Who Solved the Secretary Problem?

Who Solved the Secretary Problem?
Author(s): Thomas S. Ferguson
Source: Statistical Science, Vol. 4, No. 3 (Aug., 1989), pp. 282-289
Published by: Institute of Mathematical Statistics
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StatisticalScience
1989, Vol. 4, No. 3, 282-296
WhoSolved the SecretaryProblem?
Thomas S. Ferguson
Abstract.In MartinGardner'sMathematical
GamescolumnintheFebruary
1960issueofScientific
American,
thereappeareda simpleproblemthathas
cometobe knowntodayas theSecretary
ortheMarriageProblem.
Problem,
It has sincebeentakenup and developedbymanyeminent
and
probabilists
statisticiansand has been extendedand generalizedin manydifferent
directionsso that now one can say that it constitutesa "field"within
The objectof this articleis partly
mathematics-probability-optimization.
historical(to givea freshviewoftheoriginsoftheproblem,
touching
upon
Cayleyand Kepler),partlyreviewofthefield(listingthesubfields
ofrecent
interest),partlyserious(to answerthe questionposed in the title),and
The contentsof thispaperwerefirstgivenas the
partlyentertainment.
AllenT. Craiglectureat theUniversity
ofIowa,1988.
searchprobKeywordsandphrases:Secretary
problem,
marriage
problem,
lem,relativeranks,stoppingtimes,minimaxrules.
and the
thinkyou will findthe journeyinteresting,
conclusionsurprising.
1. INTRODUCTION
In thelate 1950'sand early1960'sthereappeareda
2. STATEMENTOF THE PROBLEM
simple,partlyrecreational,
problemknownas the
secretaryproblem,or the marriageproblem,or the
The reader'sfirstreactionto thetitlemightwellbe
dowryproblem,
thatmadeitswayaroundthematheto ask, "Whichsecretaryproblem?".Afterall, as I
maticalcommunity.
The problemhas a certainappeal.
havejust implied,thereare manyvariationson the
It is easy to state and has a strikingsolution.It
problemin its simplestform
The secretary
problem.
was immediately
takenup and developedby certain
features.
following
has
the
eminentprobabilistsand statisticians,
amongthem
Lindley(1961),Dynkin(1963),Chow,Moriguti,
Robpositionavailable.
1. Thereis one secretarial
bins and Samuels'(1964),and Gilbertand Mosteller
The numbern ofapplicantsis known.
2.
(1966). Since that time,the problemhas been exsequentiallyin
3. The applicantsare interviewed
tendedand generalizedin manydifferent
directions
randomorder,each orderbeingequallylikely.
so thatnowone can saythatit constitutes
a "field"of
4. It is assumedthatyoucan rankall theapplicants
studywithinmathematics-probability-optimization.
frombestto worstwithoutties.The decisionto
One can see fromthereviewpaperbyFreeman(1983)
acceptor rejectan applicantmustbe basedonly
howextensiveand vast the fieldhas become;moreon the relativeranksof thoseapplicantsinterin
itsexponential
over,thefieldhas continued
growth
viewedso far.
theyearssincethatpaperappeared.
5. An applicantonce rejectedcannotlaterbe reOne mainobjectiveofthepresentarticleis historicalled.
oftheproblemwiththeaim
cal,to reviewthehistory
andwillbe satisfiedwith
6. You areveryparticular
ofdetermining
whowasthefirstto solvethesecretary
nothingbut theverybest.(That is, yourpayoff
problem.The historicalreviewmaytakeus far,but I
is 1 ifyouchoosethebestofthen applicantsand
Ootherwise.)
Thomas S. Ferguson is Professorof Mathematics at
Universityof California,Los Angeles.His mailingaddress is: Department of Mathematics, Universityof
California,405 HilgardAvenue,Los Angeles,California
90024.
282
simplesoluThis basic problemhas a remarkably
can be restricted
tion.First,one showsthatattention
to theclass ofrulesthatforsomeintegerr > 1 rejects
the firstr - 1 applicants,and thenchoosesthe next
applicantwho is best in the relativerankingof the
observedapplicants.For sucha rule,theprobability,
WHO SOLVED THE SECRETARY PROBLEM?
283
/0(r),ofselectingthebestapplicantis 1/nforr = 1,
and,forr > 1,
on mathematical
probfianceproblemat a conference
Univerlemsin logisticsheldat GeorgeWashington
1988).
sityin January1950(personalcommunication,
n=E
I personallyremember
Jthapplicantis best
workingon extensionsof the
andyouselectit
(2.1)
j=nr)J=r
problemin the summerof 1959. In any case, many
peopleknewof the problemby the timeit appeared
in print.
j=r \fl/11 ( n j=rJ
Lindley(1961) seemsto be the firstto solve the
The optimalr is the one thatmaximizesthisproba- problemin a scientificjournal. He extends the
bility.For smallvaluesof n, the optimalr can easily
utilitybased on the rank
problemto an arbitrary
be computed.Of interestare the approximate
ofthe applicantselected,and considersin particular
values
oftheoptimalr forlargen. Ifwe let n tendto infinity the problemof minimizing
the expectedrankof the
and writex as the limitof r/n,thenusingt forj/n
thedifapplicantselected,rank1 beingbest.However,
and dt for1/n,the sumbecomesa Riemannapproxi- ficultproblemof finding
forlargen,
the asymptotic,
mationto an integral,
optimalstrategiesand expectedrankwas leftopen,
Robbins
and finallysolvedneatlybyChow,Moriguti,
and Samuels (1964). Dynkin (1963) considersthe
problemas an applicationof the theoryof Markov
(
i)j(
(2.2)
1)(n)
interpreted,
stoppingtimes,and showsthat,properly
x
(-) dt= -x log(x).
theproblemis monotoneso thatthe one-stagelookaheadruleis optimal.
The value ofx thatmaximizesthisquantityis easily
Then,in 1966,camethebasic paperofGilbertand
foundbysettingthederivative
withrespectto x equal
withelegantderivations
and extensionsin
Mosteller,
to zeroand thensolvingforx. Whenthisis donewe
a numberofimportant
In particular,
they
directions.
findthat
allowr choicesto obtainthe best;theyconsiderthe
problemof obtainingthe best or the nextbest;they
optimal x = lle = .367879 *.. , and
(2.3)
treatthe "full-information"
case, in whichone is al= l/e.
optimalprobability
lowedto observethe actualvalues of the applicants
Thus forlargen, it is approximately
optimalto wait
froma given
(presumedto be chosenindependently
untilabout 37% of the applicantshave been inter- distribution),
forboththebest-choice
problemand for
viewedand thento selectthenextrelatively
bestone.
theminimum-rank
problem;theyanalyzesomegame
The probability
ofsuccessis also about37%.
theoreticversionsof the problem;etc. This paper,
This derivation
mayseema littleloose,butit gives
morethanthe others,foreshadowed
the explosionof
therightanswer.Fora lucidpresentation
oftheseand
and effort
that wouldimpact
ideas, generalizations,
relatedresults,see the basic paper of Gilbertand
in 1972and thatcontinuesstrongly
thisarea starting
Mosteller(1966).
willbe mentioned
today.Someofthesecontributions
in Section5.
3. HISTORICALBACKGROUND
This, briefly,is the officialearly historyof the
to discover
secretary
problem.Sincewe areto attempt
There is some obscurityas to the originsof this
who
we
as
is
first
solved
this
shall,
customary
problem,
problem.It seemsto be generally
agreedamongworkin
in
time,
looking
historical
papers,
proceed
backward
ersin thefieldthatthefirststatement
oftheproblem
in
for
of
the
idea
hidden
the
literature.
forgotten
germ
to appear in printoccurredin the February1960
The firstpossibilityoccursin the workof Arthur
,column of Martin Gardner in ScientificAmerican,
whereit is attributed
to Fox and Marnie.A solution Cayley.
to theproblemis outlinedin theMarch1960issueof
4. CAYLEY'S PROBLEM
Scientific American and
\
-
f
attributed to Moser and
Pounder.In 1963,it appearedas a problemin The
AmericanMathematicalMonthlycontributedby Bis-
singerand Siegel(1963); in 1964,a solutionappeared
dueto Bosch.But Mostellerlearnedoftheproblemin
1955 fromAndrewGleason,"who claimedto have
heardit fromanother,"(Gilbertand Mosteller,1966).
HerbRobbinsrecallsdiscussing
theproblemin 195354 at ColumbiaUniversity,
and MerrillFlood recalls
presenting
a versionoftheproblemthathe calledthe
The distinguished
Arthur
Englishmathematician,
Cayley(1821-1895),is perhapsbest knownforhis
seminalworkin thetheoryofalgebraicinvariants.
He
was also one ofthe mostprolificmathematicians
the
worldhas ever known.His collectedworkscontain
some966 paperstouchingon manysubjectsin mathPaper
ematics,theoretical
dynamicsand astronomy.
andsolutions
#705containssome50pagesofproblems
thatCayleysubmitted
to theEducationalTimesfrom
284
T. S. FERGUSON
1871to 1894.One oftheseproblemsCayley(1875),is
as follows:
4528.(ProposedbyProfessor
Cayley)A lottery
is
arrangedas follows:There are n ticketsrepresentinga, b, c, ... pounds respectively.A person
drawsonce;looksat his ticket;and ifhe pleases,
drawsagain(outoftheremaining
n - 1 tickets);
and so on,drawingin all notmorethank times;
and he receivesthevalueofthelastticketdrawn.
Supposingthathe regulateshis drawingsin the
mannermostadvantageousto himaccordingto
the theoryof probabilities,
whatis the value of
hisexpectation?
Fromthe "Solutionby the Proposer,"we see that
thatk, n, and the
Cayleybelievesit to be understood
a, b, c, ... are known numbers. (Note that here k,
ratherthan n, represents
the totalnumberof drawings.)He solvesthe problemby whatis now known
as the methodof backwardinductionof dynamic
programming.
As an example,he takes n = 4, and
a, b, c, d = 1, 2, 3, 4, and findsfork = 1, 2, 3, 4, the
values of the most advantageousexpectationto be
10/4 38/12, 85/24, 4 resp.
fromoblivionby
Cayley'sproblemwas resurrected
Moser(1956),whoalso reformulated
theproblemin a
neaterguisewhichmaybe viewedas an approximation
to the Cayleyproblemwhenn is verylargeand a, b,
n: You observe, sequentially,
c, ... are 1, 2, ...,
randomvariables X1, ... , Xk known to be iid froma
on theinterval(0, 1); ifyoustop
uniform
distribution
then
afterobserving
youreceiveXj as yourreward.
Xj,
The optimalrule,as foundby Moser,is to stopwhen
thereare m observationsleftto be observedif the
is greaterthanEm,
value ofthe presentobservation
wherethe Emare definedrecursively
by Eo = 0 and
and Chowand Robbins(1963):RandomvariablesX1,
are observedsequentially
at a costofc > 0 per
observation;
ifyoustopafterobserving
X,, thenyou
receiveXn - nc. If recall of past observationsis
allowed,thepayoffforstoppingafterobserving
Xnis
max(X1,..., XX) - nc; such problemsweretreatedby
MacQueen and Miller (1960), Derman and Sacks
(1960) and Chow and Robbins (1961). In Karlin
(1962),the problemis solvedwitha discountrather
than a cost.See DeGroot(1970) fora fulleraccount
ofthesedevelopments.
This class ofproblemsformsa ratherdistinctsetof
problems
thatis stillbeingconfused
withthesecretary
problems.Since thereare so manyvariationsof the
basicsecretary
problem(eachofthe6 conditions
listed
in Section2 has beenmodified
byat leastone author),
I thinkit is worthwhile
to tryto definewhata secretaryproblemis. My definition
is: A secretary
problem
...
X2 *
is a sequential observationand selectionproblem in
which the payoffdepends on the observationsonly
throughtheirrelativeranksand nototherwiseon their
actual values.
Withthisdefinition
then,Cayley'sproblemis not
evena secretary
problem.We mustlookelsewhereto
see whosolvedthesecretary
problem.Proceeding
farther back in time,we come to the firstpractical
I couldfindofthesesequentialobservation
application
and selectiontechniques:the selectionof a wifeby
JohannesKepler.
5. KEPLER'S PROBLEM
Germanastronomer,
Johannes
Whenthecelebrated
Kepler (1571-1630),lost his firstwifeto cholerain
a newwifeusingthe same
1611,he set aboutfinding
of
methodical
thoroughness
and carefulconsideration
the data thathe used in finding
the orbitofMars to
be an ellipse.His first,
notaltogether
happy,marriage
Em+1= (1 - E2)/2. The correspondingequations are
discussedin Guttman(1960) forthenormaldistribu- had been arrangedfor him,and this time he was
to makehis owndecision.In a longletter
tion,in Karlin(1962) fortheexponential
distribution determined
on October23, 1613,written
and in Gilbertand Mosteller(1966) forthe inverse to a BaronStrahlendorf
afterhe had madehis selection,he describesin great
powerdistribution.
detailthe problemshe facedand the reasonsbehind
beAlthoughthereare strongpointsof similarity
tweenCayley'sproblemand the secretaryproblem, each ofthe decisionshe made.He arrangedto interviewand to choosefromamongno fewerthaneleven
thereis one important
difference.
The payoffis not
candidatesforhis hand.The processconsumedmuch
one or zerodependingon whether
select
the
best
you
or not; it is a numericalquantitydependingon the
of his attentionand energyfornearly2 years,what
intothevirtuesanddrawbacks
of
the
difference
with
theinvestigations
intrinsic
value
objectselected.This
withher
her
of
each
playsa big rolein the feelingoftheproblemand has
candidate, dowry,negotiations
of
etc.
the
advice
led to a class ofproblems,
calledvariouslythe house
parents,naturalhesitations,
friends,
an
contains
of
Koestler
The book
Arthur
huntingproblem,the problemof sellingan asset,
(1960)
of
the
and
or the searchproblem,witha literatureas largeas
entertaining insightful
exposition
process.
The bookofCarolaBaumgardt(1951) containsmuch
forthe secretary
problems.The basic problemis the
information.
supplementary
Cayley-Moser
problemwithan infinite
horizon,with
it
Suffice
to
one
will
wish
thepayoff
modified
to makecertainthat
saythatoftheelevencandidatesinterviewed,Keplereventually
decidedon thefifth.
It may
to stopin a finitetime.Forexample,Sakaguchi(1961)
WHO SOLVED THE SECRETARY PROBLEM?
be notedthat when n = 11, the function0n(r) of
valuewhenr = 5. Perhaps,
(2.1) takeson itsmaximum
ifKeplerhadbeenawareofthetheoryofthesecretary
a lotoftimeand
he couldhavesavedhimself
problem,
trouble.
oftheoretical
Ofcourse,in all practicalapplications
results,the assumptionsare neverexactlysatisfied,
and in the presentinstancethisis especiallytrueas
we can see fromKepler'sletter.For example,after
candidatenumber5 and beingstrongly
interviewing
attractedto her, Kepler listenedto the advice of
friendswho were concernedwith her lack of high
rank,wealth,parentageand dowry(she was an orphan),and whopersuadedhimto proposeto number
4. Thus, clearlyKeplerthoughthe could recallpast
applicantsin violationof assumption5 of Section2.
in the
Certainly,
Keplerwouldhave been interested
papersofYang (1974),Petruccelli(1981,1984),Rose
(1984),Ferensteinand Enns (1988) and others,who
allow backwardsolicitationwith a cost or with a
q is
q of beingaccepted.The probability
probability
not1 in Kepler'scase sincecandidatenumber4 turned
himdown.He had waitedtoo long.
That Keplerwenton afterhis failurewithnumber
in getting
the
4 showsthathe was notjust interested
best (assumption6 of Section 2). Perhaps he was
the expectedrankor some otherutility
minimizing
in whichcase thepapersofChow,Moriguti,
function,
Robbinsand Samuels(1964),Mucci(1973),Lorenzen
(1979), Frankand Samuels (1980), etc.,wouldhave
him.Since he had been marriedbefore,it
interested
is unrealisticto assumethathe knewnothingabout
women(assumption4 of Section2); he wouldhave
problemof Gilbertand
enjoyedthe full-information
Mosteller(1966) or Tamaki (1986). But it is also
unreasonableto assume that he knew everything
aboutwomen(whodoes?),so themodelswithpartial
information
and learningof Stewart(1978),Samuels
(1981), Campbelland Samuels (1981) or Campbell
(1982) are moreto thepoint.
all 11
On the otherhand,he actuallyinterviewed
candidatesand could have goneon. Perhapshe was
expectinga randomnumberof availablecandidates
(violatingcondition2 ofSection2), in whichcase he
wouldhave enjoyedreadingthe papers of Presman
and Sonin (1972), Gianini-Pettitt(1979), AbdelHamid,Batherand Trustrum(1982),and Brussand
Samuels(1987).Or perhapstherewas a costofobserand Govindarajulu
vationas in Bartoszynski
(1978),
Lorenzen (1981) or Samuels (1985) or a discount
factoras in Rasmussenand Pliska (1976). He would
in the randomarrivalmodels
be interested
certainly
ofSakaguchi(1976,1986),Cowanand Zabczyk(1978)
and Bruss (1987), in the game theoreticmodelsof
Presmanand Sonin(1975),Fushimi(1981) and Enns
and Ferenstein(1987), and in the multiplecriteria
285
ofStadje(1980),Gnedin(1983),Berezovformulation
and Gnedin(1986) or Samuelsand
skiy,Baryshnikov
Chotlos(1986). Possibly,Keplerwas concernedwith
the actual value of his brideand not just withher
amongtheothercandidates.Thiswouldmake
ranking
problemat all, but a stopping-rule
it not a secretary
problemmore like Cayley'sproblemor the search
problemdiscussedin theprevioussection.
It is clearthatmuchmoreresearchneedstobe done
to clarify
whichoftheseproblemsKeplerwas actually
one it was,therecan be no doubt
solving.Whichever
forhim.His newwife,
thattheoutcomewas favorable
whoseeducation,as he says in his letter,musttake
ranhis
borehimsevenchildren,
theplace ofa dowry,
and seemsto haveprovidedthe
householdefficiently,
necessarytranquilhomelifeforhis erraticgeniusto
flourish.
6. THE GAME OF GOOGOL
Let us returnto thequestion:Ofwhichofthemany
problemam I trying
different
versionsofthesecretary
we shouldtake as
to findthe solver?As historians,
theproblemas itfirstappeared
problem,
thesecretary
in print,in MartinGardner'sFebruary1960column
in ScientificAmerican,where it was called the game
ofgoogoland describedas follows.
Asksomeoneto takeas manyslipsofpaperas
he pleases, and on each slip writea different
positivenumber.The numbersmayrangefrom
small fractionsof 1 to a numberthe size of a
googol (1 followedby a hundred0's) or even
larger.These slipsareturnedfacedownand shuffledoverthe top of a table. One at a timeyou
turntheslipsfaceup. The aim is to stopturning
whenyoucometo the numberthatyouguessto
be the largestof the series.You cannotgo back
turnedslip.If youturn
and pickup a previously
overall slips,thenof courseyou mustpick the
last one turned.
The astutereadermaynoticethatthis is not the
simpleformof the secretaryproblemdescribedin
Section2. The actualvalues of the numbersare revealedto thedecisionmakerin violationofcondition
4. Also, there is this "someone"who chooses the
to makeyourproblemofselectnumbers,
presumably
as possible.The gameof
ing the largestas difficult
game.
googolis reallya two-person
This raisestwoquestions.First,can youguarantee
ofselectingthelargestnumberif
a higherprobability
youallowyourdecisionruleto dependon the actual
In otherwords,doesthestated
valuesofthenumbers?
solutiongivethe lowervalue ofthe game?Second,if
youaretoldhowthis"someone"is choosingthenumbersto place on the slips,can you now guaranteea
T. S. FERGUSON
286
higherprobabilityof selectingthe largestnumber?In
otherwords,does the value of the game exist,and can
we find optimal or c-optimal strategies for the sequence chooser? These questions were not addressed
in the solution presented in ScientificAmerican in
March 1960.
The statementof the problemas it appeared in The
AmericanMathematicalMonthlyin 1963 is much the
same, but somewhatmore nebulous since you are not
told where the numbers come from.Perhaps it is a
game against nature, so the second question above
does not arise. In any case, the solutionpresentedfor
the problemin 1964 contains the same oversight.
Those distinguishedstatisticians who worked on
the secretaryproblemin the 1960's were morecareful
in theirstatementsof the problemin specifyingwhat
informationcould be used in the decision rule, but
none of them attacked the above problem.Therefore,
to see who firstsolved the problem,we must proceed
into the 1970's and beyond.
Suppose you were this "someone" who mustchoose
the numbersto writeon the slips. How would you go
about choosingthe numbersto make it as difficultas
possible forme to obtain the largest? You could not
just choose the numbers 1, 2, *.., n, because then I
could wait until the number n appeared and thus
obtain the largest number with probability 1. You
could not just choose them iid fromsome fixeddistribution,because this would lead to the full-information
case solved by Gilbert and Mosteller (1966), who
showed that I can obtain the largest number with
probabilityat least 0.58016 ... , which is the limiting
value forlarge n. So, you mustchoose the numbersin
some dependentfashion.But you mightas well choose
themto be an exchangeableprocess since the numbers
are put in random order anywaybeforebeing shown
to me. Thus, you are drawnto the partial information
models of Stewart (1978), Petruccelli (1980) and
Samuels (1981).
7. PARTIALINFORMATION
MODELS
...
Let X1, *
Xn denote the values of the numbers
on the slips. These may be consideredas the parametersof our statisticalproblem.We want to finda prior
distributionforX1, ... , Xn, withrespectto whichthe
Bayes rule is the usual optimal rule based on the
relativeranks of the Xj.
In the paper of Stewart (1978), the Xj are chosen
iid froma uniformdistributionon the interval(a, f),
denoted by U(a, f), and (a, d) is chosen fromthe
three-parameterPareto distribution.Specifically,
(7.1)
(a, O) is Pa(k, loguo), and
X1,..,
Xn, given (ae, O3, are iid U(ae, f3,
wherethe three-parameterPareto distribution,Pa(k,
lo, uo) with k > 0 and 1l < uo, is the distributionwith
density
g(ae,: kgloguo)
(7.2)
= k(k + l)(uo
-
lo)"
I(<
(fa)k?+2
,u<
)
where I representsthe indicator function.This is a
conjugate familyof distributionsforthe uniformdistributions,and the posterior distributionof (a, )
givenX1, ..*, Xj is also Pareto,
(a , 3), given X1,
(7.3)
*..,
Xj,
is Pa(k + j, 1j, uj), where
= minJ1o,X1,...,Xj
and
Uj= maxfuo,X1,
*..,
Xjl.
Thus, one can interpretthe priorinformationas being
equivalentto a sample of size k froma uniformdistribution with a minimumof loand a maximumof uo.
Stewart obtained a rather strikingresult for this
priordistributionofthe Xj, as follows.First,thepayoff
is changed so that you win only if you stop on the
largestXj and ifthat Xj is greaterthan uo. Then, one
shows that
(7.4)
P (Xi
Un~I Xis*
Xi
k+ j +1
k + n + 1(24 =uj)
This implies that attentioncan be restrictedto rules
that depend only on the relative ranks of the observations including uo. In fact, the problem becomes
equivalent to the secretaryproblemof Section 2 with
n + k + 1 applicants, in which you start with k + 1
applicants already rejected,the largest having value
uO.In particular,the Bayes rule among rules that use
all the informationis the rule that rejects the first
r'- 1 applicants, and selects the next applicant who
is relativelybest (and better than uo), where r' =
max(1, r - k - 1) and r is the optimal value of r for
the secretaryproblem of Section 2 with n + k + 1
applicants. In addition,forall values of k, the probability of win under an optimal rule tends to l/e as
n -* oo!Since forsufficiently
large n, the maximumXj
will be greaterthan uo withprobabilityclose to 1, this
means that given any e > 0, there is an N and a
distributionof the form(7.1) such that for n > N if
(a, O) is chosen fromthis distributionand then the Xj
are chosen as froma uniformdistributionon (a, d),
the probabilityof win in the game of googol is less
than l/e + e.
Thus Stewart has solved googol asymptoticallyfor
n large. Unfortunately,forfixedfiniten, one cannot
find,forall e > 0, e-optimaldistributionsforchoosing
WHO SOLVED THE SECRETARY PROBLEM?
theXj amongthedistributions
he suggests.We must
lookfurther.
The paperofPetruccelli
(1980)considers
some otherdistributions
forthe Xj. If the Xj are
uniformon the interval(0 - 0.5, 0 + 0.5), thenthe
bestinvariantstoppingrulegivesa probability
ofwin
asymptoticto 0.43517... as n -* oo. If the distributions are normalwithmean g and variance1, the
situationis even worse (fromour point of view).
The best invariantrule gives a probabilityof win
asymptoticto the full-information
case, namely,
0.58016.--. Asymptotically,
you mightas well tell
youropponentwhatg youare using.
Finally,we come to the paper of Samuels (1981),
who extendsthe resultsof Stewart.Samuels shows
thatin the modelwherethe Xj are chosenfromthe
uniformdistribution
on (a, ,B),the usual rule (the
optimalrulebased on relativeranks)is minimaxfor
each n. Since the usual rule is an equalizerrule (it
givesthe same probability
of successno matterhow
theXj are chosen),we see thatit is minimaxagainst
alternatives
as well.In other
general,nonparametric
words,the usual solutionachievesthe lowervalue of
thegame.Thus,I believethatSteveSamuelsdeserves
halfof) the
creditforhavingsolved(themoredifficult
secretaryproblemas it appearedin The American
Mathematical
Monthly.
However,thisexhaustsmysearchthrough
the relevant literature.
What can be said about the game
of googol?I can finallygiveyou my answerto the
questionin the titleof this article.Who solvedthe
secretary
problem?Nobody.
Let me hastento apologizeforthisanticlimax,
and
to venturethe opinionthat the reason no one has
solvedthisproblemis thatpossiblyno one was interestedin googolas a game,or perhapsrealizedthere
was a problemyetto be solved.To remedythesituation,letus tryto findthesolutionnow.Withthehint
givenby the paperof Stewart,it turnsout notto be
hard.
I In fact,sinceweareconsidering
onlythebestchoice
we mayconsidera simplerclass ofdistribuproblem,
tionsthan that of Stewart-theone-parameter
uniformand the two-parameter
Pareto distributions.
Thus,we take
0 is Pa(a, 1), and
(8.1
(8.1) X1, *.., Xn, given0, are iid U(0, 0),
wherethe two-parameter
Pareto distribution,
Pa(a,
MO),is the distributionwith density
g(0I a,
mo)
=
aiMa/0a+lI(0
conjugatepriorforU(O,0),and containstheposterior
distributionof 0 givenX1, * **,Xj:
0, given X1, ***, Xj,
(8.3)
is Pa(a + j, mj),
where mj = max(mo,X1, *.., Xj).
Let us pretendthatwe areplayinga gameofgoogol,
I choosingtheXj and youchoosingthestoppingrule.
For a givene > 0, I will choosethe Xj accordingto
(8.1),and findan a (sufficiently
closeto zero)so that
you will win witha probabilityless than O5n+ E, where
O5nis the maximum probabilityyou can guarantee
usingstrategies
thatdependonlyontherelativeranks,
maxrqn(r).
On=
In fact,I willgiveyouan additionalslightadvantage
and stillkeepyourprobability
ofsuccessbelow'On +
e. I willsaythatyouwinifyoustopat thelargestXj,
or ifall the Xj are no greaterthan 1. This willallow
you to restrictattentionto stoppingrulesthat stop
onlyat a relatively
largestXj thatis greaterthan 1.
The probability
youwinis
P(win) = P(all Xj c 1) + P(win*),
(8.4)
wherewin*represents
theeventImaxX > 1 and you
chooseit}. The firsttermdoes not dependon your
and is easilycomputed:
strategy
00
P(all Xj < 1 1O)g(OI a, 1) do
P(all Xj < 1) =
00
(8.5)
=a
J
(1/0)n(1/o)a+1
dO
= a/(n + a).
8. A RESOLUTION
(8.2)
287
Yourproblemis to maximizethesecondtermof(8.4).
First,we findtheprobability,
conditional
at stagej,
thatmj> 1 is alreadyas largeas it willget.
P(Mi
=
Mn IXi,
=E(P(Mi
(8.6)
...
,Xi)
= Mn I a, Xi,
* * Xj)J X1, .. *
Xj)
(mi/M)n-ig(OI a + j, mj) dO
=
Jmj
=
(a + j)/(a + n)
independent of X1,
*..,
Xj. This is an analog of
Stewart'sresult(7.4). If youhave a newcandidateat
stagej, thatis ifXj = mji,it is optimalto selectit if
and onlyif
a +j
ae + n
> P(win*withbeststrategy
fromstagej + 1 on).
>
MO),
wherea > 0 and mo> 0. For simplicity,
we take
formsa
mO=1. This class of Pareto distributions
The rightside of this inequalityis a nonincreasing
ofj, sinceanystrategy
function
availableat stagej +
2 is also availableat stagej + 1. Sincetheleftsideof
T. S. FERGUSON
288
the inequalityis an increasingfunctionofj, an optimal
rule may be foundamong rules of the formfor some
r - 1: rejectthe first-r- 1 applicants and accept the
next applicant forwhichXj = mj, if any.
Using such a strategy,the probabilityof a win*may
be computedas
n
is best I select j).
P(win*) = E P(select j)P(j
j=r
The probabilitythatj is best givenyou select it is just
(8.6). The probabilitythat you select j can be found
using (8.6):
P(select j)
= P(mr1
= mjn1and mj > mjr1)
_a + j -1
= a + r-1{
a +j- i-a \
+j!
a + r-1
(a + j - 1)(a + j)
Combiningthese into (8.4) and letting0n(r,a) denote
the probabilityof a win,we find
On(r, a)
(a:
+
n) +(
1)jzr (a
+-J1)
Now, note that On(r,a) is continuousin a fora > 0,
and thatas a -O,
n(r, a) -X-n(r) forall r= 1, ** *,
0,
n, whereOn(r) is givenby (2.1). Hence, as a
maxr qn(r, a) -- maxr qn(r) =
.n-
an c-optimalmethodofchoosingthe Xj is
Therefore,
given by (8.1), wherea = a(n, c) is chosenso that
Imaxrqn(r, ao)
On
I < e.
This derivation may be considered an alternate
proofof the minimaxityresultof Samuels mentioned
in Section 7. It is interestingto note that this result
cannot be obtained using the distributionsof (7.1).
The optimal rule based on relative ranks is exactly
Stewart'srule with k = -1; but k must be positivefor
(7.1) to be a distribution;hence, we cannot approximate the case k = -1 withdistributions.(In addition,
the term correspondingto (8.5) does not go to zero.)
If it seems strange that the distributions(7.1) were
consideredbeforethe simplerdistributions(8.1), the
reason is that Stewart and Samuels treated in their
papers problems with more general payoffs,not just
the best choice problem,and needed a broader class
of distributions.Unfortunately,
(7.1) does not contain
(8.1). If there is a moral to this, maybe it is that the
simplercases should always be examined first.
NoteAddedin Proof
Steve Samuels has sent me a copy of the book, ProblemsofBest
Choice (in Russian) by B. A. Berezovskiyand A. V. Gnedin, 1984,
Akademia Nauk, USSR, Moscow. This book is devotedsolelyto the
secretaryproblemand its variations.It contains not only a review
of the field but also a careful exposition and new informationas
well. In theirdiscussionofthe partial informationmodel of Stewart,
theyuse the priordistribution(8.1) and derive (8.6). However,this
is onlyused to prove the asymptoticminimaxityresultof Stewart.
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Comment:WhoWillSolve the Secretary
Problem?
StephenM. Samuels
JustlikeJohannesKepler,whothrewa newcurve
at the solar system,Tom Fergusonhas givena differentslant to the SecretaryProblem.To its many
beginbysaying"all thatwe
practitioners
whoritually
can observeare the relativeranks,"Ferguson(citing
ineffect,
responds"let'snottake
historical
precedent),
thatassumptionforgranted."The heartofhispaper,
as I see it,is the followingFergusonSecretaryProblem:
Stephen M. Samuels is Professor of Statistics and
Mathematics at Purdue University.His mailing address is: Departmentof Statistics,Mathematical Sciences Building, Purdue University,West Lafayette,
Indiana 47907.
Given n, eitherfindan exchangeable
sequenceof
continuous random variables, X1, X2, *---, Xn, for
which,amongall stoppingrules,r, basedon theX's,
sup P{XT=
max(Xl,
X2,
XXn),
is achievedbya rulebased onlyon the relativeranks
oftheX's-or provethatno suchsequenceexists.
Fergusonhas come withinepsilonof solvingthis
problem.He has exhibitedexchangeablesequences,
foreach n and e > 0, such thatthe best rulebased
onlyon relativerankshas successprobability
within
e ofthesupremum.
But he has leftopenthequestion
ofwhether
thissupremum
can actuallybe attained.
For n = 2, the answeris easy; thereis no such
which
sequence.The following
elementary
argument,