Equilibrium solution to the lowest unique positive integer game

0.13 MB·English

by  Seung Ki Baek

Fluctuation andNoiseLetters
Vol.0,No.0(2009)000–000
(cid:13)c WorldScientificPublishingCompany
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Equilibrium Solution to the Lowest Unique Positive Integer Game
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7 SEUNGKIBAEKandSEBASTIANBERNHARDSSON
Department of Theoretical Physics, Ume˚a University,901 87 Ume˚a, Sweden
] [email protected], [email protected]
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C
Received(receiveddate)
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h Revised(reviseddate)
t Accepted (accepted date)
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[ Weaddresstheequilibriumconceptofareverseauctiongamesothatnoonecanenhance
theindividualpayoffbyaunilateralchangewhenalltheothersfollowacertainstrategy.

1 Inthis approach the combinatorial possibilitiesto consider become verymuch involved
v even for a small number of players, which has hindered a precise analysis in previous
5 works. We herepresentasystematic wayto reachthe solutionforageneral number of
6 players, and show that this game is an example of conflict between the group and the
0 individualinterests.
1 Keywords: Lowestuniquepositiveintegergame;Nashequilibrium;projectionoperator;
. socialdilemma
1
0
0 1. Introduction
1
Game theory deals with a situation where each player’s payoff is dependent not
:
v
only on her own behavior but also on other players’. Players will generally have
i
X conflictinginterestswitheachother,buttheymaygetbetteroffbyinteractingwith
r others in various ways [1]. For example, in the minority game [2], a player gets a
a point when her choice is minor among players. Thus, even if a choice happened
to be successful at previous rounds, it is hard to remain minor with that choice as
more and more players will also select it. Although it is not possible for all the
playersto win together,the averageprobabilityof winning canbecome enhancedif
they behave in proper ways.
Recently, an extreme version of the minority game was proposed as a special
caseofthereverseauction,whichiscalledthelowest uniquepositive integer(LUPI)
game [3]. This game has n >2 players, and each of them may choose one integer
from 1 to n. A player wins a point by choosing the lowest unique number. That
means, simply choosing the number 1 is not a good strategy since it is very likely
to be chosen by other players, too. In Ref. [4], this LUPI game was analyzed in
terms of the Nash equilibrium (NE): Let us imagine that every player participates
Equilibrium Solution to the LUPI game
in a game with her own strategy, which is publicly known to the other players. If
no player can improve her payoff by changing her strategy alone, the set of the
players’ strategies is called a NE. Even though this is an adequate framework to
analyzeagame,theactualenumerationofcombinatorialpossibilitiesbecomeshard
to manage as the number of players increases. Although Ref. [4] tried to find an
expressionfor generaln, many of the possible cases were missing in the probability
calculations, even for n=4. According to the suggestedformula in Ref. [4], all the
loosingbidsshouldbeabovethewinningnumberorallshouldbebelowthewinning
number. Furthermore, if they are below the winning number, they should all be
on the same number. However, there exist many more cases that would generate
the same winningnumber. We henceforthpresentawayto takeallthe possibilities
into account correctly for general n and analyze the results thereby obtained.
2. Equilibrium
Suppose that each player draws numbers from her own probability distribution,
which we will denote as her strategy. As explained above, we will find a NE in
this strategy space. Among several different NE’s in the LUPI game, we will be
concerned with one that can be prevalent among these n players, as studied in
Ref. [4]. This approach actually corresponds to the stability concept in the evo-
lutionary game theory [5]. Let us imagine that n 1 people are using a certain
−
strategy, ~p = (p ,p ,...p ), where p means the probability to choose number i.
1 2 n i
The idea is to choose ~p so that the chance of winning for the nth player becomes
independent of her strategy. We consider all the possible cases that those n 1
−
people can make. These can be symbolically represented by
n n−1
Z = p , (1)
0 i
!
i=1
X
fromwhichallthecombinationscanbeobtainedasseparateterms. Forexample,in
athree-playersgamewegetZ =(p +p +p )2 =p2+p2+p3+2p p +2p p +2p p ,
0 1 2 3 1 2 3 1 2 1 3 2 3
where the first three terms on the right-hand side mean that players 1 and 2 chose
the same number 1, 2 or 3, respectively. The other three terms correspond to the
caseswhenplayers1or2havechosentwodifferentnumbers. Together,these terms
representallthepossibleoutcomesofthegame(asseenfromthethirdplayer)since
they add up to the probability 1 due to the normalization condition np =1.
i i
Now, starting from Eq. (1), we extract all the cases where there is a winner,
P
among the n 1 players, at number 1. These are characterized by all the terms
−
which contain only one p . In other words, if we write Eq. (1) as a polynomial in
1
p :
1
Z =A(p ,...,p )+B(p ,...,p )p +C(p ,...,p )p2+...,
0 2 n 2 n 1 2 n 1
those cases are expressed by the second term and one can easily find it by B =
Baek and Bernhardsson
dZ /dp . In other words, we find the cases with no winner at number 1 as
0 1|p1=0
dZ
0
Z = Z p
1 0 1
− dp
1(cid:12)p1=0
(cid:12)
n n(cid:12)(cid:12)−1 n n−2
= p (n 1)p p .
i 1 i
! − −  
i=1 i6=1
X X
 
Ifwearefurthertoexcludethecaseswherethereisawinneratnumber2,weapply
the same operation on Z as follows:
1
dZ
1
Z = Z p
2 1 2
− dp
2(cid:12)p2=0
(cid:12)
n n(cid:12)(cid:12)−1 n n−2
= p (n 1)p p
i 1 i
! − −  
i=1 i6=1
X X
n−2   n−3
n n
(n 1)p p +(n 1)(n 2)p p p .
2 i 1 2 i
− −   − −  
i6=2 i6=1,2
X X
   
Generalizingthisidea,wecanfindcaseswherethereisnowinneruptoanarbitrary
number i by the following recursion relation:
dZ
i−1
Z =Z p . (2)
i i−1 i
− dp
i (cid:12)pi=0
(cid:12)
(cid:12)
Formally, we may define Li as the linearity pro(cid:12)jection operator with respect to pi.
Let us denote a generic polynomial of p ,...,p as Q, which may have an index
1 n
to indicate more than one polynomial. The operator L can be algebraically repre-
i
sentedasL [Q]=p E D [Q],whereE iseliminationofp bysubstitutingzeroand
i i i i i i
·
D is differentiation with respect to p . In addition, a and b mean some coefficients
i i
independent of p . We then have the following relations for these operators:
i
L [apn]=ap δ
• i i i 1,n
L L [Q]=L [Q]
i i i
• ·
L L [Q]=L L [Q]
i j j i
• · ·
L [aQ +bQ ]=aL [Q ]+bL [Q ]
i 1 2 i 1 i 2
•
E L [Q]=L E [Q]
j i i j
• · ·
L [Z ]=0, if i j
i j
• ≤
where δ is the Kronecker delta. It immediately follows that
1,n
k
Z = (1 L ) [Z ]. (3)
k i 0
( − )
i=1
Y
Equilibrium Solution to the LUPI game
Let us now calculate the probability for the nth player to win the LUPI game by
choosing number i. Then within the other n 1 people, there should be no winner
−
uptoi 1andnooneshouldchoosethenumberi. The lastconditioncanbeeasily
−
imposedbythesametrick. Thatis,wesimplysubstitutezeroforp toexcludeevery
i
case where p appears. Therefore, the probability of our concern is simply written
i
asc (p~)=E [Z ]= i−1(1 L ) E [Z ]andthenthplayer’sexpectedpayoff
i i i−1 j=1 − j · i 0
is n o
Q n
W(~π;~p)= c (p~)π ,
i i
i=1
X
where ~π =(π ,π ,...,π ) is the nth player’s strategy.
1 2 n
For example,
n−1
n
c = Z = p =(1 p )n−1, (4)
1 0|p1=0  i − 1
i6=1
X
 n−1 n−2
n n
c = Z = p (n 1)p p
2 1|p2=0  i − − 1 i
i6=2 i6=1,2
X X
= (1 p )n−1 (n 1)p (1 p p )n−2,  (5)
2 1 1 2
− − − − −
c = Z
3 2|p3=0
n−1 n−2
n n
= p (n 1)p p
i 1 i
  − −  
i6=3 i6=1,3
X X
  n−2   n−3
n n
(n 1)p p +(n 1)(n 2)p p p
1 i 1 2 i
− −   − −  
i6=2,3 i6=1,2,3
X X
= (1 p )n−1 (n 1)p (1 p p )n−2  
3 1 1 3
− − − − −
(n 1)p (1 p p )n−2
2 2 3
− − − −
+ (n 1)(n 2)p p (1 p p p )n−3. (6)
1 2 1 2 3
− − − − −
From the normalization condition, p = 1 n−1p , indeed we have only n 1
n − i=1 i −
degrees of freedom in choosing p~ so that
P
n−1
W(~π;p~)= (c c )π +c .
i n i n
−
i=1
X
The NE solution, ~p , should satisfy c (p~ ) = constant for every i so that W
NE i NE
cannot be better by changing ~π. That is, the chance of winning is the same on
all numbers. For n = 3, all the above calculation coincides with that presented in
Ref. [4]. Note that we haven 1 degreesof freedomand n 1 equations so we can
− −
solvethese simultaneously. Figure1(a)showssolutions for somen values, obtained
by using the Newton method. In Fig. 1(b), we plot its scaled version, considering
that the horizontal axis naturally scales with n, and the vertical axis roughly with
1/n.
Baek and Bernhardsson
0.3 3
n=9 n=9
(b)
10 10
0.2 11 2 11
pi 12 n pi 12
0.1 1
(a) Uniform
0 0
0 2 4 6 8 10 12 0 0.2 0.4 0.6 0.8 1
i i/n
Fig1. (a)NEstrategies,p~NE,withvaryingthenumberofplayers,n. (b)Scaledplotswithrespect
ton. Asnincrease,thesolutiondeviates morefromtheuniformsolution(dotted).
100
n=6 1
9
10-1 12 Maximum
ci10-2 ∞ n W 0.5 (b) UnifoNrmE
(a) Ref. [3]
Ref. [4]
10-3 0
0 2 4 6 8 10 12 14 3 6 9 12
i n
Fig2. (a)Theexpected chanceofwinningbychoosingi,whentheothern−1playersfollowthe
uniform solution defined as pi = 1/n. (b) Comparison of outcomes scaled with n. The uniform
solutiongivesanoutcomethatrapidlyconverges tothetheoretical maximumW =1/n(dotted),
while the NE outcome remains below that. The other two curves represent outcomes from the
strategies suggestedinpreviousworks.
Setting c =c from Eqs. (4) and (5), we find
1 2
(1 p )n−1 (1 p )n−1 =(n 1)p (1 p p )n−2. (7)
2 1 1 1 2
− − − − − −
Sincetheright-handsideshouldbepositive,wecanconcludethatp <p forevery
2 1
finite n, which means the equilibrium solution cannot be uniform. Its implication
is remarkable: Let us imagine that everyoneemploys the same strategy p~. In other
words, ~π =p~. The question is what the strategy ~p should be in order to maximize
the expected payoff, W(p~;~p). For example, for n=3, we find
W(p~;p~)=(1 p )(1 p )(1 p ),
1 2 3
− − −
which is maximized at a uniform solution, ~p = (1/3,1/3,1/3). This is clearly
∗
differentfromthe NEsolution, ~p =(2√3 3,2 √3,2 √3)[4]. The expression
NE
− − −
is not so simple for n > 3, but solving W(p~;~p) = 0 under p = 1, we still
∇p~ i i
see that the uniform solution gives the best outcome to the population on average.
P
Although weconfirmedthis only for n 12,this resultis plausible sincethe riskof
≤
overlappingchoices will be minimized by the uniform solution. On the other hand,
we already know that the equilibrium strategy cannot be uniform. Indeed, if all
the people follow ~p = (1/n,...,1/n), a player’s chance of winning rapidly decays
∗
with her chosen number i [Fig. 2(a)]. For example, one may use the formulation
abovetoobtainc =e−1(1 e−1)i−1forlargen,whichcoincidewiththemultinomial
i
−
analysisinRef.[3]. Therefore,eachplayerismotivatedtodeviatefromthisuniform
Equilibrium Solution to the LUPI game
solutionbychoosinglownumbersmorefrequently. Inshort,thereexistsadilemma
between the best and the equilibrium strategy. Note that since each play only has
one winner, the theoretical upper limit for the gain per play is 1/n. Let us then
comparethesethreecases: theequilibriumoutcome,thebestpossibleoutcome,and
the theoreticalupperlimit[Fig2(b)]. While the uniformsolutionquicklyconverges
to the theoretical maximum, the NE solution remains suboptimal in this plot. For
comparison, we also plot outcomes from the strategies in previous works: Ref. [3]
suggested a strategy with p = p = 1/2 which gives W(p~,p~) = 21−n, while in
1 2
Ref. [4] the solutionis approximatelywritten as p =2−i for i<n with p =21−n.
i n
Note that the latter one actually performs better than presented in Ref. [4], since
we have taken all the possibilities into consideration. In any case, these become
smaller than the outcome from our NE solution.
As Eq.(3)indicates, the number oftermsto considerincreasesexponentiallyas
n grows, making the exact enumeration of probabilities intractable. Although our
currentcomputationalresourcesallowustoarriveonlyupton=12,ourprocedure
providesasystematicwaytotakeallthecombinatorialcasesintoaccountforanyn,
inprinciple. We believethatthe behaviorsshowninFig.2giveaclue to anticipate
p~ in the limit of large n. Specifically, we sketch a way to perform this task: The
NE
NE solution is obtained when c = c for all i and j. This means that we can
i j
set c to a certain constant c for i = 1,...,n. The point is that c has only one
i 0 i
more variable, p , than c , so that we can attack each variable p one by one.
i i−1 i
For example, Eq. (4) with c = c gives p (c ) = 1 c1/(n−1). Substituting this
1 0 1 0 − 0
value into Eq. (5) and with c = c gives a value for p (c ), and so on. If the
2 0 2 0
correct NE-value is chosen so that c = c then the equilibrium solution, ~p
0 NE NE
is obtained. When there exist several real solutions for p then the correct root
i
can be found using the restriction 0 < p < 1 and the normalization condition
i
i p < 1. The procedure for a three-players (n = 3) game, where we have
j=1 j
c =c =28 16√3 0.287 (Fig. 2b), goes as follows:
P0 NE − ≈
c =c =(1 p )2 p 0.4643.
1 0 1 1
− ⇒ ≈
c =c =(1 p )2 2p (1 p p )
2 0 2 1 1 2
− − − −
[inserting p ] p 0.2684.
1 2
⇒ ⇒ ≈
c =c =(1 p )2 2p (1 p p ) 2p (1 p p )+2p p
3 0 3 1 1 3 2 2 3 1 2
− − − − − − −
[inserting p and p ] p 0.2673 0.0246i.
1 2 3
⇒ ⇒ ≈ ±
The fact that the last solution is complex indicates that the value of c is not
0
exactly correct. Adding more digits decreases the imaginary part while the real
part moves closer to the correctsolution (e.g., c =0.287187gives p =0.267949
0 3
−
5.04978 10−4i). Sincethereisnogeneralformulaforrootsofpolynomialsofdegree
×
larger than four, it seems that this procedure should be carried out numerically.
Figure 3 shows the behavior of c for i = 1 to 4 and n = 9. The solution of p is
i i
obtained in each step by finding the crossing point c = c (where p has been
i 0 i−1
inserted into c ).
i
To sum up, the benefit of this procedure is that if we know the correct NE-
value, c = c , then we can calculate the probabilities one by one up to any i
0 NE
Baek and Bernhardsson
1
c
1
c
2
c
3
c
4
c
NE
ci
0.1
0 0.2 0.4
p
i
Fig 3. The expressions ci as functions of pi for n = 9 players. A Nash equilibrium solution is
obtained when ci = cNE for all i ≤ n. p1 is found by the condition c1 = cNE. Inserting this
value into c2 allows for a determination of p2 as the crossing point between c2 and cNE, and so
on. OrthevalueofcNE canbeobtainedinthiswaybyfindingc0 whichfulfillsthenormalization
n
condition Pi=1pi =1. The two dotted horizontal lines represent the interval within which cNE
mustlieifweusetheexpressionsuptoc4 insteadofcn.
we want or are capable of doing within our computational resources, for any n. In
manysituationsitmight,forexample,beenoughtoknowthesolutionuptocertain
i above which p ’s are negligible. The method of solving for all i simultaneously
i
instead results in an all-or-nothing situation. Of course, the down side is that we
need to know c in advance. However,recalling that c =W W(p~ ;~p ),
NE NE NE NE NE
≡
we see the limiting behavior lim nc = lim nW in Fig. 2(b). Putting
n→∞ NE n→∞ NE
such a c = c into this procedure may thus yield p~ for large n. The solution
0 NE NE
obtained in this way up to i = 4 for n = 9 and c = 0.0985 c (see Fig. 3) is
0 NE
≈
p~=(0.2515,0.2349,0.2087,0.1643,...). Thiscanbecomparedtotheresultobtained
fromsolvingforallisimultaneously,whichisp~=(0.2515,0.2348,0.2086,0.1641,...).
ThismethodcouldalsobeusedtofindtheactualNEsolution,withoutknowing
c , by iterating the procedure until a self-consistent solution is found for all i.
NE
If this is again beyond our computational power, one could at least determine an
interval in which c = c lies. Let’s say that we use expressions for c up to
0 NE i
i = j. If c is too small then the normalization gets violated by j p > 1. If
0 i=1 i
c is too large then the crossing points result in probabilities that are too small to
0
P
add up to one in total, as long as p is a monotonically decreasing function of i.
i
Thus, the largest possible sum of all n probabilities is obtained if p is uniform for
i
i j, so the sum j p +(n j)p must be larger or equal to one. The interval
≥ i=1 i − j
obtained for n =9 and j =4 is marked by the two horizontal, dotted lines in Fig.
P
3 (c = 0.078 and c = 0.146). This interval becomes narrower as higher
0,min 0,max
values of i are addressed. Note that when c is changed all the curves c for i > 1
0 i
also change since we get new p ’s with j <i to be inserted into c .
j i
3. Discussion
We have shown a way of taking all the combinatorial possibilities into account for
the LUPI (lowestunique positive integer)game andproposedaprocedureto find a
Equilibrium Solution to the LUPI game
Nashequilibriumforageneralnumberofplayers,n. Itturnsout,however,thatthe
number of terms to consider grows exponentially in n, and that the computational
capability of our personal computer can only solve the problem up to n = 12
players. To dealwith this problemwe alsosuggestanalternativewayofsolvingfor
the equilibrium strategy for each value of i at a time, by inserting an estimate of
the average expected payoff of each player. This procedure allows the problem to
be solved sequentially instead of solving the whole problem simultaneously.
Wealsohavefoundthatauniformsolutiondistributesnearlyoptimaloutcomes
equally among all the players. However, this would be driven to a suboptimal
equilibrium solution, so we may regard this n-person game as an example of a
social dilemma to manifest a conflict between group and individual interests.
Acknowledgments
S.K.B.acknowledgesthesupportfromtheSwedishResearchCouncilwiththeGrant
No. 621-2002-4135.
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[3] Q. Zeng, B. R. Davis, and D. Abbott. Reverse auction: the lowest unique positive
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[4] A.P.Flitney. Commentson‘reverseauction:thelowestuniquepositiveintegergame’.
Fluct. Noise Lett., 8:C1–C4, 2008.
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