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
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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