) = 0 and a(k) = -1 In particular, k is an ODD integer. We now reverse the signs of each of the first k terms; that is we replace a(i) by -a(i) for each i = 1, 2, 3, ..., k and leave unchanged the remaining terms. The resulting sequence a'(1), a'(2), .. , a'(2n) is a sequence of (n+1) +1's and (n-1) -1's. This process is reversible: Given a sequence of (n+1) +1's and (n-1) -1's there is a first instance when the number of +1's exceeds the number of -1's (since there are more +1's than -1's). Reversing the +1's and -1's up to that point results in an unacceptable sequence of n +1's and n -1's. Thus there are as many unacceptable sequences as there are sequences of (n+1) +1's and (n-1) -1's. The number of sequences of (n+1) +1's and (n-1) -1's is the number (2n)!/[(n+1)!(n-1)!] = C(2n,n-1) (2n)! Therefore U(n) = ----------- (n+1)!(n-1)! (2n)! (2n)! Therefore A(n) = --------- - ----------- n! n! (n+1)!(n-1)! (2n)! 1 1 = -------- [ ---- - ------] n!(n-1)! n (n+1) (2n)! 1 = --------- [ -------] n!(n-1)! n(n+1) 1 (2n)! = ------ [ ---------] (n+1) n! n! = 1/(n+1) C(2n,n) and so, returning to the problem of the journeys from home to work, the total number of acceptable routes above the diagonal is C(n) = 1/(n+1) C(2n,n) where C(n) means the nth Catalan number. There are many different interpretations of this theorem. One such is that of the theatre tickets. Example ------- There are 2n people in line to get into a theatre. Admission is $1. Of the 2n people, n have $1 bills and n have $2 bills. The box office rather foolishly begins with an empty cash register. In how many ways can the people line up so whenever a person with a $2 bill buys a ticket, the box office has $1 in order to give change. If we regard the people as indistinguishable and identify a $1 with a +1 and a $2 bill with -1, then the answer is the number C(n) = 1/(n+1) C(2n,n) of acceptable sequences. The probability that there is always change available is this number divided by the the total number of possible sequences, C(2n,n). Therefore the required probability = 1/(n+1) Applying this to your problem of 20 people in the queue, 10 with $20 bills and 10 with $10 bills, we put n=10 into the above expression and obtain the probability 1/11. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ Date: 05/30/2003 at 08:25:48 From: Rangarajan Subject: Thank you (Dollar Change Problem) Sir, Thanks a lot for your reply. Thanks to you, I have been introduced to the new concept of Catalan numbers. The solution also was written in an easy-to-understand format. Once again thanks a lot. I hope you continue this great service forever. Yours truly, Rangarajan V. |
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