How Often Should You Beat Your Kids?

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This note is a follow-up to the note "How to Beat Your Kids at Their Own Game, ... play what is obviously the optimal strategy-guessing randomly (or, if you prefer,.
How Often Should You Beat Your Kids? Author(s): Don Zagier Source: Mathematics Magazine, Vol. 63, No. 2 (Apr., 1990), pp. 89-92 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/2691064 . Accessed: 21/05/2011 10:34 Your use of the JSTOR archive indicates your acceptance of JSTOR's Terms and Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp. JSTOR's Terms and Conditions of Use provides, in part, that unless you have obtained prior permission, you may not download an entire issue of a journal or multiple copies of articles, and you may use content in the JSTOR archive only for your personal, non-commercial use. Please contact the publisher regarding any further use of this work. Publisher contact information may be obtained at . http://www.jstor.org/action/showPublisher?publisherCode=maa. . Each copy of any part of a JSTOR transmission must contain the same copyright notice that appears on the screen or printed page of such transmission. JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected].

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VOL.

89

63, NO. 2, APRIL 1990

How OftenShould You Beat Your Kids? DON ZAGIER

University of Maryland College Park, MD 20742

A resultis provedwhichshows,roughlyspeaking,thatone shouldbeat one's kids everyday exceptSunday. This noteis a follow-up to thenote"How to Beat YourKids at TheirOwn Game," by K. Levasseur[1], in whichthe authorproposesthe followinggame to be played children:Starting witha deck consisting againstone's two-year-old of n red cardsand n black cards(in typicalapplications,n = 26), thecardsare turnedup one at a time, each playerat each stagepredicting the colorof the card whichis about to appear. The kid is supposedto guess"Red" or "Black" randomlywithequal probability (this solves the problemof constructing a perfectrandomnumbergenerator),while you play what is obviouslythe optimalstrategy-guessingrandomly(or, if you prefer, alwayssaying"Black") wheneverequal numbersofcardsofbothcolorsremainin the in the majority.Levasseur the colorwhichis currently deck and otherwisepredicting analyzesthegameand showsthaton theaverageyouwillhave a scoreof n + (VFW 1)/2 + O(n- 1/2),whilethekid,of course,willhave an averagescoreof exactlyn. We, however,maintainthatonlythe mostdegenerateparentwould play againsta formoney,and thatour concernmusttherefore two-year-old be, not by how much you can expectto win,but withwhatprobability you will win at all. Our principal result is that this probabilitytends asymptotically to 85.4% (more precisely:to This shows with what unerringinstinctLev1/2 + 1/ F8) as n tendsto infinity. asseur'smotherselectedthe game-the high85% loss rate will instillin the young oftheirparents,whilethe 15% win progenya due respectfortheimmensesuperiority rate will maintaintheirinterestand preventthemfromsuccumbingto feelingsof hopelessnessand frustration. The analysisbeginsas in Levasseur'sarticle:each of the (2n) possibleorderingsof the cards into red and black elementscorrespondsto a path p movingdownwards and leftwardsfroman initialvalue (R, B) = (n, n) to a finalvalue (R, B) = (0,0) of the pair (R, B), where R and B denote the numbersof red and black cards If thispathmeetsthe diagonalR = B a totalof m(p) times, remaining, respectively. wheretheinitialpointat (n, n) is countedbut thefinalpointat (0, 0) is not,thenthe expectedwin of the parentis m(p)/2. Indeed, at each meetingpoint the parent withan expectedscoreof 1/2 and hence an expectedwin overhis guessesrandomly, child of 0; betweeneach pair of meetingpoints,the parentwill consistently guess "Red" or consistently "Black," dependingon whetherp is now below or above the diagonal,and will be rightexactlyone moretimethanhe is wrong,gainingexactly halfa pointoverhis randomly guessingchild.Levasseurshowsthatthe averagevalue of m(p), as p ranges over the set ;n of paths as described above, is exactly 4f/(2n )- 1, leading to the resulton the expectedwin stated above. To solve the we mustanswertwo questions: problemwe have set ourselves, of winning?and -(i) fora givenvalue of m(p), whatis the probability will m(p) takeon a givenvalue m, 1 < m < n? -(ii) withwhat probability We answerthe secondquestionfirst.

90

MATHEMATICS MAGAZINE

Let Nm(ln)denote the numberof paths p E 9)n with m(p) = m. For n = 0 this equals 1 if m = 0 and 0 otherwise, but forpositiven we musthave m > 1 since the initialpointof the pathis countedas a meetingwiththe diagonal.If a path p e 9n meetsthediagonalmorethanonce,i.e., if m(p) > 1, thenthefirstmeetingpointwill ifwe picksucha k, be at somevalue (k, k) of(R, B) with1 < k < n - 1. Conversely, then the numberof paths p e 9Znwith m(p) m and having(k, k) as theirfirst meetingpoint will be equal to the productof N1(n- k) (the numberof ways of descendingfrom(n, n) to (k, k) withoutmeetingthe diagonal on the way) and Nm-(k) (the numberof waysof descendingfrom(k, k) to (0,0) withexactlym - 1 further meetings).Hence

Nmf n)

n-1

Y2,N( ni-k) Nm( k)

(m > 1).

k=1

It followsthat the generatingfunctionX,4'(x)= l ,NmNm(ln)x'is the productof Xl(x) and X,' -I(x), and hence that X,,j(x) = Xl(x)'. This formulaholds also for m = 0 since NO(n)= 0 for all positiven. On the otherhand, the sum of all the functionsX,4'(x) is thegenerating function whose nthcoefficient is the totalnumber ofpathsin ,n,i.e.,( an) Hence 00

1

i

,/~(x)

-

00

E 1X(X)=

rn=O

m=O

1

X.(x)

n=O

(2n)Xn= n

r -4

so Xj( x) = 1 -

Xm(x) = ( - Fl 1-4x )

14~x,

Usingthe well-known Taylorexpansionof thisfunction, we find: Nm(n) = 2mm

( l < m< n) .

r--

the probability Therefore, fora randompath p e gn to have m(p) prob{ m(p)m}

N (n) jmr=N() (n )

m

(1 1

n) 2n)

n

... 2n) ...

=

m is givenby n

)

2n)

For m ofthe orderof VW (the rightorderaccordingto Levasseur'sanalysis),thiswill

Distributionof m forn large

m

O

n

n 3 6UR 2 n 5FG 4 n FIGURE 1

91

VOL. 63, NO. 2, APRIL 1990

equal to (m/2n)e-m be asymptotically we have forthe totalprobability m}=O

1). As a test,when n is large

00

00

L

/4n (cf. FIGURE

prob{m(p)=m}

E

m=O

m e-m2/4n 2n~~~~~~~~~o

00x eX

dx

n

-

00

- ex2/4n

=

1

~~~~~~~0

and forthe expectedvalue of m thevalue 00

00

m m e-m2/4n

m

E prob{m(p)=m}

M=~~~00

n1=0

t2-

44

dx

x-- e-/4n Ix

dt=Vn,,

in accordancewithLevasseur'sresult. We now turnto the firstof the two questionsabove. For the reasons already explained,foran orderingof cards givenby a path p e ;n withm(p) = m, of the to points on p not on the diagonal one will guess 2n - m turnscorresponding of exactlyn - m times,whilethe probability exactlyn timesand incorrectly correctly to pointson the diagonalis at one of the m turnscorresponding guessingcorrectly 50% each time.Hence one's totalnumberof correctguesseswill be describedby a bell-shapedcurvecenteredaroundthe expectedvalue n + am and witha widthof

(cf. FIGURE 2). On the otherhand,if or (foralmostall paths p) VW the orderof x4m, n + k times,thenone's chanceofbeatingtherandomlyplaying one guessescorrectly kid is

2rn

2nnk-1 L

22n

and since 2 -2n( equal to 1

1

2?+2r7

k

e-r Y. =O

2

/n

1

2

+

k

(2n)

by Stirling'sformula,this is approximately

1/rn e r/n

X2nX))

- + 2-2n

1

k

feu

2

_1

/ndu

+

lk/#n

-

Probability distribution ofone'sown of number guesses correct

Probability distribution ofkid's of number correct guesses O(r4m) n-2V2A n-VC

n

tn+Vn (distributed according to Fig. 1)

FIGURE2

n+2Vn

eu

2

du.

92

MATHEMATICS MAGAZINEE

of winningwhen the probability Since k/ xn is almostalwaysverynearto 'm/ x/W,

m(p) = m is verynearly equalto

1

1

+

?

thatm(p) Multiplying thisby theprobability ofwinning probability

P

Y

g

?

du. =

m as computedabove,we findfinally

mE-O~0 x2n

2+ | +2

2xe

x( X

/4

n(

/4(- xe- x

2 n/l

1

00 1 +e-uI I

2

2/

- +

2(fOO2

I O (2u)d le-U2(2e-u

1

1

2+

2 /'

f

1

-f

2

2

2

m/2F

Ie-u

1

m/2r

e

F

-U2

du)dx

Ee2~

l/

fx/2

j

du)

2

e-u du dx 0)

xe-x/4dxl

\

du

)du

as claimed.This is verynearly6/7, so the resultof our paper can be conveniently but neveron Sunday. implementedby beatingone's kidson weekdaysand Saturdays, REFEREN C E 1. KennethM. Levasseur,How to Beat YourKids at TheirOwn Game,thisMAGAZINE61 (1988), 301-305.

A Note on the Five-CircleTheorem JORDAN B. TABOV

Instituteof Mathematics BulgarianAcademy of Sciences P.O. Box 373 1090 Sofia,Bulgaria

In his paper [1] H. Demirstatedand proved THE FIVE-CIRCLE THEOREM. Let P and Q be twopointson theside BC of a triangle ABC in the order B, P, Q, C. If the trianglesABP, APQ, AQC have congruent incircles,thenthetriangles ABQ, APC have congruent incircles.

He also asked fora geometricproofof thistheorem. Here we give such a proofforthefollowing moregeneral