c 2003 Society for Industrial and Applied Mathematics

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1 SIAM J DISCRETE MATH Vol 7, No, pp 7 c 23 Society for Inustrial an Applie Mathematics SOME NEW ASPECTS OF THE COUPON COLLECTOR S PROBLEM AMY N MYERS AND HERBERT S WILF Abstract We exten the classical coupon collector s problem to one in which two collectors are simultaneously an inepenently seeking collections of coupons We fin, in finite terms, the probability that the two collectors finish at the same trial, an we fin, using the methos of Gessel Viennot, the probability that the game has the following ballot-like character: the two collectors are tie with each other for some initial number of steps, an after that the player who first gains the lea remains ahea throughout the game As a by-prouct we obtain the evaluation in finite terms of certain infinite series whose coefficients are powers an proucts of Stirling numbers of the secon kin We stuy the variant of the original coupon collector s problem in which a single collector wants to obtain at least h copies of each coupon Here we give a simpler erivation of results of Newman an Shepp an exten those results Finally we obtain the istribution of the number of coupons that have been obtaine exactly once singletons ) at the conclusion of a successful coupon collecting sequence Key wors coupon, Stirling number, lattice path AMS subject classification 5-2 DOI 37/S Introuction an results The classical coupon collector s problem is the following Suppose that a breakfast cereal manufacturer offers a souvenir coupon ) hien in each package of cereal, an there are ifferent kins of souvenirs altogether The collector wants to have a complete collection of all souvenirs What is the probability pn, ) that exactly n boxes of cereal will have to be purchase in orer to obtain, for the first time, a complete collection of at least one of each of the kins of souvenir coupons? The answer to that question is well known eg, [5, p 32]) to be pn, )! { } n ) n, where the { n k} s are the Stirling numbers of the secon kin We stuy, in this paper, a number of other aspects of this problem as well as a generalization of it to a two-player game First, suppose we have two coupon collectors, rawing coupons simultaneously, an each seeking to obtain a complete collection of coupons We ask for the probability that the two games are complete at the same time The answer is given by 26) below That answer is expresse in finite terms, owing to the close form evaluation of the orinary power series generating function for the squares of the Stirling numbers of the secon kin, containe in 25) Receive by the eitors February 25, 22; accepte for publication in revise form) December 2, 22; publishe electronically September 7, 23 Department of Mathematics, University of Pennsylvania, Philaelphia, PA Current aress: Mathematics an Computer Science Department, St Joseph s University, Philaelphia, PA 93 amyers@sjueu) Department of Mathematics, University of Pennsylvania, Philaelphia, PA wilf@ centralcisupenneu)

2 2 AMY N MYERS AND HERBERT S WILF Next, we consier the following two-person game Again two coupon collectors are simultaneously rawing coupons at ranom This time we are intereste in a ballot-like problem: What is the probability that the player who first complete a collection the winner) was never behin ie, never ha fewer istinct coupons) at any intermeiate stage of the play? Here we give a complete answer to a slightly easier question, namely the following: What is the probability that, after an initial segment of play in which the players are tie, one of them takes the lea an keeps the lea strictly until the en The answer is in 226) below an is obtaine by the Gessel Viennot theory of nonintersecting lattice paths In each of these cases the answer can first be written as an infinite series whose coefficients involve various proucts of Stirling numbers What is interesting, though, is that in all such cases we are able to express the answers in finite terms Inee, one of our main results here is the observation that infinite series whose coefficients involve various powers an proucts of Stirling numbers of the secon kin can reaily be evaluate in finite terms In section 4 we return to the original collecting problem of obtaining at least one copy of each coupon, but now we stuy the variant of the problem in which a single collector wants to obtain at least h copies of each coupon We obtain the generating function 45) for the probability that exactly n trials are neee, the exact value of the average number of trials 4), an the asymptotic behavior 45) of these quantities as n Finally, in section 5 we stuy the number of coupons that have been collecte only once, at the en of a collection sequence We fin the istribution function 54) for this number an show that the average number of these singletons is just the harmonic number H +/2+ +/ 2 The two-person collecting competition 2 Simultaneous completion We fin now the probability of simultaneous completion of two inepenent coupon collecting sequences Eviently this is pn, ) 2! 2 { } n 2 2) 2n, n n which expresses the answer as an infinite sum We can rewrite this as a finite sum by fining a finite expression for the generating function for the squares of the Stirling numbers of the secon kin, { } n 2 F k x) ef x n, k analogously to the well-known generating function for these numbers themselves, { } n x n x k 22) k x) 2x) kx) n k n k The easiest way to o this is via the stanar explicit formula for these Stirling numbers, viz { } n k ) k 23) ) k r r n k n) k k! r ef r k A k,r r n k, r

3 SOME NEW ASPECTS OF THE COUPON COLLECTOR S PROBLEM 3 where we have written 24) A k,r )k r r k k! ) k r It follows that 25) F k x) ef n k { } n 2 x n k n k x n k A k,r A k,s r n k s n k r,s k x k A k,r A k,s rsx) n k r,s x k k r,s A k,r A k,s rsx n k x < k 2 ) Thus for the simultaneous completion probability we obtain, from 2), 26) n pn, ) 2!2 2 r,s A,r A,s rs 2 by 25), where the A s are given by 24) This sequence of probabilities, for, 2,, begins as, 3, 7, 9 9, , , , ,, ie, as, 33333, 574, 989, 7249, 5664, 46622, 39586, 22 Neck-an-neck then always ahea We encoe a sequence of n raws as a path ω with n vertices in the lattice L consisting of vertices i, j) an eges {i, j), i +,j)}, {i, j), i +,j+)} for all i, j The first coorinate of a vertex in the path gives the number of raws, or steps, an the secon coorinate gives the number of istinct coupons the collector has at that step Thus ω starts at, ) inicating the collector has coupons at raw, procees to, ) the collector has coupon after raw), an ens at n, ), n the collector has a complete collection at step n) We write ω, )ωn, ), where ω is a path from, ) to n, ), to inicate that ω starts at the vertex, ), continues with the first vertex, ) in ω, then follows ω through to n, ), an finally ens with the vertex n, ) We assign a weight of i/ to each horizontal ege {i, j), i +,j)} in the lattice L This is the probability that at the j + )st step, the collector raws one of the i istinct coupons alreay collecte at step j We assign a weight of i/ to each northeast ege {i, j), i +, j + )} The probability that the collector raws the particular sequence of coupons encoe by the path ω is given by the prouct of the weights on the eges of ω We let P ω) enote this probability Suppose one collector, the winner, collects all istinct coupons for the first time at step n At step n the winner ha istinct coupons) Let ω be the

4 4 AMY N MYERS AND HERBERT S WILF lattice path which encoes the winner s sequence of raws Let ω 2 encoe the other collector s raws We compute the probability p) that ω an ω 2 are ientical until some point at which the winner takes the lea an the other collector never catches up To o this, we begin by supposing ω is ientical to ω 2 until step k, at which point both collectors have istinct coupons The argument splits into two cases, namely k n 2 an k n In both cases, at step k + the winner collects one aitional istinct coupon while the other collector oes not After step k, the two paths never intersect again The winner collects all istinct coupons for the first time at step n Suppose the other collector has 2 istinct coupons at this point The probability we seek is 27) p) n n k 2 ω,ω 2) P ω )P ω 2 ), where the innermost sum ranges over all pairs ω,ω 2 ) escribe above 23 The case k n 2 Write ω αω n, ), where α enotes a lattice path from, ) to k, ), an ω enotes a path from k +, +)ton, ) Similarly, set ω 2 αω 2, where α is as above an ω 2 is a path from k +, )to n, 2 ) Note that ω an ω 2 are nonintersecting paths in the lattice L In terms of these we have P ω )Pα) /)P ω )/) an P ω 2 )Pα) /)P ω 2 ) Hence from 27) we fin for the combine probability of all pairs if k n 2, p) k n 2 28) n 2 n k 2 ω,ω 2) n 2 n k 2 [ P α) ) P ω ) ) ) ) α P α) 2 )] [ P α) ω,ω 2) ) ] P ω 2 ) P ω )P ω 2 ) At this point we have translate a question about coupon collecting into a problem involving nonintersecting paths in a lattice We have set the stage for application of the Gessel Viennot theorem [3] This result concerns pairs of nonintersecting lattice paths with no constraints on vertices or eges in the paths For this reason we have written ω an ω 2 in terms of ω an ω 2 The theorem refers to an arbitrary set L, which we will take to be the lattice efine earlier, an a weight or valuation) v, which we take to be P The theorem equates a sum of weights of paths with the eterminant of a matrix a ij ) i,j l The entries of this matrix are efine by a ij ω vω), where ω ranges over all paths from A i to B j The theorem requires that two given sequences, A,A 2,,A l ) an B,B 2,, B l ), of vertices in L, the sets Ω ij, i, j l, of all paths in L between A i an B j, an the weight v satisfy both the finiteness an crossing conitions The finiteness conition requires the set of paths in Ω ij with nonzero weight be finite The crossing conition requires that paths in Ω ij an Ω i j, i<i an j<j, with nonzero weight share a common vertex Both conitions hol for the paths we consier Theorem 2 Gessel Viennot) Suppose L, v, A,A 2,,A l ), an B,B 2,,B l ) satisfy both the finiteness an crossing conitions Then the eterminant of the matrix a ij ) i,j l is the sum of the weights of all configurations of paths ω,ω 2,,ω l ) satisfying the following two conitions:

5 SOME NEW ASPECTS OF THE COUPON COLLECTOR S PROBLEM 5 i) The paths ω k are pairwise nonintersecting, an ii) ω k is a path from A k to B k In other wors, ) et {a ij } l i,j vω )vω 2 ) vω l ) ω,ω 2,,ω l ) Application of this theorem to our problem requires the computation of only a 2 2 eterminant! Let A k +, + ), A 2 k +, ), B n, ), an B 2 n, 2 ) Then 29) ω,ω 2) [ a a P ω )P ω 2 ) et 2 a 2 a 22 where a ij is the sum ω P ω) over all paths ω from A i to B j 24 Paths from A to B In this section we compute the probability P ω)ofan arbitrary path ω from a vertex A a,b ) to a vertex B a 2,b 2 ) as well as the sum over all such paths Such a path contains b 2 b northeast eges {i, j), i +,j+)} an a 2 a ) b 2 b ) horizontal eges The weights assigne to northeast eges in orer from left to right are b, b+ b2,, The weight assigne to a horizontal ege epens on its coorinates Consier the ege {i, j), i +, j)} This ege inicates the collector has j istinct coupons at step i an raws one of the same j coupons at step i + The probability of this weight of the ege) is j Thus the probability of a path ω from A to B is P ω) 2) ) e b b ) b + b )! a2 a b 2 )! b ) e b +) e2 b 2 ) e b 2 b +, ], ) e2 b ) + b ) ) eb2 b 2 b2 + where e e,e 2,,e b2 b +) is an orere partition, a composition, ofa 2 a ) b 2 b )intob 2 b + nonnegative integer parts With this we compute the sum of the probabilities of all paths from A to B 2) ) e P ω) b b ωa B e b 2 b )! a2 a b 2 )! e ) b + ) ) eb2 b b2 + ) e2 b ) + b ) e b +) e2 b 2 ) e b 2 b +, where the sum is over all compositions e e,e 2,, b2 b +) ofa 2 a ) b 2 b ) into b 2 b + nonnegative integer parts This is the coefficient of x a2 a) b2 b) in the series expansion of b x) b +)x) b 2 x),

6 6 AMY N MYERS AND HERBERT S WILF so we can fin a simpler formula for it by looking at the partial fraction expansion 22) where ωa B b b ma mx) ma B m )b m m b a b a)! B m mx, ) b a m a From this an 2) we obtain { b )! b2 P ω) a2 a b 2 )! [xa2 a) b2 b) ] 23) b )! a2 a b 2 )! b )! a2 a b 2 )! b 2 mb mb B m m a2 a) b2 b) b 2 mb )b2 m ma2 a b 2 b )! B m mx b2 b m b 25 Evaluating the eterminant We use the results of the previous section to evaluate the eterminant in 29) To compute a, we substitute A A k +, + ) an B B n, ) into 23) This yiels 24) a n k 2 )! In a similar manner we obtain 25) a 2 )! n k 2 )! m + 2 m + ) m m n k 2 2)! ) 2 m m n k 2 )! } ) ) 2 m ) 2, m a 2 n k 2 ) m m n k 2 ) 26) )!, )! m m )! 2 ) 2 m m n k ) 2 27) a 22 n k 2 )! 2 )! m m Using 24) 27) we compute the eterminant of our 2 2 matrix [ ] a a et 2 )! )! 28) a 2 a 22 2n 2k 3 2 )! 2 ) ) mlm)n k 2 )+2 l )m2 l2 ) 2 )! 2 )! l m m 29) ef l et,, 2,k,n) Substituting 29) in 29), we obtain 22) P ω )P ω 2 ) et,, 2,k,n) ω,ω 2)

7 SOME NEW ASPECTS OF THE COUPON COLLECTOR S PROBLEM 7 26 The initialcommon segment In the previous section we evaluate the eterminant in 29) In this section we compute the sum α P α)2 in 28) Recall α is a path from, ) to k, ) Equation 2) gives the probability of an arbitrary path from A to B Substituting A, ) an B k, ) gives the probability P α)! k )! e 2 e2 e of an arbitrary path α from, ) to k, ) It follows that α,) k, ) 22) 222) P α) 2! 2 2k )! 2! 2 2k )! 2 [xk ]! 2 2k )! 2 2e 2 2e2 2e e + +e k { m 2! 2 )! 2 2k 2 )! ef init,,k) 2 x) 2 2 x) 2 x) C m m 2k 2 as in 22)) ) ) m 2 + m 27 The case k n Suppose now that the two walks are ientical up to the point n, ) Since step n is the finish, the next step for the winning player will be to n, ) an for the losing player to n, ) These last steps have respective probabilities / an / Hence the probability of the complete pair of walks in this case is the probability of two ientical walks from, ) to n, ) which is given by 22) with k, ):n, )) multiplie by )/ 2 28 Putting it together We now substitute 29) an 222) into 28) to obtain the probability of all pairs of paths that we are consiering, 223) 224) p) n 2 n k ef Σ +Σ 2 ) m init,,n ) n m 2k ) ) init,,k)et,, 2,k,n) It turns out that the sums over the inices 2,n,kcan all be carrie out in explicit close form Hence we can obtain an expression which is in finite terms for the total probability First, the sum on 2 in Σ above can be one in close form since 2 ) 2 ) ) 2 2 t ) ) + t }

8 8 AMY N MYERS AND HERBERT S WILF Next, the remaining sum over the inices n an k in the first summation Σ is 225) ψ, r, s, t) ef n 2 n k r 2k t n k s n k 2 2n r r 2 +3r 2 ) r 2 ) 2 s 2 t r 4 st) r 2 2 )+str 2 2 st) st) 2 r 2 )r 2 st)r 2 s if r 2 st, otherwise The sum over n in Σ 2 is trivial, an so there remain no infinite sums in our final expression for the probability p), which is 226) 2! 2 ) )! 2 2 )! r,s,t ) r s t 2 s t) + r r ) t ) ) ψ, r, s, t) s 4 )!2 ) 2 2 r ) r )! +r 2 r 2 + δ,, where ψ is given by 225) This is the probability that the game is of the type we escribe, namely where the players are tie for some initial segment of trials an then the player who pulls ahea remains ahea always, expresse as a finite sum albeit a complicate one!) More precisely, the values of p) can be calculate, as rational numbers, with O 4 ) evaluations of the above summan The exact values of p) for, 2, 3, 4, 5, are {, 2 3, 43 7, , } 92246, As ecimals, the values of {p)} are {, 66667, 6429, 4334, 29667, 277, 66, 2748, 55, 8988} 29 One collector never behin In contrast to the problem of staying ahea as soon as the tie is broken, which we have solve in the preceing sections, the problem in which the ultimate winner has never been behin is unsolve Suppose the winner collects all istinct coupons for the first time at step n, at which point the other collector has <istinct coupons We iscuss the probability b) the winner has never been behin We use b for ballot since this version of the problem has a istinct ballot-problem flavor see, eg, []) Let w be the lattice path which encoes the winner s sequence of raws Let ω 2 encoe the other collector s sequence of raws Then b) is the probability that ω 2 oes not cross ω Tosayω 2 oes not cross ω means, for each horizontal coorinate i share by vertices i, j )inω an i, j 2 )inω 2, we have j 2 j In the case j 2 j, we say ω an ω 2 intersect at i, j )i, j 2 ) Thus we seek all pairs ω,ω 2 ) such that ω is a path from, ) to n, ) incluing the vertex n, ), ω 2 is a path from, ) to n, ) for, an ω 2 oes not cross ω Such a pair ω,ω 2 ) is illustrate by Figure 2 Note that ω an ω 2 may intersect several times The probability we seek is b) n ω,ω 2) P ω )P ω 2 ),

9 SOME NEW ASPECTS OF THE COUPON COLLECTOR S PROBLEM 9 n, ) n, ) Fig 2 The winner is never behin i, j ) i, j ) i 2, j 2) i, j ) A tail, plus a frame, equals i, j ) i, j ) Fig 22 A kite i 2, j 2) where the innermost sum ranges over all pairs escribe above Look again at Figure 2 A pair ω,ω 2 ) appears to form a chain of flying kites anchore to the groun at, ) The highest kite has two ribbons attache to its tip Their loose ens are at n, ) an n, ) Each kite consists of a frame together with a tail See Figure 22 A frame from

10 AMY N MYERS AND HERBERT S WILF i,j )toi 2,j 2 ) consists of a pair of paths from i,j ), the lower tip of the frame, to i 2,j 2 ), the upper tip, which intersect only at the enpoints A tail from i,j )to i 2,j 2 ) consists of two ientical paths between these enpoints The length of a tail is the number of vertices in the tail minus one, ie, the number of eges A pair ω,ω 2 ) such that ω 2 oes not cross ω forms an alternating sequence of tails an frames, beginning with a tail Note that tails may have length zero The upper tip of the final frame in this sequence is the common enpoint for two paths which intersect only at this common enpoint these are the ribbons escribe above) One path ens at n, ), this is the top ribbon, an the other ens at n, ), the bottom ribbon Below we compute the probability of a frame from i,j )toi 2,j 2 ), a tail from i,j )toi 2,j 2 ), an a pair of ribbons with common initial point k, ) an terminal points at n, ) an n, ), respectively Let f i2,j2) i ) enote the probability of a frame from i,j ),j )toi 2,j 2 ) Note for f i2,j2) i ), we must have i,j ) 2 i +2, j 2 >j, an j 2 j i 2 i Assuming these conitions, we write 227) f i2,j2) i ),j ) P α)p β), α,β) where α, β) is a pair of paths from i,j )toi 2,j 2 ) intersecting only at the enpoints such that β oes not cross α That is, α forms the upper ege of the frame, an β forms the lower ege) We convert the sum above into a eterminant using the Gessel Viennot theorem Evaluation of the eterminant gives f i2,j2) i ) j j 2 j )! 2,j ) 2i2 i) j 2 j )! 2 j 2 )! 2 j 2 ) ) ) l+m lm) i2 i 2 j2 j j2 j l j )m j 2 ) m j l j l,mj We compute the probability t i2,j2) i,j ) ) of a tail from i,j )toi 2,j 2 ) in a manner analogous to the computation of α P α)2 in section 26 We obtain t i2,j2) i ) j )! 2,j ) 2i2 i) 2j 2 )! j 2 )! 2 j 2 ) ) ) j2 j m 2i2 i) 2j2 m + j 2m)! j 2 + m j m m Finally we compute the probability r,,,k,n) of a pair of ribbons with common initial point k, ) an terminal points n, ) an n, ) The probability is given by a eterminant similar to the one in 29) In the present case, we have in place of an in place of 2 Thus r,,,k,n) et,,,k,n) 3 Winning margin Now we look for the probability istribution of the number of istinct coupons that the secon player has collecte at the moment the first

11 SOME NEW ASPECTS OF THE COUPON COLLECTOR S PROBLEM player completes the collection Let g, ) enote the probability that the secon player has collecte exactly istinct coupons at that moment The probability that the first player finishes after exactly n trials is pn, ); see ) The probability that the secon player has exactly istinct coupons after n trials, given that the first player has just complete the collection at that time, is ){ n }! n if < Thus for <our istribution g is given by g, ) { } ){ }! n n! n n n! 2 { }{ } n n )! 2n n! 2 )! 2 r r s A,r A,s rs 2 δ,, by 23) A table of the probabilities {g6,j)} 5 j is as follows: 3,793,8444,8454, These o not sum to because the secon player might have complete a collection at some time before the first player i 4 The ouble ixie cup problem, of Newman an Shepp, revisite Here we consier a ifferent generalization of the coupon collector s problem Let integers h, be fixe Again we are sampling with replacement from kins of coupons, but now T is the epoch at which we have collecte at least h copies of each of the coupons for the first time For example, my h siblings an I might each want to have our own copy of every one of the available baseball cars) We stuy the expectation, the probability generating function, an the asymptotic behavior of the expectation of this generalize problem These questions were investigate by Newman an Shepp [4] an the asymptotics were refine by Erős an Rényi [2] It is interesting to note that this problem is equivalent to one about the evolution of a ranom graph Suppose we fix n vertices, an then we begin to collect from among n kins of coupons If we collect a particular sequence, say, {c,c 2,c 3,}, then we a the eges c,c 2 ),c 3,c 4 ) That is, we a an ege each time we choose a new pair of coupons Our problem about collecting at least h copies of each kin of coupon is thereby equivalent to the question of obtaining a minimum egree of at least h in an evolving ranom graph In this section we will not a anything new to the asymptotics of this problem Instea we claim only a erivation simpler than the original an an explicit generating function, which gives a nice roa to the asymptotics We eal only with generating functions in one variable, whereas in [4] multivariate generating functions were use We obtain not only the expectation of the time to reach a collection that has at least h copies of each kin of coupon, but also the complete probability istribution of that time Our thanks to E Bener an to a helpful referee for pointing this out

12 2 AMY N MYERS AND HERBERT S WILF For n fixe, consier a sequence of n rawings of coupons that constitutes, for the first time at the nth rawing, a complete collection of at least h copies of each of the kins of coupons There are possibilities for the coupon that completes the collection on the nth rawing There are n h ) ways to choose the set of earlier rawings on which that last coupon type occurre On the remaining n h rawings we can efine, as usual, an equivalence relation: two rawings i, j are equivalent if the same kin of coupon was rawn at the ith an the jth rawings The number of such equivalence relations is equal to the number of orere partitions of a set of n h elements into classes, each class containing at least h elements We will enote this latter number by )! { n h }h, where the { } n k s count the unorere partitions of an n-set into k h classes of at least h elements each The number of sequences of n rawings for which we achieve a complete collection for the first time at the nth rawing is therefore ) } n h { n h )! Since there are n possible rawing sequences of length n, the probability that T n is p n! ){ } n n h 4) n h an the probability generating function is P h x) ef p n x n ){! n n h n h n n ) xd { } n h x ) n 42)!, h h n where D / x It remains to fin the orinary power series generating function of the { } n k s h The exponential formula immeiately gives us their exponential generating function as 43) n h, h } x n h { } n x n k h n! ) k e x x xh k! h )! We can convert this into an orinary power series generating function by applying the Laplace transform operator to both sies, which yiels n { } n k h s n+ k! e sx x ) k e sx e x x xh x, h )!

13 SOME NEW ASPECTS OF THE COUPON COLLECTOR S PROBLEM 3 or finally 44) n { } n t n k h k!t ) k e x/t e x x xh x h )! Now if we substitute 44) into 42) we obtain the probability generating function of the generalize coupon collector s problem in the form 45) P h x) h 2 { ) xd )x h e }e t/x t t th t h h )! In the above, ) xd h is the ifferential operator that is efine by xd h ) fx) x ) h )! x x ) x 2 x ) x h fx) However, it is easy to establish, by inuction on h, the interesting fact that 46) xd )x h e t/x t)h h h )! e t/x Hence we have prove the following evaluation Theorem 4 The probability generating function for the coupon collecting problem in which at least h copies of each coupon are neee is given by 47) P h x) h )! ) t h e t/x e t t th t h )! 4 Two examples Let s look at the cases h, the classical case, an h 2, where we want to collect at least two specimens of each of the kins of coupons If h, then 47) takes the form P x) e t/x e t ) t If we expan the power of e t ) by the binomial theorem an integrate termwise, we obtain ) ) j P x) x j jx, j which is precisely the partial fraction expansion of the classical generating function 22) To see something new, let h 2 Then 48) P 2 x) te t/x e t t ) t Again, by termwise integration this can be mae fairly explicit, but since the most interest attaches to the expectation, let s look at the average number of trials that

14 4 AMY N MYERS AND HERBERT S WILF are neee to collect at least two samples of each of coupons This is P 2), which after some simplification takes the form P 2) 2 t 2 ) 49) e t + t)e t ) t t From this we can go in either of two irections: an exact evaluation or an asymptotic approximation By termwise integration it is easy to obtain the following exact formula, which is a finite sum, for T 2, the average number of trials neee to collect at least two of each of the kins of coupons: 4) T 2 2 m,j ) m m ) m j ) j + 2)! m +) j+3 For 2, 3, 4, 5 these are 2, /2, 347/36, 2259/864 To facilitate comparison with the classical h ) case, we show below, for, a table of the expecte numbers of trials neee when h, 2 : T : T 2 : Asymptotics Now we investigate the asymptotic behavior of 49), for large, to compare it with the log behavior of the classical case where h Theorem 42 If there are ifferent kins of coupons, an if at each step we sample one of the kins with uniform probability, let T h enote the average number of samples that we must take until, for the first time, we have collecte at least h specimens of each of the kins of coupons Then for every h, we have T h log ) Consier first the case h 2 In 49) we make the substitution 4) e u + t)e t, where u is a new variable of integration We then fin that 42) P 2) 2 tu)e u u, where tu) is the inverse function of the substitution 4), which is well efine since the right sie of 4) increases steaily from to as t increases from to The main contribution to P 2) comes from values of u near u, an when u is near we have tu) log u + Olog log u) Following the arguments in [6, sect 22], we see that P 2) of 42) has the same asymptotic behavior as c 2 log u)e u u <c<), an in [6] this is shown to be 2 log log

15 SOME NEW ASPECTS OF THE COUPON COLLECTOR S PROBLEM 5 Now we consier the asymptotic behavior of the expecte number of trials for general values of h From 47) we see that this expecte number of trials can be written in the form 43) 2 h )! { t h e t t th h )! Again we make the change of variable 44) e u +t + } { +t + t2 2 ) t22 th + + e t h )! ) } th + + e t t h )! in the integral, an it takes the remarkably simple form compare 42)) P h) 2 tu)e u u, where tu) is the inverse function of the substitution 44) Again the main contribution to the integral comes from small values of u, an when u is small an positive we have tu) log u +h ) log log u)+ Using the metho of section II2 of [6] once more, we fin that 45) T h log +h ) loglog + o)) ) We remark that in the case of 2 coupons, the correct expecte number of trials to obtain two of each coupon is 64 trials, the approximation log is 75, an the approximation log +h ) loglog is 393, each roune to the nearest integer 5 The number of singletons In view of the asymptotics in the preceing section we realize that at the moment when a coupon collector sequence terminates with a complete collection, most coupons will have been collecte more than once, an only a few will have been collecte just once We call a coupon that has been seen just once a singleton We will now look at the istribution of singletons In more etail, let j be the number of singletons in a collecting sequence that terminates successfully at the nth step We first want the joint istribution fn, j) of n an j, ie, the probability that a collecting sequence halts successfully at the nth step an has exactly j singletons at that moment We claim that 5) fn, j)! n n j ){ n j j Inee, the last coupon to be collecte can be chosen in ways; the other j singleton coupons can be chosen in j ) ways an can be presente in an orere sequence in j )! j ) ways This orere sequence can appear among the first n trials in n j ) ways, an the remaining n j trials constitute an orere partition of n j elements into j classes, no class having fewer than two elements, which can be chosen in j)! { } n j j ways If we multiply these together an ivie by 2 n, the number of n-sequences, we obtain the result 5) claime above } 2

16 6 AMY N MYERS AND HERBERT S WILF Next we compute the probability that a complete collecting sequence contains exactly j singletons, whatever the length of the sequence may be That is, we fin F j) n fn, j), where f is given by 5) We have, after using the generating function 44), F j) ){ }! n n j 52) n j j n 2 {! { t t + j ) )} } e xt 53) j)! j e x x) j x j t t / But using the fact that, analogously to 46), we have t t + j ) ) e xt xj j t j )!t j e x/t, we can simplify the expression for F j) to ) 54) F j) j x j e x x) j e x x j, 2, 3,), j which is the esire istribution of the number of singletons in a successfully terminate coupon collecting sequence Now if we multiply by j an sum over j, we ll get the average number of singletons that appear in a complete collection of coupons This is, after some termwise integration, j) ) 2 m +)+ ) m m m +2) 2 m +) m If we expan the summan in partial fractions, viz j) ) 2 ) m m m + m +2 + ) m +2) 2, m then each of the three sums inicate can be expresse in close form, in two cases by using the ientity ) n ) k k x + k 55) x x+n), k irectly, with x an x 2, an in the thir case by ifferentiating 55) wrt x an using the result with x 2 The ientity 55) is itself certifie, after multiplying by the enominator on the right, by the WZ proof certificate Rn, k) kx+k)/n+ )k n )) What results is that j) H, the th harmonic number That is, the average number of singleton coupons in a complete collection sequence of coupons is the harmonic number H n REFERENCES [] L Comtet, Avance Combinatorics, D Reiel, Dorrecht, The Netherlans, 974 [2] P Erős an A Rényi, On a classical problem of probability theory, Magyar Tu Aka Mat Kutató Int Közl, 6 96), pp English Russian summary)

17 SOME NEW ASPECTS OF THE COUPON COLLECTOR S PROBLEM 7 [3] I Gesselan G Viennot, Binomial eterminants, paths, an hook length formulae, Av Math, ), pp 3 32 [4] D J Newman an L Shepp, The ouble ixie cup problem, Amer Math Monthly, 67 96), pp 58 6 [5] H S Wilf, generatingfunctionology, 2n e, Acaemic Press, Boston, 994 [6] R Wong, Asymptotic Approximations of Integrals, Acaemic Press, Boston, 989