Table Of Content

MAXIMAL PERCENTAGES IN PO´LYA’S URN ERNSTSCHULTE-GEERSAND WOLFGANGSTADJE 4 1 Abstract. We show that the supremum of the successive percentages 0 2 ofredballsinPo´lya’surnmodelisalmostsurelyrational,givethesetof values that are taken with positive probability and derive several exact t c distributional results for theall-time maximal percentage. O 2010 Mathematics Subject Classification: 60C05. 9 ] R 1. Introduction P . The classical urn of Eggenberger-Pólya [3] contains initially r 1 red h ≥ t balls and b 1 black balls. In the course of the drawing process in each a ≥ drawone ball is taken from theurn(whereeach ballin the urnhasthe same m chance of being drawn), and this ball and another d 1 balls of the same [ ≥ color are put into the urn. The theory of Po´lya urn schemes is expounded 4 in [8], where also an extensive bibliography can be found. v Let R , n 1, denote the number of red balls in the urn after the nth 5 n ≥ 0 draw and set R0 = r. The ratio Zn = Rn/(nd+r+b) gives the percentage 8 of red balls among all balls in the urn at “time” n. Let 1 . S = supZ 1 r,b n 0 n≥0 4 be the supremum of all successive percentages of red balls during the entire 1 drawing process. : v InthispaperweshowthatS isattainedalmostsurelyandthatP(S is rational)= r,b r,b i X 1, thereby settling in the affirmative a conjecture of Knuth (posed in the an- r swer to problem 88 of pre-fascicle 5a to volume 4B of The Art of Computer a Programming [7]). The support of S (i.e., the set of values that are taken r,b with positive probability) is also given. Moreover, we derive exact formulas for certain values of the distribution function of S , mainly in the case r,b d= 1. For general d 1 we show that ≥ ∞ a+1 P(S > (t 1)/t) = r,b − n(t 1)+a+1 nX=0 − nt+a B(n(t 1)+a+1+(r/d),n+(b/d)) − , (cid:18) n (cid:19) B(r/d,b/d) Key words and phrases. Po´lya’surn;binomialrandomwalk;all-timemaximalpercent- age; exact distribution. 1 2 ERNSTSCHULTE-GEERSANDWOLFGANGSTADJE where (r +b)/r t N, a = (b(t 1) r)/d and B(α,β) is the Beta ≤ ∈ ⌊ − − ⌋ function. For d = 1 we have 1 ln2 for t = 2 − P(S 1/t) =  1 t−2 1,t−1 ≤  (t 1) 1 1 (t 2) for t > 2 − (cid:18) − t(cid:19) − − h i.  and 2t 3 t 2 P(S = (t 1)/t) = − H(1 (1/t)) − H(1 (2/t)) 1,1 − t − − t − t 2 − − t 1 − P(S (t 1)/t) = (1 (1/t))H(1 (1/t)), 1,1 ≤ − − − where H(x) = Ψ(x+1)+γ. Here Ψ = Γ′/Γ denotes the Digamma function and γ is Euler’s constant. Another exact result in form of an infinite series is (again for d =1) ∞ (nt)! ((n+1)(t 1))! P(S > (t 1)/t) = (t 1) − . t−1,1 − − (nt n+1)! ((n+1)t)! nX=0 − In the course of our derivations we obtain a remarkable equidistribution propertyofthesupremumM(p) = supn∈Nn−1(B1+ +Bn)ofthebinomial ··· random walk generated by i.i.d. Bernoulli variables B (i.e., P(B = 1) = i i 1 P(B = 0) = p (0,1)). For t N with pt 1 it is shown that i − ∈ ∈ ≤ P(M(p) (1/(k+1),1/k]) = p/(1 p) for every k 1,...,t 1 ∈ − ∈ { − } P(M(p) (p,1/t]) = (1 tp)/(1 p). ∈ − − 2. Used facts and related work In Po´lya’s urn scheme let X = 1 if a red ball is drawn in the nth draw n and X = 0 otherwise, and let X = (X ,X ,...) be the full sequence of n 1 2 these red ball indicators. The following facts about Po´lya’s urn are well-known [2, 4]. (1) Z converges almost surely to a random variable Z, which has a n Beta(r/d,b/d) distribution on (0,1). (Here and in the sequel Beta(α,β) denotes the Beta distribution with parameters α,β > 0). (2) Conditionally on Z = z , X ,X ,... are independent, 0,1 -valued 1 2 { } random variables with P(X = 1) = z. i Thusthepercentage Z of redballs eventually tendstoarandomvalueZ, n and if Z “exits” to Z = z the red ball indicators behave like independent n 0 1variableswith“successprobability”z. Onecanthereforetrytoaverage − known results for the Bernoulli sequence to obtain results for Po´lya’s urn. MAXIMAL PERCENTAGES IN PO´LYA’S URN 3 Forasequenceof 1,1 -valuedBernoullivariablesU ,U ,...withP(U = 1 2 i {− } 1) = 1 P(U = 1) = p (0,1) the supremum i − − ∈ M = supn−1(U +...+U ) 1 n n≥1 of the averages of their partial sums was considered in [9]. It was shown that this supremum is attained and that (1) P(M (2p 1,1] Q)= 1 ∈ − ∩ (2) P(M =x) > 0 for each x (2p 1,1] Q. ∈ − ∩ Moreover, explicit formulas for the distribution of M were derived. Note that M is one of the rare examples of a naturally occurring random variable taking every rational number in some interval with positive probability. In the sequel we combine these results to study the all-time maximal percentage of red balls in Po´lya’s urn. 3. Existence and Possible Values of Maxima LetX bethePo´lyaredballindicatorsequence. Wehavealreadyremarked that one may view this sequence as a randomized Bernoulli sequence. To make this note self-contained we include a short proof. Theorem 3.1. Let Z and Y = (Y ,Y ,...) be defined on some probability 1 2 space such that: (a) Z is Beta(r/d,b/d)-distributed. (b) Conditionally on Z = z , (Y ,Y ,...) is an i.i.d. sequence of 0,1 - 1 2 { } valued random variables with P(Y = 1) = 1 P(Y = 0) = z. i i − d Then Y = X. Proof. Let (y ,...,y ) 0,1 n, s = y +...+y . Then 1 n n 1 n ∈ { } P(Y = y ,...,Y = y Z = z) = zsn(1 z)n−sn. 1 1 n n | − Therefore, P(Y = y ,...,Y = y ) 1 1 n n 1 1 = z(r/d)−1+sn(1 z)n+(b/d)−1−sndz B(r/d, b/d) Z − 0 B((r/d)+s , n+(b/d) s ) n n = − . B(r/d, b/d) Hence a straightforward calculation (using the elementary properties of the Beta function) yields P(Y = 1) = r/(r+b) and 1 r+s d P(Y = 1 Y = y ,...,Y = y ) = n , n+1 1 1 n n | nd+r+b so that the finite-dimensional distributions of Y coincide with those of X. Thus, Y =d X. (cid:3) Since the red ball indicators in Po´lya’s urn scheme are a randomized Bernoulli sequence they have similar properties. 4 ERNSTSCHULTE-GEERSANDWOLFGANGSTADJE Proposition 3.2. In the situation of Theorem 3.1 let S = n Y and n i=1 i Zn = (r+dSn)/(nd+r+b). Thenlimn→∞Zn = Z a.s. andlimsuPpn→∞(Sn− nZ)= a.s. ∞ Proof. Let z (0,1). We only need to prove the assertion conditionally on ∈ Z = z. SinceY is, conditionally on Z = z, asequenceofi.i.d. 0 1variables − with probability z for the value 1, we have P(Z z,limsup(S nz)= Z = z) =1. n n −→ − ∞ | n→∞ Note that P(Z z Z = z) = 1 follows from the strong law of large n −→ | numbers, and P(limsup (S nz)= Z = z)= 1 follows (e.g.) from the law of the iterated long→a∞rithmn−. ∞ | (cid:3) Let M = sup Z . We now use a variant of the proof of the corre- r,b n≥1 n sponding property of M in [9] to show that this supremum is almost surely attained. Proposition 3.3. (a) The supremum in the definition of M is almost r,b surely attained. (b) P(M is rational) = 1. r,b Proof. Again we condition on Z = z. Since limsup (S nz) > b+1 n→∞ n − a.s. we have that a.s. infinitely often (r+dS ) (nd+r+b)z > d(b+1)+ n − r(1 z) bz 1+(r+b)(1 z) > 0, i.e., Z > z for infinitely many n a.s., n − − ≥ − and since Z z a.s. the supremum is a.s. attained. It is then obviously n rational. −→ (cid:3) Define r+i d j Q = max n N,i ,...,i N 0 , 1 n n1≤j≤n r+b+jd (cid:12) ∈ ∈ ∪{ } (cid:12) (cid:12) i j for all j 1,...,n ,0 i i ... i . j 1 2 n ≤ ∈ { } ≤ ≤ ≤ ≤ o Clearly, Q is the set of all possible maximal percentages for finite sequences ofdrawsfromtheurn. Itfollows fromProposition3.3thatP(M Q)= 1. r,b ∈ Now we show Proposition 3.4. P(M = q)> 0 for all q Q. r,b ∈ Proof. Fix an element q of Q and corresponding m N and nonnegative ∈ integers i i ... i such that i j for all j 1,...,m and 1 2 m j ≤ ≤ ≤ ≤ ∈ { } r+i d j q = max . 1≤j≤mr+b+jd Let the maximum be attained at n 1,...,m . Then we have ∈{ } r+i d r+i d n j q = = max . r+b+nd 1≤j≤nr+b+jd MAXIMAL PERCENTAGES IN PO´LYA’S URN 5 Let E be the event that R = i d for j = 1,...,n and M = q (thus on j j r,b this event the percentage after the nth draw is the largest among the first n ones). It remains to show that P(E) > 0. Write P(E Z = z) in the form | P(E Z = z) = P R = i d for j = 1,...,n and j j | (cid:16) r+i d+(Y′+...+Y′)d r+i d sup n 1 k n Z = z k∈N r+b+nd+kd ≤ r+b+nd (cid:12) (cid:17) (cid:12) = P(R = i d for j = 1,...,n Z = z) (cid:12) j j | r+i d+(Y′+...+Y′)d r+i d P sup n 1 k n , × (cid:16)k∈N r+b+nd+kd ≤ r+b+nd(cid:17) where Y′,Y′,... is an i.i.d. sequence with P(Y′ = 1) = 1 P(Y′ = 0) = z 1 2 i − i which is independent of Y and Z. The first factor on the righthand side is obviously positive. Next, the inequality r+i d+(Y′+...+Y′)d r+i d sup n 1 k n k∈N r+b+nd+kd ≤ r+b+nd holds if and only if Y′+...+Y′ r+i d sup 1 k n . k∈N k ≤ r+b+nd By [9], this latter inequality occurs with positive probability if P(Y′ = 1) < (r+i d)/(r+b+nd). 1 n It follows that P(E Z = z) > 0 for z (0,(r+i d)/(r+b+nd)). Hence, n | ∈ min[1,(r+ind)/(r+b+nd)] P(E) P(E Z = z)β (z) dz > 0, r/d,b/d ≥ Z | 0 where β (z) is the density of Beta(r/b,b/d). The proof is complete. (cid:3) r/b,b/d Remark 3.5. Proposition 3.4 can also be formulated in the following form. Let supp(U) denote the support of the random variable U. Let q (i,n) = r,b (r+id)/(r+b+nd) for i,n N 0 . Then ∈ ∪{ } supp(S )= q (i,n) 0 i n,q (i,n) r/(r+b) (3.1) r,b r,b r,b { | ≤ ≤ ≥ } and supp(M ) = supp(S ) supp(S ). (3.2) r,b r,b+d r+d,b ∪ (3.2) is obvious, and (3.1) is proved in the same way as Proposition 3.4 (considersequencesofdrawsbeginningwithX = ... = X = 0,X = 1 n−i n−i+1 ...,X = 1). n In particular, for d =1 we obtain supp(S ) = [r/(r+b),1) Q. r,b ∩ 6 ERNSTSCHULTE-GEERSANDWOLFGANGSTADJE 4. Some Closed-Form Results on Maximal Percentages for the Binomial Random Walk In this section we only consider the case d = 1. We start with two exact results for the binomial random walk, which may be of independent interest. Although they are consequences of the t-ballot theorems they apparently have not been formulated in this form before. Inthesequelfix0 < p < 1, q = 1 pandletY ,Y ,...bei.i.d. 0 1random 1 2 − − variableswithP(Y = 1) = p. IntroducetheirpartialsumsS = Y + +Y i n 1 n ··· and define, for r,b N 0 , ∈ ∪{ } r+S n M (p)= sup and M(p)= M (p) r,b 0,0 r+b+n n≥1 r+S n L (p) = inf and L(p) = L (p). r,b 0,0 n≥1r+b+n Since P(M (p) x) = P(L (q) 1 x) for 0 x 1 the distribution r,b b,r ≤ ≥ − ≤ ≤ function of L (q) can be obtained from that of M (p). b,r r,b Clearly M (p) x if and only if S nx bx (1 x)r for all n N. r,b n ≤ − ≤ − − ∈ Thusifx Q (0,1),sayx = s/tforpositiveintegerssandtwithgcd(s,t) = ∈ ∩ 1, the value of P(M (p) s/t) depends on s,t only through the integer r,b ≤ value m = sb (t s)r. − − For m Z we define a = P(tS ns m for all n N).; + m n ∈ − ≤ ∈ Remark 4.1. For two integers s,t satisfying (s,t) = 1, p < s/t < 1 the following can be shown (along the lines of [9]): (a) The sequence a ,a ,a ,... has a rational generating function; it is 0 1 2 given by q( s−1a zj)/(pzt zs +q). This function is regular for j=0 j − z < 1. P | | (b) The denominator f(z)= pzt zs+q has only simple roots: exactly − s 1 roots z ,...,z inside the unit disk, z = 1, and exactly t s 1 s−1 s − − roots z ,...,z outside the unit disk. s+1 t Thus a can be written as a linear combination of 1 and the (m + 1)th m negative powers of the roots of f(z) outside the unit disk. However, explicit expressions are hard to come by. Let t 2 be an integer. In the sequel we give expressions for the proba- ≥ bilities P(M(p) (t 1)/t) and P(M(p) 1/t). ≤ − ≤ For the first case we need the t ary tree function T (z). Its power series t − is given by ∞ nt zn T (z) = . t (cid:18)n(cid:19)n(t 1)+1 Xn=0 − This power series converges for z < 1(1 1)t−1 and for z = 1(1 1)t−1 | | t − t t − t and represents there the unique solution of the implicit equation T (z) = t 1 + zT (z)t. (Thus 1/T is the inverse function of y yt−1(1 y) for t t 7→ − MAXIMAL PERCENTAGES IN PO´LYA’S URN 7 1 y < 1/t.) We define | − | R (p)= pT (qpt−1). t t We recall the following fact from path counting combinatorics (“t-ballot numbers”). (See e.g. [6], Problem 26 of Section 7.2.1.6., for an equivalent statement.) Lemma4.2. Leta N 0 . Thenumberofpaths, withstepsin (0,1),(1,0) , ∈ ∪{ } { } from (0,0) to (n,n(t 1) + a) whose points lie on or below the line y = − a+x(t 1) is the coefficient [zn]T (z)a+1. We have t − a+1 nt+a [zn]T (z)a+1 = . t n(t 1)+a+1(cid:18) n (cid:19) − Proposition 4.3. If m = b(t 1) r 0, we have − − ≥ P(M (p) (t 1)/t) = 1 R (p)m+1 (4.1) r,b t ≤ − − (4.2) p R (p)+qR (p)t−1 for m = 0 t t − P(M (p) = (t 1)/t) =  (4.3) r,b −  1 R (p) R (p)m for m 1 t t − ≥ Proof. (1) Consider P(M (p) >((cid:0)t 1)/t).(cid:1)Start a lattice path R = r,b n − r + S ,B = b + n S in (r,b) by stepping (1,0) if Y = 1 , (0,1) if n n n i − Y = 0. Since M (p) > (t 1)/t, there must be a smallest j b such that i r,b − ≥ R = j(t 1)+1,B = j and (t 1)B R for i ℓ = jt r b+1. ℓ ℓ i i − − ≥ ≤ − − Equivalently, j b is the first position where the lattice path (n S ,S ) n n − − starting at (0,0) steps at time ℓ for the first time above the line y = (t − 1)x+m. Call this event A . Clearly the last step is Y = 1, appended to a j ℓ path from (0,0) to (j b,j(t 1) r)= (j b,(j b)(t 1)+(t 1)b r) − − − − − − − − of the type considered in the lemma above. Since each path from (0,0) to (j b,j(t 1) r+1) has the same probability pj(t−1)−r+1qj−b, we get − − − P(A ) = [zj−b]T (z)m+1pj(t−1)−r+1qj−b j t = pm+1[zj−b]T (z)m+1p(j−b)(t−1)qj−b. t Since M (p)> (t 1)/t is the disjoint union of the A , the first assertion r,b j { − } follows. (2) For k 0 let b = P(tS < n(t 1)+k) for all n N). Conditioning k n ≥ − ∈ with respect to Y then shows that b = qa and b = a for j 1. 1 0 t−2 j j−1 ≥ Since P(M (p) = (t 1)/t) = a b the second assertion follows from r,b m m the result in (1). − − (cid:3) Remarks 4.4. (a)ThepathcountingargumentaboveisduetoKnuth,who used it to determine P(S > (t 1)/t) in Po´lya’s urn (see next section). 1,1 − Alternatively the results could be obtained using the theorem from [9] given below, but in a more laborious way. (b) In the derivation no restrictions on p (other than 0 < p < 1) were imposed, the formulas above are thus also valid for p (t 1)/t. In fact, by ≥ − 8 ERNSTSCHULTE-GEERSANDWOLFGANGSTADJE the remark before Lemma 4.2 we have R (p) = 1 for p (t 1)/t and the t ≥ − corresponding probabilities in Proposition 4.3 evaluate to 0 (as they must do in view of the strong law of large numbers). (c) R (p) p for t . More precisely, lim q−1p−t[R (p) p]= 1. t t→∞ t −→ −→ ∞ − Examples 4.5. (a) Let t = 2, p < 1/2. Then R (p) = min(1,p/q) and we 2 recover the well-known facts that for m = 0 P(M(p) > 1/2) = p/q and P(M(p) = 1/2) = (q p)p/q. − and that for m 1 ≥ P(M (p) 1/2) = (p/q)m and P(M (p) = 1/2) = (p/q)m(q p)/q. r,b r,b ≥ − (b) Let p = 1/2. Then R (p) = 1 and 2 R (p) 0.618034,R (p) 0.543689,R (p) 0.518790 3 4 5 ≈ ≈ ≈ R (p) 0.50866,R (p) 0.504138. 6 7 ≈ ≈ For example we have S 6 S 2 n n P sup > 0.504138 and P sup > 0.381937 n 7 ≈ n+1 3 ≈ (cid:16)n≥1 (cid:17) (cid:16)n≥1 (cid:17) Let us now turn to the second probability P(M(p) 1/t). We use the ≤ following result of [9]. Theorem 4.6. Let s,r Z 0 , s > 0, r < s. The polynomial ∈ \{ } | | pz2s zs+r +q (4.4) − has exactlys r simple roots z ,...,z outside the unitdisk. If p < r+s, − s+r 2s−1 2s we have 2s−1 r+s P(M(p) ) = (1 z−1) (4.5) ≤ 2s 1 − i − i=Yr+s 2s−1 2s−1 r+s P(M(p) = ) = (1 z−1)+p (1 z ). (4.6) 2s 1 − i − i − i=Yr+s i=Yr+s Remarks 4.7. (a) In [9] this theorem is proved under the additional as- sumption (r,s) = 1, but it remains valid if (r,s) = k > 1. To see this let u = zk. As a function of u the polynomial (4.4) has (s r)/k simple − roots u outside the unit disk, say u ,...,u , by the theorem above. If i 1 (s−r)/k η is a primitive kth root of unity, then as a function of z the polynomial (4.4) has exactly the s r simple roots z = ηj u 1/k,j = 0,...k 1 and i,j i − | | − i= 1,...,(s r)/k,outsidetheunitdisk,andsince k−1(1 z−1) = (1 u−1) − j=0 − i,j − i and k−1(1 z ) = (1 u ) the formulas above rQemain valid. j=0 − i,j − i (b) IQn [9] also formulas for the corresponding a in terms of the roots of m (4.4) were provided, but we do not use them here. MAXIMAL PERCENTAGES IN PO´LYA’S URN 9 Proposition 4.8. Let p < 1/t. If m = b (t 1)r 0,...,t 1 , we have − − ∈ { − } P(M (p) 1/t) = (1 tp)q−(m+1) (4.7) r,b ≤ − P(M (p) = 1/t) = (1 tp)q−(m+1)p. (4.8) r,b − For m t the a can be computed by the recursion m ≥ qa = a pa . (4.9) k k−1 k−t − For p 1/t the probabilities in (4.7) and (4.8) are 0. ≥ Proof. (1) Let m = 0 and set s = t, r = (t 2) in Theorem 4.6. The − − polynomial pz2t z2+q has 2t 2 roots z ,...,z outside the unit disk, 2 2t−1 − − anda = 2t−1(1 1). Equivalently, since +z , z are roots, we have a = 0 i=2 −zi i − i 0 t−1(1 Q1)wherey ,...,y aretheroots of pyt−1 y+q outsidetheunit i=1 −yi 1 t−1 − dQisk. Let y = 1/(1 u). Then we can write a = t−1u where u ,...,u − 0 i=1 i 1 t−1 are the non-zero roots of the polynomial g(u) =Qp (1 u)t−1 +q(1 u)t. − − − Thus qa = g′(0) = 1 pt. Furthermore, since 2t−2z = q/p we find 0 − − i=2 i − from the second formula P(M(p) = 1/t) = a0 qaQ0 = pa0. − (2) By conditioning on Y we find that a = qa for k = 0,...,t 2, and 1 k k+1 a = pa +qa . The rest is straightforward. − (cid:3) k k−t+1 k+1 Remarks 4.9. (a) In the case m = 0 and p 1/t, we find a curious ≤ equidistribution property: the maximum M(p) lies in any of the intervals (1/(k+1),1/k] (for 1 k t 1) with the same probability p/q (and it lies ≤ ≤ − in the interval (p,1/t] with probability 1 (t 1)p/q). − − (b) The resultfor a can also beshown to follow from a path counting result 0 (Barbier’s theorem): the number of paths with step set (0,1),(1,0) from { } (0,0) to (k,n) with n > tk that never touch the line y = tx except at (0,0) is equal to (n tk) n+k /(n+k). − n (cid:0) (cid:1) 5. Distributional results for Po´lya’s urn Returning to Po´lya’s urn, recall that S = sup Z = max r , M r,b n≥0 n {r+b r,b} and let I = inf Z = min r , L . We now give some formulas for r,b n≥0 n {r+b r,b} the distribution function of S and I for special values. r,b r,b We introduce the generalized harmonic number function 1 1 H(x) = = Ψ(x+1)+γ, x R 1, 2, 3,... n − n+x ∈ \{− − − } nX≥1(cid:16) (cid:17) where Ψ is the Digamma function (the logarithmic derivative of the Γ- function) and γ is Euler’s constant. For positive integers p,q with p < q Gauss has shown (see e.g. Problem 19 of Section 1.2.9 in [5]) that q π p 2pn n H(p/q) = cot π ln(2q)+2 cos π ln sin π . p − 2 (cid:18)q (cid:19)− (cid:18) q (cid:19) (cid:18)q (cid:19) X 1≤n<q/2 Thus H(p/q) can be expressed in terms of finitely many elementary func- tions. 10 ERNSTSCHULTE-GEERSANDWOLFGANGSTADJE Let us first consider the distribution function of S at (t 1)/t. The r,b − obvious route to results is to condition on Z, use the formulas for the binomial random walk and integrate the power series of Tm term-by term, using t the Beta integrals. The general solution is given in the following proposition. Note that in the case d > 1 we have to determine the probabilities a = P(tdS nds m for all n N). Since clearly a = a , (for m,d n m,d ⌊m/d⌋ − ≤ ∈ r R, r denotes the largest integer not exceeding r) this can be reduced ∈ ⌊ ⌋ to the computation of the a , i.e., the case d = 1, which we dealt with in m Section 4. Proposition 5.1. Let m = b(t 1) r 0 and a = m/d . Then − − ≥ ⌊ ⌋ ∞ a+1 P(S > (t 1)/t) = r,b − n(t 1)+a+1 nX=0 − nt+a B(n(t 1)+a+1+(r/d),n+(b/d)) − . (cid:18) n (cid:19) B(r/d,b/d) Ford= 1onegetsseriesofrationalfunctionsofn,whichcan(inprinciple) be evaluated in closed form in terms of the Digamma function Ψ. We only give two examples for the results of these calculations: Proposition 5.2. For d= 1 and 2 t N we have ≤ ∈ 2t 3 t 2 P(S = (t 1)/t) = − H(1 (1/t)) − H(1 (2/t)) 1,1 − t − − t − t 2 − − t 1 − P(S (t 1)/t) = (1 (1/t))H(1 (1/t)). 1,1 ≤ − − − Essentially this has already been shown by Knuth ([7], Problem 88). The situation for b = 1,r = t 1 is of special interest because in this case − p (t) = P(I (t 1)/t) (the probability that the urn content process + t−1,1 ≥ − staysabovetheliney = (t 1)x)aswellasp (t) =P(S (t 1)/t)(the − t−1,1 − ≤ − probability that it stays below the line y = (t 1)x) can both be considered. − Let q (t) =1 p (t) = P(S > (t 1)/t). One finds that − − t−1,1 − − Proposition 5.3. For d= 1, ∞ (nt)! ((n+1)(t 1))! q (t) = (t 1) − . − − (nt n+1)! ((n+1)t)! nX=0 − Special values are q (2) = ln2 0.6931 − • 4 ≈ q (3) = π√3 0.8061 − • 27 ≈ 9 27 q (4) = ln2+ π 0.8576 − • 32 128 ≈ q (5) 0.8874,q (6) 0.9068,...,q (20) 0.9726. − − − • ≈ ≈ ≈