Matematicko-fyzikálny časopis

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1 Matematicko-fyzikálny časopis Josef Kaucký ote on the Banach's Match-Box Problem Matematicko-fyzikálny časopis, Vol. 2 (962), o., Persistent URL: Terms of use: Mathematical Institute of the Slovak Academy of Sciences, 962 Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access to digitized documents strictly for personal use. Each copy of any part of this document must contain these Terms of use. This paper has been digitized, optimized for electronic delivery and stamped with digital signature within the project DML-CZ: The Czech Digital Mathematics Library

2 MATEMATTCKO-FYZIKALY ČASOPIS SAV, 2,, 962 OTE O THE BAACHS MATCH-BOX PROBLEM JOSEF KAUCKY. Brno A certain mathematician* always carries two match boxes, each of which contains initially matches. Every time he wants a match, he selects a box at random. Inevitably a moment occurs when, for the first time, he finds a box empty. At that moment the other box may contain r = matches. We wish now to find the corresponding probabilities. Let u r be the probability of the evenf'when the first box is found empty, the second box contains r matches". This means that out of the first + ( r) matches drawn, are from the first one. Therefore u r is the probability of successes in 2 r trials, each of which has the probability /2, so that Feller calculates the mean value f 2 Y / n2-0,,2 V. (I) I '», = /- (2) While he is not able** to calculate this expectation in a direct way, he proceeds in an indirect way, namely using the fact that all the probabilities u r add to unity I«r =. (3) He says, this relation is not easily verified. The present paper contains two proofs of the last equation. The first proof (section 2.) uses the method of mathematical induction. The second proof (section 3.) is based on an elementary analytical method. Using the last method wc derive in the section 4. the value of D. * See [] p. 08. ** loc. cit. p

3 With the substitution we sec that the equation (3) 2 - r = k r?o" r ",So( ) 2 -~ may be put into the form ' + k\ - ( ' k = 0 \ 2 k = 2. (4) It is evident that this relation holds for = and we will show that if (4) holds for, then this equation holds also for +. But if we replace by /V + in (4), we have + / i /-V+!. +k\_l _?J + i(' )t = o \ k k = 0 \ or S +l = 2S < (5) where _ / + k\ SH-ÍO k = 0 \ ow making use of the intentity k f 2 T - + k\ ( + k\ / + + k " - ) + ( * ) - ( * we obtain "J; / + + k\ 2 k and so it is only necessary to find the values of both the sums on the right side of this equation. But after some small modifications we have: first "+ ( + k\ J * ( + I + k\ J.Mt-lJV 2 4 V A: J 2 k _j_r _/2 + 2\ -!_ - 2^«+i y + l j 2 + l j- _ L ±_- 2 l" 2^ + A x ~ T Af+ - T"ivTT^ )"2 TTr ~ = Í2 + J 2 +Í ' _ 29

4 the second sum gives»± l ( + k\ I _ e (I + M * )t - S» + { + \J 2 " + > so that by substituing both the values in (6) we get the required relation (5). The assertion (4) is thus established. We introduce for abbreviation the notation (n) k = n(n - )... (n - k + ) so that the binomial coefficient ( U ' K ] may be written in the form Л (")*. kj k\ and for the k-th derivative of x n we have ow we go out from the expansion (xt } = («)**" A_= I*- which holds for x \ <. If we differentiate both sides of this equation times, we obtain the relation _Ľ ( x " Ţ } = V ( n + k l[í-x) Лl Лř -iгľk- Äłł (7) The derivative on the left side may be easily computed by use of the Leibniz's theorem. In effect, with respect to 30 _f_y fc ' k] i + *y (i-x/ +i

5 we have succesivly J_ \! V - * i(>'w" * Í _L í( y vу»u-vу. l^!v! Є 0 \vt' vnv " ''' (\-x) + - iö f 0 \v/ ( - x ) w+ -' The value of this derivative for x = /2 is / x" V*Ч s -.y(n = 2 + l U\l-x) =4 ІДv Two cases are important for the following: namely n = and n = 2 +. Using the identity we have for n = i(v)= 2 " ^(T^rL-iO For // = 2 + follows./,«-v., ' (» +.) \\\-xj J x=, 2" uto. and there remains to calculate the sum on the right side of this equation. Replacing by 2 + in the identity (8) we have = X '2 + -Л( 2 "ГYT( 2 Г)= V u = 0 \ = " (2+\\ 2+ / v = /V + \ * / ( 2 + \ M v ) + v Jr +l \2+l-v 4І 2 Г)4ҐГ)- =4ťГ),8> 3

6 so that Substituing this value in (0) we obtain ^2+ X \(.V) l\l-xf U Л = 2*. ow turning back to the equation (7) we have the following relations y ( + k\ J i + M ) 2 k k^ + l\ ) 2 k = y (2+\ +k\_j = *f V ) 2" + + * and with these results we obtain finally the required relation (4). amely ( + k\ y M ) A J 2 k _» //V + fc\ _ y ( + k\ J = * = 2*. It is necessary to remark that we can obtain the equation (9) more easily as follows. As -;)-<-< + r' we have for x < the expansion (l-.r)-< w+, =J o ("V )(-^ = A(" + ^ and from this for x = /2 we find (9). 32

7 By the same method as we have used in the preceding section to proof the relation (4), we can calculate the mean value (2). There is f 2 - v\ Џ= r lг'= Â ľ { ) ДГ - «"-<-*) )2-r 2 / ^ + k * ( + k\ _ " ( + k\ k_ ~'l Іo\ ) 2 k 2 Іi\ ) 2 k or with respect to (3) Í [ + k\ k M» = -^l\ ) 2 k (U) so that it remains to evaluate the second term of this expression. For this purpose we derivate once more the equation (7). Multiplying the resulting equation by x we obtain n \(+\) \\\ - x - o (^/C )(^-^ + ^)^+fc - (2) The derivative on the left side may be calculated again by use of the Leibniz's theorem. We obtain x t x V +)! V - x h-ir.: M w* Jt_ / y(+^ ' ^<+ -'' ) I A -0 \ v r"' v i - x x + 'Í + = ж i : («)..(" +-. )! /V!.,fn\ = 0 V r '" V ' í\ ( - лf y\ + 2 ' =<->:?; o +l /n\ x" +l ~ v o VV ( - *) + 2-v In the following we shall need the values of this derivative for x = /2 and n = = lv, n = 2 +. :) Matematioko-lyzikálny čas. XII. 33

8 For n = and x = /2 we get x l! \ - x, \(+\y x / 2(lV+l)Z o^) = (lv+l)2 л ' +. For n = 2 + and x = /2 there is.2+ \( + ) X X! \ - x + + Ł (2 +.. V 2* Å JV+J 22* +,2+Ì + (Л. + )2 + «w 2 «; ow if we return to the equation (2) we have first as result the relation!,rr)-h-^: Further it follows (i: ( + l)2" + " + Ч - " + l 2 _ " /2 + + fc\ + l_+_fc _ * ~~ --- \ /V / ~>+l+k ~ --- /v fc = o \ / 2 = /v + On base of these results we can now calculate /n. There is /( = ІV - Y ( + k )J< _ ү 2",M H 2 T *-ki (Л г + )2 + -(Л' + )2 + І /2 + = + k\ k + k\ k k ) 2 + / /2 Let us only re marque that the equation (3) can be obtained in a shorter way from the last equation of the section 3. Derivative and multiplication by x gives '~~2/V.Y(-+l)(l-A-)-, ' V+,) ' = t ( + k )kx k jfc = \ therefrom we obtain for x = /2 the equation (3). 34 K

9 REFERECES [II Feller W., An introduction to ProbabШty Theory and its Applications I, ew York 950. Receivcd Oktober 5, 96. Kabinet matematiky Slovenskej akadémie vied v Bratislave O JISTÉ BAACHOVË ÚLOZE Z POČTU PRAVDË PO DO BOSTI Josef Kaucký Výtah V podstatč jde o tuto úlohu ([], str. 08, 76). Máme dvě krabice, z nichž každá obsahuje IV predmetú. Predměty v obou krabicích jsou stejné. Potrebujeme-li predmět, zvolíme nejprve úpln libovolnč jednu z obou krabic a z ní teprve předmět vyjmeme. Označme u r pravdèpodobnost tohoto jevu jedna krabice je prázdná a druhá obsahuje r předmétù", přičemž r 0,, 2,..., IV. Tento případ po určitém počtu tahů" jistě nastane. Snadno se zjistí, že je,w Ñ Г )-ф- r -- ' = 0,,2,...,*. () Fellcr počítá strední hodnotu Џ = Z ru r (2) Г = a protože, jak sám říká, není s to počítat // přímo, postupuje nepřímo užívaje té okolnosti, že součcí všech pravděpodobností u r dává jedničku I«,. = i. (3) r = 0 O této rovnici Feller ríká, že ji nelze snadno dokázat. Pľedcházející práce obsahuje dva dükazy vztahu (3). A sice ve 2. odstavci je podán dükaz úplnou indukcí, ve 3. odstavci je tato rovnice dokázána jednoduchou analytickou metodou, pri níž se vycházi ln I k^ / vytvorující funкce pro binomicкé кoefìcienty k V posledním odstavci je této metody užito k výpočtu střední hodnoty//. 35