Vol. 10 No 5 - Pi Mu Epsilon

3 73 PI MU EPSILON JOURNALThis can be written more compactly by lettingFrom the Taylor series for e x about x = 0,converges to 1 - 1/e as n - m. It should also be noted that 0 2 a^a< 1 for all k andn, and that for each k lim = 1. These facts taken together would suggestn--(- l)k+lthat for n large, pQ = -, and therefore that lirn p = 1 - lle. Ak=1 k! n-00careful proof of this result is somewhat technical, and is deferred until the end ofthis note.If the 2n chairs are arranged in a circle, a couple occupying chairsnumbered 2n and 1 would then be in adjacent chairs. The event that at least onecouple occupies adjacent chairs in this situation can be thought of as occurring inone of two mutually exclusive ways: (1) at least one couple occupies adjacentchairs other than the two numbered 2n and 1, or (2) a couple occupies chairs 2nand 1 with no other couple occupying adjacent chairs. The first of these eventshas probability pn. For the second event we note that the probability of being inthe adjacent chairs 2n and 1 is l/(2n - I), and given that a couple is seated therethe probability that no other couples occupy adjacent chairs among the remaining2(n - 1) seats is 1 - p,,.). Therefore, the probability, fin, of at least one coupleoccupying(A)adjacent chairs when the chairs are arranged in a circle is given byPn = pn + (1 - fiom which it follows that fin also approachesBYRNE, MATCHING PROBLEM 3741 - 1/e as n - m.It is interesting to note that the congruence of both pn and 3 to 1 - 1Ie*is related to that of -( ' , which itself is the answer to a weii knownk=1 k!probability matching problem: if n married couples are randomly paired up for adance, what is the probability that at least one pair is a married couple? Thisproblem is sometimes cast in terms of matching hats to men, letters to envelopes,or positions in two decks of cards (see for example [I], [2] p. 107, and [4]). We5 à converges to e, we can choose Nl so thatnow present the proof that lim pn = 1 - lle. Let e > 0 be given. SincenÑk=O k!1- e 1=1 e (_1)k+lk=0 k kq+l k! 3 k=, k!2 - < - for all n t N). Since -converges to 1 - lle, we can choose N2 so that1~(- 1lk+l- for all n t N2. Let N3 = max(Nl, N2).T - ( l - : ) lSince lim = 1, for each k = 1, 2 ,..., N3, we can chooser so thatn--1-a^ c e#e for all n s n,;. Let N4 = max(n,, %,..., s). Now supposen s max(N3, N4). We haveThe first term on the right side of (6) issince 1 ~