Content uploaded by Gisèle Umbhauer
Author content
All content in this area was uploaded by Gisèle Umbhauer on Jul 20, 2023
Content may be subject to copyright.
1
The puzzling three-player beauty contest game: play 10 to win.
Gisèle Umbhauer
May 2023
Abstract
In this paper, we study the 3-player beauty contest game. This 3-player guessing game has the
same Nash equilibrium than the usual (large) N-player beauty contest game but it has also nice
specific properties. To highlight these properties, we study classroom experiments on 2-player,
3-player and large N-player guessing games, both from a theoretical and behavioral point of
view. The spirit of the paper is the spirit of the French newspaper Jeux et Stratégie which, in
the early eighties, proposed the beauty contest game to his fun of logic readers. As a matter of
facts, we wonder if it is possible to win the 3-player guessing game. So we show that, despite
the 3-player beauty contest game has no weakly dominant strategy, it is possible to play it in a
way that leads to win with a large probability, provided the parameter a is lower than 0.75. And
we argue that playing 10 for a=0.6 ensures a large probability to win.
Keywords: beauty-contest game, 3-player game, guess, nombre d’or, dominance, win area,
behavioral heuristic.
JEL Classification : C72, C9
1. Introduction
In 1995, Nagel initiated a huge literature both on the N-player beauty-contest game and on
level-k reasoning. In the classic version of the beauty-contest game, also called guessing game,
N players simultaneously choose a number in the interval [0,100] and the winner is the player
whose number is closest to a times the mean of all proposed numbers. In case of a tie, the
winners share the prize. Many things have been written on the (large) N-player guessing game
and some papers focused on the very special 2-player guessing game (among them Grosskopf
& Nagel, 2008, and Costa-Gomez &Crawford, 2006). Yet, at least to our knowledge, only few
authors, among them Ho et al (1998) and Breitmoser (2012), are interested in N-player guessing
games with N small but different from 2.
This is quite astonishing, given the strong logical differences between the large N-player
guessing game and the 2-player guessing game. For N=2, 0 is the weakly dominant strategy
that always wins the game, given that aX/2 is always closer to 0 than to X, for parameters
. So it is enough to play 0 to win the game, without guessing anything about the behavior of
the other player. In other terms, the 2-player guessing game is not a guessing game, by contrast
to the large N-player guessing game, where the winning strategy strongly depends on the way
umbhauer@unistra.fr, BETA-University of Strasbourg, 61 Avenue de la Forêt Noire, 67085 Strasbourg Cedex,
France
I thank Jalal El Ouardighi for insightful comments on statistics and I also thank the third-year class students (year’s
class 2022/2023) at the Faculté des Sciences Economiques et de Gestion, University of Strasbourg, France, who
played the games.
2
other players play the game. Usually, as is well known, in the large N-player beauty contest
game, guessing leads to iterated dominance and level-k reasoning, at least if people play the
game for the first time. In the large N-player guessing game, 0, which is both the Nash
equilibrium strategy and the only strategy obtained by (infinite) iterated dominance, is generally
not the winning strategy. The winning strategy, at least if players are not trained in this game,
is often closer to 50a2, the level-2 behavior strategy, in that many players simply play 50a, the
level-1 behavior amount.
Given the gap between the very easy winning behavior in the 2-player guessing game and the
uncertain winning strategy in the large N-player guessing game, one might be interested in a
game in-between, namely the 3-player guessing game. This game reveals to be quite different
both from the 2-player game and from the large N-player game. In facts, the 3-player beauty
contest game has nice special properties, that can be exploited in a way to win the game with a
large probability, provided the parameter a is not too large (lower than ¾).
So we work on this game, with a special focus on . The spirit of the paper is the spirit
of the French newspaper “Jeux et Stratégie”, already mentioned by Moulin (1984, 1986) and
nicely detailed in Nagel et al. (2017). The newspaper - see Ledoux (1981)- proposed the large
N-player guessing game to its readers, fun of logic, who all wanted to win the prize. And this
is exactly what we want to do in this paper: we aim to win the 3-player beauty contest game,
at least for .
To do so, we do a theoretical and a behavioral study. We start, in section 2, by presenting and
analyzing two classroom experiments around the 3-player beauty contest game. Then we show
theoretical properties of this game. In section 3, we fix a value Y for one opponent’s guess and
show that, for all the guesses Z of the other opponent, there exists an interval of winning guesses
X. This interval draws attention to a threshold parameter value, . Section 4 puts into
light the stronger impact of non-iterated dominance for (by contrast to N large) and how
this fact helps playing the 3-player guessing game. In section 5, like Breitmoser (2012), for
each number X, we look for the area of couples of opponents’ guesses such that X is a winning
strategy. In section 6, we show why playing 10 is a good way to play in the 3-player game with
, whether the players are trained or not in guessing games. In both classroom
experiments, playing 10 leads to win with a probability exceeding 0.81. We conclude, in section
7, with a rather counterintuitive result: if one expects that the opponents play integers, then it is
better to play an integer too, rather than playing an integer plus or minus a small quantity.
2. Two classroom experiments
In the classic large N-player beauty contest game with parameter a – now called Guess(N, a)
throughout the paper-, a large number N of players simultaneously propose a number in
, and the winner is the player who is closest to a times the mean of the N proposed
numbers. The prize is shared among the winners in case of a tie.
The rules of 3- player guessing games and 2-player guessing games are those of Guess(N,a),
with N=3 and N=2.
We study two classroom experiments on a 3-player guessing game with parameter a – called
Guess(3,a) - in two different contexts. The first experiment was run at the University of
3
Strasbourg during the first lecture in a third-year class in game theory, in the academic year
2022-2023, so the students did not know the concepts of dominance and Nash equilibrium (at
least in games with infinite pure strategy sets). In total 241 students participated to this first
experiment. In a within-subject framework, the 241 students first played the usual large N-
player guessing game with parameter a=3/5, Guess(241,3/5), before playing the 3-player
guessing game with parameter , Guess(3,3/5). Several examples (with N=5) were
explained to the students before the beginning of the experiment, to be sure that the rules of the
games were understood. The students were invited to explain their choices by writing.
We ran a second experiment a few weeks later, in the same third-year class in game theory,
when the students knew the concept of dominance and Nash equilibrium. They were also
informed about the winning strategy (namely 19-which is not far from 50a2) in Guess(241,3/5),
one of the two games they played a few weeks ago
1
. In this second experiment, 199 students,
in a within-subject framework, first played the 2-player guessing game with parameter
, Guess(2,3/5), before playing again Guess(3,3/5). In the 2-player guessing game, a player
who plays 0 always wins the game (or is one of the two winners). In order to help the students
to discover this fact, we informed them, before they played Guess(2,3/5), that this game has a
winning strategy. By so doing, we wanted to avoid that the students simply replicate (or best
replied to) the numbers chosen in Guess(241,3/5) a few weeks before (we namely wanted to
avoid that they play 3/5.19). The students viewed this experiment as an exercise they had to
solve. And they had enough time to find the winning strategy. So clearly the students were not
in the same context than the players in Grosskopf & Nagel (2008), who were not informed on
the existence of a winning strategy. And it worked: 20.6% of the students played 0, and many
students played low numbers (27.14% played 1 or below, 30.65% played 5 or a lower number),
in that they discovered, often after many numerical trials, that the winning number is always
the lowest proposed one. These percentages are far from the 9.85% of participants playing 0 in
Grosskopf and Nagel’s (2008) experiment on Guess(2,2/3). After having played Guess(2,3/5),
the students were invited to play again Guess(3,3/5), the game they played a few weeks before.
We did not talk about the existence of a winning strategy -because it does not exist- but we
hoped that, given the proximity of N=2 and N=3, the students aimed to find such a strategy. We
also gave them two contrasted examples, one in which the winning number is the lowest one
(the three proposed numbers were 17, 30 and 53) and one in which the winning number is the
middle one (the three proposed numbers were 14, 31 and 72). With this second example, we
wanted to avoid that the students who played 0 in Guess(2,3/5) simply replicate their behavior
in Guess(3,3/5). The students had again enough time to make many numerical trials. We hoped
that what they learned in Guess(2,3/5) would help them to better exploit some special properties
of Guess(3,3/5), yet it only partly worked. The students were again invited to justify their
choices by writing.
Figures 1a and 1b, respectively Figures 2a and 2b, display the way the students played in the
first, respectively in the second experiment.
1
Yet they were not informed about the Nash equilibrium in the guessing games and we gave no information about
the 3-player beauty contest game Guess(3,3/5) they played a few weeks ago.
4
As expected, the students played Guess(3,3/5) differently in both experiments. This is
confirmed by a Kolmogorov Smirnov test (p-value =1.786x10-8) and all other tests. When they
play Guess(3,3/5) after having played Guess(241,3/5), the mean proposed value is 31.71,
whereas it shrinks to 23.76 when they play Guess(3,3/5) after Guess(2,3/5). A simple look at
Figure 1b and Figure 2b shows that the distribution in Figure 2b shifts to the left in comparison
to the distribution in Figure 1b.
Figure 1a: Guess(241,3/5), 1st experiment Figure 1b: Guess(3,3/5) 1stexperiment
Figure 2a: Guess(2,3/5), 2nd experiment Figure 2b: Guess(3,3/5) 2nd experiment
In both experiments, the students, while playing Guess(3,3/5), are influenced by the game they
played just before. So, in the first experiment, the mean proposed value in Guess(241,3/5),
31.92, is close to the mean proposed value in Guess(3,3/5), 31.71. The same is true in the second
experiment: the mean proposed value in Guess(3,3/5), 23.76, is only slightly larger than the
mean value proposed in Guess(2,3/5), 22.72.
In the first one-shot experiment, both distributions are similar according to the Kolmogorov-
Smirnov test (p-value 0.9262), and this is confirmed by a Khi-2 test and a Wilcoxon-Pratt
signed rank test. Yet this fact does not mean that the students play Guess(3,3/5) and
Guess(241,3/5) in the same way. Among the 241 students, only 17.43% choose the same
number in both games, 42.74% choose a lower number and 39.84% choose a higher one. To
put it more precisely, according to the students’ choices and explanations, the students, when
switching from Guess(241,3/5) to Guess(3,3/5), understand that their own choice has more
impact on the mean in Guess(3, 3/5) than in Guess(241,3/5), but this leads to two opposite
shifts. The students think that, to win, they have to propose a number between the two
opponents’ numbers (remember that we only gave examples with , so the students are
not compelled to observe that their conjecture can be wrong). This induces some students
81412
3127
44
28
151718 4 8 0 7 5 2 1 0 0 0
0to5
6to10
11to15
16to20
21to25
26to30
31to35
36to40
41to45
46to50
51to55
56to60
61to65
66to70
71to75
76to80
81to85
86to90
91to95
96to100
Guess (241,3/5) 241 students
mean 31.92, standard deviation 16.84
61
4121923
36
712 111 2 4 0 0 0 0 1 0 0 6
0to5
6to10
11to15
16to20
21to25
26to30
31to35
36to40
41to45
46to50
51to55
56to60
61to65
66to70
71to75
76to80
81to85
86to90
91to95
96to100
Guess(2,3/5) 199 students
mean 22.72, standard deviation 21.55
19 614
2626
50
161813
29
6 6 3 4 2 1 0 1 0 1
0to5
6to10
11to15
16to20
21to25
26to30
31to35
36to40
41to45
46to50
51to55
56to60
61to65
66to70
71to75
76to80
81to85
86to90
91to95
96to100
Guess(3,3/5) 241 students, mean
31.71, standard deviation 17.49
23
1519
48
26
34
11 4 3 2 0 9 0 0 0 1 1 0 0 3
0to5
6to10
11to15
16to20
21to25
26to30
31to35
36to40
41to45
46to50
51to55
56to60
61to65
66to70
71to75
76to80
81to85
86to90
91to95
96to100
Guess (3,3/5) 199 students
mean 23.76, standard deviation 17.21
5
playing low, respectively large numbers in Guess(241,3/5), to propose a larger number (up to
50 units larger), respectively a lower number, in Guess(3,3/5). So among the 106 players
playing less than 30 in Guess(241,3/5), 66 choose a larger number in Guess(3, 3/5) and only 27
choose a lower number. And among the 105 students playing more than 30 in Guess(241,3/5),
the behavior is reversed: 67 choose a lower number in Guess(3,3/5) and only 23 choose a higher
one. There are also more students playing between 45 and 55 (17.43%) in Guess(3,3/5) than in
Guess(241,3/5) (10.79%) and some students do not hesitate to shift from numbers below 10 in
Guess(241,3/5) to 50 in Guess(3,3/5). In general, the students view Guess(3,3/5) as being more
random and difficult to play than Guess(241,3/5).
In the second experiment, despite the means are close in Guess(2,3/5) and Guess(3,3/5), the
distributions are different, as confirmed by a Kolmogorov Smirnov test (p-value 0.0014). Many
students understand that that there is a strong difference between the 2-player game and the 3-
player game, which explains that they do not adopt the same behavior in both games. Thanks
to the two given examples and thanks to their own many trials, they observe that, in
Guess(3,3/5), it is not enough- in contrast to what happens in Guess(2 3/5)- to play a very low
number to win and that it is sometimes better to play a number between the two opponents’
ones, but a not loo large one. This explains that among the 83 students playing less than 20 in
Guess(2,3/5), 71.08% choose to increase their number – and only 6.02% choose to decrease it-
when switching to Guess(3,3/5), but only 6.02% choose to increase it above 30. This also
explains the shift from a mode on 0 in Guess(2,3/5) to a mode on 20 in Guess(3,3/5).
Anyhow, in the four games, the students’ behavior is very dispersed. The standard deviations
are large: 16.84 in Guess(241,3/5), 17.49 in Guess(3,3/5) in the 1st experiment, 21.55 in
Guess(2,3/5) and 17.21 in Guess(3,3/5) in the 2nd experiment. In all games, there is a large range
of played numbers. Almost all integers from 0 to 50 are played by at least one or two persons
in the two games in the first experiment, and almost all integers from 0 to 40 are played by at
least one or two persons in both games in the second experiment. The very strong standard
deviation in Guess(2,3/5) is namely due to the fact that some students find the winning strategy
(20.6% play 0 and 30.65% play a number lower than or equal to 5) and the other students do
not (hence they rather play like in a large N-player guessing game-12.06% play 30). Moreover,
some students play 100, trying to use their strong power to incite the opponent to play 100. To
summarize, even if having some training in game theory, even if having enough time to make
many trials, the students view 2-player and 3-player guessing games as being rather difficult.
As said above, they perceive the 3 – player game as being more difficult than the classic large
N-player guessing game: they say that the mean behavior of two opponents is more difficult to
guess than the mean behavior of a large number of opponents which they often -wrongly-
estimate around 50. And in some degree, they are right: as is well known in theory of mind,
what matters in the N-player beauty contest game is not the truth, but what others think it is. So
even if only 10.79% of the players play around 50 (between 45 and 55) in Guess(241,3/5), if a
student believes that most students expect this fact and accordingly play 30 = 3/5.50, then, by
best-replying to this fact, hence by playing around 3/5.30, he is not far from winning the game
(19 is the winning value in the experiment). By contrast, the fact that the students know that
6
the central limit theorem is inappropriate in front of two opponents
2
, deprives them from a
starting behavior they can best reply to. So guessing becomes indeed more difficult.
3. Intervals of winning strategies and threshold parameter
Before going into the 3-player guessing game, we make a clarification: throughout the paper,
given a triple of propositions , we say that wins the game if it is the number closest
to
, but also if it is one of the numbers closest to
. So “winning proposition”
does not mean “best-reply”: as a matter of fact, if, for example, and X and Y are
both at the same distance from
, then we say that X and Y are winning strategies in
that they share the prize, but a best reply would be to play .
This is an important point. By saying that a player wins the game as soon as he is among the
winners, we also give weight to the triples where the three players play a same number
3
. So
some players may focalize on a possible common behavior, a fact that is not in the spirit of the
guessing game. Yet this should not change a lot the general behavior in the game. For example
iterated elimination of weakly dominated strategies is not affected by this change (a100 still
weakly dominates all the numbers larger than a100, which generates the infinite iterated
elimination process up to 0). And level-k reasoning is not much affected either. The triples
where everybody plays a same number X different from 0 are very unstable, in that, as soon as
one player switches to , the other players switch also away from X in order to not lose the
game. In other terms, with a level-k reasoning, if one expects that all the others play X, playing
aX is a much more stable best response than playing X (because aX, contrary to X, is also a best
response to a context where most of the players play X except some of them who switch to
). So iterated elimination of weakly dominated strategies and level-k reasoning should not be
affected. Moreover, the set of triples where a player who plays X shares the prize with other
players, is at the frontier of the set of triples where he is the only winner, and so it is of smaller
dimension than the set of winning triples.
Let us see things from a pragmatic point of view. Do not forget that we want to give hints to
win the game. When somebody announces the winner(s) of a game, the point that matters for a
participant is to be among the winners, not so much to be the only winner. To say things
differently, the gap between a loser and a winner is very huge, regardless of what is won by the
winner. And this is particularly true when a player only shares the prize with at most 2 other
players (which is the case in the 3-player game). What is more, given the dispersed behavior
distributions in both experiments, the probability for a player to meet 2 players playing the same
amount than himself is quite small. So the only context to take into account is the one in which
a player may share the prize with one other player. Yet, in this context, if the prize is a
sufficiently large amount of money -unless the game is not worth being played-, then, taking
risk and Von Neumann Morgenstern utility into account, we can say that most players assign
to the half of the prize a utility which is much larger than half of the utility assigned to the prize.
2
The central limit theorem is also inappropriate in front of many students but the students do not know this fact.
3
In the large N guessing game, this also leads to focalize on N-tuples where K players play a number X and the
(N-K) other players play a number Y different from X, so that X and Y are at the same distance from
. In the 3-player game, these triples only exist for a strictly larger than ¾. If so, when one player plays X
and two other players play the larger value Y equal to , then the three players win the game.
7
And, to conclude on this point, should a player be only interested in the triples where he is the
only winner, this is not problematic. Focusing on the triples where is the only winner
just amounts to deleting the frontiers of the set where it is the only winner.
We now go into the 3-player guessing game.
A first observation is that the 3-player guessing game is a real guessing game. Despite 3 is close
to 2, there is a main discontinuity between the 2-player guessing game and the 3-player guessing
game. Whereas, in the 2-player game, 0 always wins the game, regardless of the opponents’
choice and regardless of the value of the parameter a (<1), in the 3-player game the winning
strategy depends on the numbers chosen by the two other players and also on the value of a.
Insofar, in contrast to the 2-player game, the 3-player game is a true guessing game, like the
usual large N-player game.
Yet, given the proximity of 3 and 2, many students, after having discovered the winning strategy
in Guess(2,3/5), seriously hoped to also find a winning strategy in Guess(3,3/5), despite it does
not exist. And they were right by so doing, because the 3-player game, in contrast to the large
N-player game, has special properties that can be exploited to establish a strategy that often
wins the game.
By playing Guess(3,3/5) in the 2nd experiment, the students, helped by the two given examples,
often adopt one of the two following contrasted points of view: either they are convinced that
playing a low number- namely 0-remains the best strategy to play, or they become convinced,
like in the first experiment, that aiming to be between the two opponents’ numbers is the best
behavior. And all students are right in part. Let us call X a player’s strategy and let us suppose
that the 3 proposed numbers are X, Y and Z. We fix throughout the paper. X, with
, is a winning strategy if Y is not too different from Z, even if Y and Z are quite large. By
contrast, if the two opponents play distant numbers (like 14 and 72 in the given example), then
it is better to play a number in-between (like 31 in the example).
These observations give rise to proposition 1:
Proposition 1:
a) If then any strategy in
is a winning
strategy.
b) If
and Z>Y, then any strategy in
is a winning
strategy.
c) If Y=Z, then any strategy in
is a winning strategy.
Proof of proposition 1
For X≤Y<Z, Z can never be the winning strategy.
Let us assume the contrary. So
Hence , which is not possible, given that and .
So the winning strategy is either the lowest proposition or the middle proposition.
: X is winning if
, hence
which requires that
hence
.
8
is winning if
hence
which requires
.
And when , then X is a winning strategy if X≤Y and
hence
Proposition 1 shows that, for a given couple there is a range of values X that win the
game. This fact is also true in the 2-player game, but completely new in comparison to the large
N-player game. In Guess(N,a), a player, to win the game, has to be very close to a times the
mean value, because many played numbers can be scattered around the winning value. By
contrast, in the 2-player game, the whole interval is the set of winning values if the
opponent plays Y. So the notion of winning interval is a common point between the 3-player
game and the 2-player game. Yet, by contrast to Guess(2,a), in Guess(3,a), the nature and the
width of the interval of winning values both depend on the distance between Y and Z and on
the value of the parameter a.
It derives from proposition 1 that is a threshold parameter, as can be seen in Figures 3a,
3b and 3c. In these figures we draw the interval of winning values X, for a fixed value Y, when
Z goes from Y to 100. The special case , to which the students are confronted, is
illustrated in Figure 4.
For , there is an interval of values Z such that simply playing less than Y (the lowest
opponent number), including 0, leads to win the game (see Figure 3a and Figure 4). This fact
agrees with the students’ conjecture that 0 is most often the winning strategy and with their
0 Y
100 Z
100
Y
0
Figure 3a: a<3/4, areas of winning X
Figure 3b: a=3/4, areas of winning X
Figure 3c: a>3/4, areas of winning X
Figure 4: a=3/5, areas of winning X
0 Y 2Y 100 Z
100
Y
0
100-Y
0 Y 1.5Y 3Y 100 Z
100
Y
200/3 -Y
0
0 Y
100 Z
100
Y
0
9
conjecture that playing a number lower than the two opponents’ numbers is the best way to win
the game. Yet this is right only if the two opponents play sufficiently close values: for ,
as long as , 0 remains a winning strategy and it is enough to play less than Y to win
the game.
For a larger than ¾, by contrast, there is a range of values, the values Y and Z such that
, such that, to win the game, it is enough to choose a number between the
two opponent numbers Y and Z (see Figure 3c).
This may explain that some students, in the first and in the second experiment, are convinced
that they just have to play a number between the two opponents’ guesses to win the game. This
way of thinking is especially observed in the first experiment, where the students play
Guess(3,3/5) after having been confronted to a large number of players, and without having
experienced the strong power of 0 in Guess(2, 3/5). So, by playing Guess(3,3/5), many students
simply think that, when there are 3 numbers, the middle one is surely closest to the winning
strategy, given that the winning strategy is linked to the notion of mean (as it is the mean
multiplied by a). Hence playing in the middle becomes a new heuristic of behavior. In the
second experiment, the students who aim to propose a number between the opponents’ guesses
base their way of doing on the second given example (and many personal trials).
Yet, given that 3/5 <3/4, the students are only partly right. For , a player wins by
playing a number X between Y and Z when Z is far from Y,
, but X has to be below
, which is strictly lower than Z. By contrast, for , provided
, all
the numbers between Y and Z win the game. Yet this area only exists as long as
,
hence for
.
The above results show that is a threshold value in the 3-player beauty contest game.
As long as , the 3-player guessing game shares nice similarities with the 2-player game,
namely the fact that 0 remains a winning strategy as long as Z is sufficiently close to Y (
, but for , 0 is no longer a winning strategy, except if at least one opponent
plays 0 too.
4. Dominance is more powerful
In the 2-player game, dominance is extremely powerful, given that 0 is the weakly dominant
strategy. In the N-player game, by contrast, non-iterated dominance only allows to eliminate
the strategies larger than a100, which is much less powerful. In the 3-player game, the result is
in-between.
Proposition 2
All the strategies larger than a100/(3-2a) are weakly dominated.
Proof of proposition 2
We show that X weakly dominates any number Y larger than X, if and only if X is the winning
strategy when confronted to any numbers Y and Z
We first prove the necessary condition:
10
Suppose that X does not win when confronted to a couple of opponents’ guesses such
that . In that case X does not dominate Y, because Y does better, in that it shares the
prize, when confronted to the couple . This is true regardless of the value .
Now suppose that X wins when confronted to any couple of opponents’ guesses with
. This means that , for any couple such that
. So this condition must hold for the most restrictive values i.e. for Z=100 and Y
almost equal to , so it becomes
, i.e.
We now show that the condition
is sufficient for X to weakly dominate any larger
number S. Suppose the contrary. This means that there exist couples , with
(without loss of generality), such that X loses when confronted to , and S wins. S is
necessarily lower than Z, unless it cannot win and , unless X wins (because X wins against
any couple with ). So there only remains one possible configuration for X to
lose and S to win: . Yet this means that
and
, hence and which is
impossible for
This result is interesting for two reasons.
First, given that it allows to eliminate all the strategies larger than
(100/3 for a= 3/5), non-
iterated dominance is much more powerful than in the large N-player guessing game where it
only allows to eliminate the numbers larger than 100a (larger than 60 for ). The
difference between
and 100a is quite attractive, at least if a is lower than ¾, as illustrated
in table 1.
a
Undominated strategies in
Guess(N,a), without and after one
iteration
Undominated strategies in
Guess(3,a), without and after
one iteration
Undominated strategy
in Guess(2,a)
3/5
[0, 60] [0, 36]
[0, 33.33] [0, 11.11]
0
2/3
[0, 66.67] [0, 44.44]
[0, 40] [0, 16]
0
0.7
[0, 70] [0, 49]
[0, 43.75] [0, 19.14]
0
0.75
[0, 75] [0, 56.25]
[0, 50] [0, 25]
0
Table 1: intervals of undominated strategies, without iteration and after one iteration in Guess(N,a),
Guess(3,a) and Guess(2,a).
Second, dominance becomes closer to standard level-k reasoning. The mean value played by
(standard) level-0 players is usually supposed to be around 50. Standard level-1 players play
a50, and more generally standard level-k players play ak50. Standard level-1 behavior is often
observed in beauty-contest games with untrained players, but standard level-1 and level-2
behavior are also often observed with more trained players. For example, in the first experiment,
both in Guess(241,3/5) and in Guess(3,3/5), the mode is (more than 15% of the
students are standard level-l players in Guess(3,3/5)). But even more trained students adopt a
standard level-1 and level-2 behavior: more than 13% of the students play 30 and more than
18% of them play 20 -a value close to a250- in Guess(3,3/5) in the second experiment. As
argued by Breitmoser (2012) playing a50 or a250 makes no sense in Guess(3,a), because a true
level-1 behavior leads to a whole range of winning strategies and not to the value a50 (see next
11
section). Yet many students, when playing Guess(3,3/5), are unable to exploit the specificity of
N=3, and therefore prefer behaving like standard level-1 or level-2 players. And other students,
even if they understand the specificities of Guess(3,3/5), fear that their opponents are unable to
catch them. So they still guess that many of their opponents just play a50, in which case the
standard level-2 behavior is a good way to play.
Insofar, the stronger power of dominance helps by making the compared amounts more similar.
For example, in Guess(3,3/5), the opponents’ expected behavior is similar, whether they are
supposed to be rational -hence able to discover that they should not play more than
-, or whether they are supposed to be standard level-1 players who play .
Given that 33.33 is close to 30, for both kinds of conjectures, it becomes reasonable to not play
more than a number around 11 (in that 11.11 weakly dominates any larger number when the
opponents play a value in ). This is not true in the usual large N-player guessing game
where weak dominance leads to 60 and standard level-1 behavior to 30. In other terms, by
playing around 11 in Guess(3,3/5), you win with a larger probability, because you win against
opponents with different kinds of rationalities.
To summarize, in the large N-player beauty contest game, iterated dominance does the same
iterations than level-k reasoning, but starts with 100, whereas level-k reasoning starts with 50
(iterated dominance leads to whereas level-k reasoning leads to
). So a player who works with dominance in a context where many
players do a level-k reasoning generally loses the game (he systematically plays too large
numbers). By contrast, in the 3-player game, standard level-k reasoning still leads to the
numbers , but dominance more quickly converges to similar numbers in
that the multiplicative factor now decreases to
. So, if many persons still do a standard
level-k reasoning, a player who reasons with dominance in the 3-player game may more easily
win than a player who reasons with dominance in the large N-player guessing game.
5. Uniform distribution and optimal guesses
In order to help a player to guess, we now turn to another way to exploit proposition 1.
Proposition 1 gives the interval of winning guesses X when the opponents play a couple
Rather than looking for the winning X in front of a given couple of opponents’ guesses ,
we now look for all the couples , with , such that X is a winning guess. We call this
area the win area of X. Insofar we follow Breitmoser (2012) and we get the same optimal win
areas.
Let us stress the point that the notion of win area makes no sense in the large N-player game,
given that the win area is very small for any number. And it makes no sense in the 2-player
game because the win area of 0 covers the total area of the opponent’s guesses. By contrast, it
makes sense in the 3-player game, because, on the one hand, it differs in function of the played
number and, on the other hand, it may be quite large for some guesses, especially if a is not too
large (lower than ¾). So playing a number with a large win area may help winning the game
with a large probability, and remember that this is exactly what we aim to do.
Breitmoser (2012) argues that a “serious” level -1 player should in fact play the value X that
maximizes the win area. As a matter of fact, a level-1 player should best reply to a level-0
behavior, where a level-0 behavior does not consist in playing 50, but consists in playing
12
randomly on [0, 100]. It derives from this fact that a level-1 player has to suppose that the two
opponents’ guesses are randomly scattered on the set and to choose the
value X that maximizes its win area.
So we compute, for each amount X, the area of couples in , that leads
X to win the game (or to be among the winners).
Proposition 3 (identical to Breitmoser 2012)
Let us fix . X wins the game if :
- either and
hence and and
- or and
hence and and
For , the value X* that leads to the largest win area is
. The
corresponding win area is
Proof of proposition 3: see below
The win areas differ depending whether a is lower or larger than ¾. And gives rise to
two different cases (we are not really interested in given that it becomes hazardous to give hints
to win for ).
Given that we want to win the game, it is surely not a good idea to suppose that the couples
are uniformly distributed on , in that the opponents also aim to win
the games. So it makes sense to calculate the win areas by assuming that the opponents have
some minimal rationality. For example, we can suppose that nobody plays more than a100,
given that everybody- if trained a little- can understand that the mean cannot be larger than 100,
so that it makes no sense to play more than a100
4
.
More generally, some behavioral observations allow to reduce the set of couples to
, where M can take different values. Mathematically, this changes nothing as
regards the win areas except that 100 has to replaced by M and
by
. Observe
also that we only study the values X lower than – more generally
- given that the other numbers are weakly dominated by -respectively
-, so do less often win the game and therefore lead to a smaller win area.
It is also more or less inadequate to suppose that the opponents’ guesses are uniformly
distributed on but we come back to this fact later on. For the moment we
comment the optimal win areas when the couples are uniformly distributed on
.
For the win areas are given in Figure 5a and 5b. Given that Figures 5a and 5b only
focus on the couples with , the win area is 2 times the area on the graphic, i.e.:
4
Yet remember that some players may wish to bring the three players to the same value so that everybody wins
the game. Insofar they can play 100, because 100 is a focal value. The number of players with this aim is however
quite small in general.
13
Maximizing this expression leads to the optimal
. The optimal corresponding win
area is
.
So the optimal win area decreases in a (it goes from 80.77% for to 66.67% for
). For , it covers 75.44% of the total area of the opponents’ possible guesses, which
is a large percentage of the total area, and . The division factor
is fast
decreasing in a, so goes from 6.5 for to 3 for . The strong division factor 4.75
helps to focus on a small set of values, and the winning probability of 75.44% helps to be
confident in the chosen value (under the uniformity assumption). As a matter of fact, according
to the expected rationality of the two opponents, we may set , if the opponents are
supposed to be able to apply naïve dominance (hence do not play more than a100); we may
also set , if the opponents are supposed to not play more than the standard level-0
amount, if the opponents are supposed to not play more than the standard level-1
amount 30, or, by contrast, , if the opponents are supposed to be rational enough to
not play weakly dominated strategies. Given the strong division factor 4.75, for M going from
30 to 60, the set of optimal values is reduced to [6.32, 12.63] which is a rather small interval of
values in comparison to the set of possibilities [0, 100]. So, if it can be assumed that the
behaviors are uniformly distributed on [0, M], the good amount to play in Guess(3,3/5), leading
to win with a probability around 3/4, is between 6.5 and 12.5.
Things are more complicated for a larger than ¾. For these values, the notion of win area, even
if it can be assumed that the played values are uniformly distributed on [0, M], does not allow
to prognostic a value to play, because the size of the optimal win area becomes small. Two
configurations are possible, given in Figures 6a and 6b. The good configuration depends both
on X and on a. Figure 6a holds for
(the two lines
and do not intersect in ), whereas Figure 6b holds for
(the two lines intersect in .
Figure 5a: win area for
a<0.75a<0.75
Figure 5b : win area for
Z
X
M Y
0
M
X
Z
X M-X M Y
0
M
X
14
In Figure 6a, the win area is:
This area holds for
and leads to X*=
. The optimal win area becomes
X* has to check so these results only hold for .
Observe that the range of optimal X* becomes large for different values of M, given that the
division factor
of M is decreasing, going from 3 for to 1.88 for
. Moreover the optimal win area is strongly decreasing in a, going from 2/3 for
to only 53.31% of the total area for .
For example, for , the maximal area is obtained for but it is only equal
to 63.49% of the total area. Moreover the division factor is only 2.57 so the range of optimal
X* becomes large for different values of M. If we take (because everybody should
understand that it makes no sense to guess more than 80), then X* becomes 31.11, an amount
that is not far from the standard level-2 reasoning amount 32 and from the amount obtained
with one step of iterated (true) dominance (32.65). But if we suppose that the two opponents
play at most 50, because they aim to be between the two other opponents or will not play more
than a standard level-0 player, then X* shrinks to 19.44. The range [19.44, 31.11] is very large.
Combined with the fact that the win area for the optimal X* is only 63.49% of the total area, it
clearly becomes difficult to give a hint to win.
This explains that we focus on the smaller class of games with a lower than ¾. This also
explains that we give hereby the results for a larger than 0.9078 only for sake of completeness.
In Figure 6b, the win area is:
Figure 6a : win area for a>0.75, X>
Figure 6b : win area for a>0.75, X<
Z
0
X
M
Y
0
M
X
Z
0
X
M Y
0
M
X
15
=
The optimal X* is equal to
and the optima win area is MX*
X* has to check X*< so these results only hold for .
Let us now talk about the uniformity of the behavior distributions, as the results in proposition
3 help guessing only if we can suppose that the opponents’ guesses are uniformly distributed
on . Clearly, the distributions for Guess(3,3/5) in our two classroom
experiments- see Figures 1a and 2b- are not uniform, but they have some uniform
characteristics. With untrained students, by setting (because only 11.2% students play
more than 50), we observe that only 8 integers in [0, 50] are not played in Guess(3,3/5) and that
29 integers in [0, 50] are played by a number of persons going from 1 to at most 5. With trained
students, by setting (in that only 9.55% students play more than 40), we observe that
only 6 integers in [0, 40] are not played, and that 28 integers in [0, 40] are played by a number
of persons going from 1 to at most 5. So, despite the distributions are not uniform, if we exclude
some peaks on isolated values (20, 25, 30, 40 and 50 in the first experiment, 0, 15, 20, 25 and
30 in the second experiment) the students’ guesses are rather uniformly distributed on the
integers in . This is namely due to the fact that Guess(3,3/5) is perceived as a complicated
game: the behavior rules “I should play a number between the numbers proposed by the two
opponents” and “I should rather play a low number” are very vague and therefore explain that
the guesses are scattered on a whole interval in a rather uniform way.
6. 10, a “nombre d’or” for
What can we deduce up to now? First, that is much too hazardous to give a hint when .
Fortunately, usually a is often below this threshold, one of the most standard value being 2/3.
For a<0.75, it becomes possible to suggest a small range of values that can lead to a good
probability to win the game, at least if it can be assumed that the guesses are rather uniformly
distributed on an interval Given that is reasonable to expect that the two opponents
should not play more than 100a, and that it would be a too strong assumption to suppose that
both play less than the standard level-1 behavior a50, the guesses giving rise to the optimal win
areas (proposition 3) are in the interval [
. The range of the interval is increasing
in a (it goes from 3.85 (interval [3.85, 7.69]) for to 12.5 (interval [12.5, 25]) for
and is equal to 6.32 (interval [6.32, 12.63]) for .
Dominance, if the strategy set of the opponents is reduced to would lead to play a
number below
, yet
(20 for ) is always larger than
so does not help
to reduce the interval and may even broaden it.
Moreover, uniformity of the distribution is a strong assumption. For , with untrained
students, 2 values are played by more than 15 persons, 30 (37 students) and 50 (25 students).
With trained students, 2 values too are played by more than 15 persons, 20 (37 students) and
30 (26 students). So, in order to give a hint, we have to focus on more specificities of the game.
To do so, we now only focus on the specific parameter a= 3/5.
16
It appears that 10 is like a “nombre d’or” in the 3-player guessing game with parameter
.
First, 10 is the mean of the set of values that weakly dominate the larger numbers when
the two opponents are supposed to play less than 60 (the largest naïve dominance amount).
10 is only slightly larger than the mean of the interval [
=[6.31, 12.63].
10 is only slightly lower than the number that maximizes the win area when it can be assumed
that the population is uniformly distributed on , an assumption that makes sense in
Guess(3,3/5), especially in the second experiment, where many students think that to win it is
good to play a number between the two opponents’ ones, but not a too large one, in that the
students also know that low numbers often win the game.
10 weakly dominates all the larger numbers if it can be assumed that the other persons will not
play more the standard level-1 amount 30.
What is more, 10 wins the game each time at least one of the two opponents plays the standard
level-1 amount 30, provided the other opponent does not play more than the naïve weakly
dominating strategy 60. So 10 performs nicely even in a nonuniform setting when we expect
that at least one of the opponent plays like a standard level-1 player.
This can be seen in Figure 7a (we adapt Figure 5a by setting , and ).
For , is the standard level-1 behavior a50, that is to say 30. The blue lines
are in the win area. They show that, as soon as one opponent plays 30, 10 wins the game
regardless of the other opponents’ amount, between 0 and 60.
This is quite uncommon. Usually, the specific value
that checks
, i.e.
, is seldom an integer, and the interval for which
wins the game when one of the two opponents plays the standard level-1 behavior a50 and the
Z
10 15 30 60 Y
0
30
15
10
60
Figure 7b: areas A and B for a=3/5 and X=10
Figure 7a: win area for a=3/5, X=10 and M=60
The blue lines are in the win area: if one
opponent plays 30, 10 wins the game regardless
of the other opponents’ amount, between 0 and
60.
Z
10 15 23.33 30 56.67 60 100 Y
0
30
15
10
60
B
A
100
50
17
other plays a value in the interval , is different from . This is very specific to
, in that only for .
By the way let us observe that, for any parameter a lower than ¾, numbers around the number
that equalizes a50 and
, hence around
, are worth of interest. As a matter of fact,
is included in [
regardless of the value , and it is not stupid to expect
that at least one of the opponent plays the standard level-1 amount. And if so, it is interesting
to look for a value that ensures to win regardless of the other opponent’s chosen number in a
large enough interval with no hole, the interval [0, 75-25a].
Last but not least 10 is an integer. Moreover, 10 is an easy integer, a multiple of 5 and of 10;
these multiples are always focal values and often played by the participants. As a matter of fact,
whereas the multiples of 5 only constitute 20% of the integers in [0,100] and a set of dimension
0 in [0,100], in the first experiment, 59.75% students play a multiple of 5 in Guess(3,3/5), and,
in the second experiment 67.84% students play a multiple of 5 in Guess(3,3/5) (similar
observations hold for Guess(241,3/5) and Guess(2,3/5)). And the values played by 12 persons
or more are all multiples of 5 (20, 25, 30, 40 and 50 with untrained students, 0, 15, 20, 25, 30
with trained students). This fact is not uncommon in experiments
5
and this matters because we
will show in the next section that it is better to play an integer, rather than a little more or a little
less than it, when we expect that the opponents also play integers. Moreover, given that 10 wins
against any couple with , 10 wins against the numerous
couples of opponents playing (0, 15), (15, 20), (15, 30), (20, 30), (20, 40), (25, 30), (25, 40),
(25, 50), (30, 40), (30, 50) and (40, 50) .
But for the moment let us study the true performance of 10 in our two experiments, with
untrained students (first experiment) and trained students (second experiment). To do so we
confront 10 to any couple of guesses that can be extracted from the students’ behavior (i.e.
241x120 couples in the first experiment, 199x99 couples in the second experiment). We look
for all the couples (Y,Z) such that 10 does not win against (Y,Z). These couples are either in
area A or in area B (see Figure 7b); in area A, the two opponents choose low values so that 10
is too large to win. In area B, one opponent plays a large value, larger than 30, and the other
plays a value larger than 10 (but not too large), so that 10 is too low to win the game. Depending
on whether we play with trained or untrained students, the most “dangerous” area for 10 will
not be the same.
In the first experiment, area B is more dangerous for 10 than area A. As a matter of fact, when
players are untrained, they sometimes (wrongly) think that the best thing to do, in a 3-player
guessing game-, is to play a number between the two opponents’ guesses, which can lead them
to play around 50, and more generally a number in [20, 50] (68.88% of the students play in this
way). Namely 25 students among 241 play 50 (more than 10%) whereas only 11 among the
same students play 50 in the large N-player guessing game (which proves that the students more
feel the necessity to guess between the (two) opponents’ guesses in a 3-player game than in a
large N-player game). Yet these players may be problematic for 10. If an opponent plays 50, 10
loses the game as soon as the other opponent plays a number larger than 10 and lower than
100/3-10=23.33. Yet 1/5 of the students (49 students) play a number in this range of values; so
5
See for example classroom experiments on the traveler’s dilemma in Lefebvre and Umbhauer (Cairn International
forthcoming)
18
10 loses the game in front of these 25x49= 1225 couples of opponents (the horizontal red line
in Figure 7b). More generally, with untrained persons, area B corresponds to 4448 couples
because there are 100 students among 241 (41.49%) who play Z larger than 30 and because 10
loses the game when these students are coupled with a student who plays Y with
. These couples represent 15.38% of all possible couples. Fortunately, with
untrained students, area A is rather deserted by the players. Untrained students do not often play
less than 10 (only 9.54% of the students play less than 10), so area A only counts 837 couples
(only 2.89% of all the possible couples). So, at the end, 10 performs well, given that it wins
with probability 0.8173.
In the second experiment, area B is much less dangerous for 10, in that trained students do less
often play more than 30: only 17.09% of the trained students play in this way in contrast to
41.49% of the untrained students. Namely only 2 students play 50, given that they know that
they can win by playing a number that is not between the two opponents’ amounts. So only 2
types of students are dangerous for 10. Those who play 60 (9 students stick to naïve dominance)
and the 3 persons who play 100 (for wrong reasons). So only 1953 couples (9.91% of the
possible couples) are in area B. By contrast, area A becomes more dangerous, in that there are
now 14.07% of the students who play less than 10 and 21.11% who play less than 15. It becomes
dangerous to play 10 namely because 12 students (6.03%) play 0. In front of an opponent
playing 0, 10 loses the game if the other opponent plays a number lower than 15 and larger than
or equal to 0 (see the vertical red line in Figure 7b). This line represents 426 couples. So area A
counts 1523 couples (7.73% of the total couples). In facts, area A counts less couples than area
B but the percentages are more equilibrated than in the first experiment (2.89% and 15.38% for
areas A and B in the first experiment, respectively 7.73% and 9.91% in the second experiment).
And, at the end, 10 performs very well in the second experiment, in that it wins with probability
0.8236.
So what can we deduce? First that playing 10 performs well in both experiments. The more the
players are trained, the more area A grows but area B shrinks, and the reverse holds when players
are untrained. At the end, the sum of the areas A and B does not much change and remains low.
But what about the performance of other numbers? Does 10 better perform than 8, 9, 11, 12,
13, …?
10 is the best way to play in the experiment with trained students.
Switching to a lower value like 9 leads to a success rate of 78.08% because the number of
couples in area B strongly increases (+1000), and the number of couples in area A only
smoothly decreases (-150). Area B namely strongly increases because it shifts to the left (see
Figure 8a) so that 9 systematically loses against all couples with a player playing 10, and the
other player playing from 28.5 to 100 (see the vertical red line in Figure 8a) which represents
630 additional couples. And things get worse for all numbers lower than 9.
By contrast, by playing more than 10, like 11 for example, area A strongly increases, namely
because 11 loses against the 1325 couples where one player plays 10 and the other plays from
10 to 31.5 (see the vertical red line in Figure 8b). In facts, area A increases by 1650 couples –
so does more than double when switching from 10 to 11-, whereas area B only shrinks by about
100 couples. It follows from these facts that 11 wins in only 74.48% of all possible
19
confrontations. And things get worse for all the integers larger than 11 (for example 12 and 15
respectively win with probability 0.7321 and 0.6739).
We give in table 2 the results obtained for integers from 6 to 17 (other integers do strictly worse
than 10): we give the percentage of couples in area A and area B for the different integers, as
well as the percentage to win the game.
By construction, area A can only grow and area B can only decrease when the studied integer X
increases, so the percentage of couples in area A increases from the left to the right of table 2
and the percentage of couples in area B decreases from the left to the right. Given that for 10,
the sum of the percentages of couples in areas A and B is equal to 17.64, it derives that all
numbers lower than or equal to 6 or larger than or equal to 12 do strictly worse than 10 and
need not to be studied. For information, in this experiment with trained students, 0 wins in
51.80% of all possible confrontations, so has a better success rate than 19.
By contrast, with untrained students, 10 and 13 are in a pocket handkerchief (the success rate
is 81.73% for 10 and 81.75% for 13), 14 performs a little better than 10 (success rate 81.96%)
and 2 numbers win clearly more often than 10, the integers 12 and 15 (the success rates are
respectively 83.69% and 83.71%). The high scores of 12 and 15 are namely due to the 25
students playing 50. 15, when confronted to 50, only loses when the third played value is larger
than 15 and lower than 18.33 (only 9 students play such a number), whereas 10 loses when the
third number is larger than 10 and lower than 23.33 (49 students play in this way). So 10 loses
in 25x49=1225 confrontations while 15 only loses in 25x9=225 confrontations. In facts, area B
decreases by 2686 couples when switching from 10 to 15 (4448 couples in area B for 10, only
1762 couples in area B for 15). Yet 10 performs much better than 15 when confronted to at least
100
27
9 10 56.67 57.67
B
28.5
16.5
15
30
33
10 11
31.5
Figure 8a : area B for X=9 increases
by the additional grey area
Figure 8b : area A for X=11 increases
by the additional grey area
A
6
7
8
9
10
11
12
13
14
15
16
17
18
19
A %
2.68
3.60
5.07
6.96
7.73
16.11
17.92
21.43
25.30
26.52
37.62
39.90
41.81
45.07
B%
20.88
17.57
15.94
14.96
9.91
9.41
8.87
8.62
8.12
6.09
5.96
5.49
5.00
4.48
SR %
76.44
78.83
79.00
78.08
82.36
74.48
73.21
69.95
66.58
67.39
56.42
54.61
53.19
50.44
Table 2: percentages of couples in area A and area B and success rate for integers from 6 to 17
in the experiment with 199 trained students.
20
one player playing less than 15. So, for example, when confronted to one of the four players
playing 12, 10 wins provided the third number is lower than 33.33, whereas 15 loses (because
12 wins) in all these confrontations, which represents 4x152=608 couples. This namely explains
that area A increases by 2112 couples when switching from 10 to 15 (837 couples in area A for
10, 2949 couples in area A for 15). Yet, at the end, 15 performs better than 10.
All the other numbers do worse than 10. We give in table 3, for each integer from 6 to 19, the
percentages of couples in areas A and B and the success rate.
For the same reasons than for table 2, given that the sum of percentages of couples in areas A
and B is 18.27 for 10, all the guesses larger than or equal to 19 or lower than or equal to 9 can
only do worse than 10 and therefore need not to be studied.
We can add that 0’s success rate is 46.35%. This score is of course lower than in the experiment
with trained students but it is larger than perhaps expected. As a matter of fact, if the distribution
were uniform, 0’s success rate would be 33.33%. Yet the distributions are not really uniform
and we know that 0 is a best guess as long as the opponent’s guesses Y and Z (with are
such that . Many couples share this property, among them (20, 20)
but also (30,30) or (34, 34. This largely explains the success rate of 0.
We can also observe that the progression of the success rates in not regular. With untrained
students for example, 10 does better than 11, 11 does worse than 12 but 12 does better than 13,
13 does worse than 14 and 14 does worse than 15, but 15 does better than 16 ….
This is partly due to the nonuniform features of the distributions, namely to the fact that some
numbers are played by many students, and others are not played at all. This induces large or
small changes of area A and/or area B (even if monotonically increasing and decreasing in X),
so that the sum of couples in areas A and B may rise or decrease in X. So, if X is often played,
one consequence is that area B for contains many couples that do not exist in area B for
X. As a matter of fact, area B for X contains the couples , with and , such
that . So, for it contains all the couples with
and . So it contains many couples with that do
not exist in area B for X. In the same time, if X is often played, then the number of couples in
area A may strongly increase when switching from X to , because area A for
contains all the couples (, with , which may be numerous
(because X is often played). This may help X to win more often than and , and this
fact is for example observed for 15 in both experiments with trained and untrained students,
where 15 is respectively played by 15 and 8 students.
6
7
8
9
10
11
12
13
14
15
16
17
18
19
A %
0.97
1.24
1.68
2.46
2.89
4.20
4.68
7.63
9.44
10.19
15.22
16.69
17.46
23.80
B%
26.59
23.44
20.53
18.28
15.36
14.40
11.61
10.60
8.59
6.09
4.78
4.10
3.33
2.90
SR%
72.41
75.30
77.77
79.25
81.73
81.38
83.69
81.75
81.96
83.71
79.98
79.19
79.19
73.27
Table 3: percentages of couples in area A and in area B and the success rate for integers from
6 to 17 in the experiment with 241 untrained students.
21
A similar phenomenon is at work when a number X is played by nobody. Then, when switching
from X to , area A will not be enlarged by the couples with
given that nobody plays X. And when switching from X to , area B, which contains
the couples , with and , such that is
not enlarged by the left because the couples checking and the
couples checking are the same. This may explain that X does not
necessarily make a better score than and .
The above observations for example (also) contribute to explain that 15 has a better success rate
than 14 in both experiments, and that 10 has a better success rate than 9 in both experiments, 9
and 14 being the only integers not played in both experiments. It also explains why 9,
respectively 14, has a worse success rate than 8 and 10, respectively 13 and 15, in the
experiment with trained students.
At the end it derives from the two curves in Figure 9 that, as long as we only compare the
performance of integers, 10 remains the good value to play (with a success rate larger than
81.7% in both experiments). 12 and 15 perform less well in the experiment with trained students
(73.21% for 12 and only 67.39% for 15), namely because there are much less trained students
who play 50. By contrast, 10 does well in both experiments despite the fact that trained students
and untrained ones do not play in the same way. In other terms 10 resists to different kinds of
logics.
Figure 9: success rates for integers from 6 to 19 with the untrained 241 students (blue curve)
and with the 199 trained students (grey curve).
7. Play an integer
In this concluding section, we draw attention to the following fact. Very often, the players
playing the beauty contest game play an integer, despite they can choose any real in .
In the classroom experiment with trained students, all the 199 students play integers in
Guess(2,3/5) and only two of them do not play integers in Guess(3,3/5). In the classroom
experiment with untrained students, only four among the 241 students do not play integers, both
0,00
10,00
20,00
30,00
40,00
50,00
60,00
70,00
80,00
90,00
6 7 8 9 10 11 12 13 14 15 16 17 18 19
Success rates with trained and untrained students
22
in Guess(N,3/5) and in Guess(3,3/5). Moreover, we can add, as said in section 6, that many of
them play multiples of 5, which is not uncommon in many experiments.
So intuitively, a spontaneous reaction would be to say that the best way to play is to not play
an integer but rather a real just below or above this integer. Yet this would not be a good idea.
Let us prove this rather counterintuitive result.
We show it for 10 in the experiment with trained students, but the same proof holds for any
integer in both experiments.
First, it would be bad to play a little more than 10, say 10.01 for example. In facts, area B is the
same for 10.01 and 10, but area A is larger for 10.01 than for 10.
We recall that area B, for 10, is the set of couples , with , , and
. Area B, for 10.01, is the set of couples , with , , and
. Yet, given that the students only (most often) play integers, the
first played number larger than 30 is also the first played number larger than 30.03 and the set
of integers Y (and even integers divided by 2) such that 1 is also the set of
integers (and even integers divided by 2) such that . So area B
counts the same number of couples, for 10 and 10.01.
By contrast, area A becomes larger by construction, when switching from 10 to 10.01, for two
reasons. As a matter of facts, area A for 10 is the set of played couples with and
. So the first reason is that switching to 10.01 leads to switch to the couples
. So Z can be equal to , and there may exist students who
play when Y and Z are mainly integers. Moreover, and this is the main reason, given
that we now work with , we have to add all the couples and
which can be numerous (as in our experiment). So area A will increase.
It follows from above that, in the second experiment with trained people, 10.01 wins the game
only with probability 0.7495 (whereas 10 wins with probability 0.8236), so 10.01 does worse
than 6, 7, 8, 9 and 10 in this experiment).
Perhaps more surprisingly, it is better to play 10 than a real just below 10, say 9.99 for example.
This is due to the fact that area A is not affected by this change when most people play integers
(or integers divided by 2) but area B increases.
As a matter of fact, as regards area A, we now switch from the couples with and
to the couples and . Yet, given that the
students mainly play integers, is equivalent to , and the played Z lower than
14.985+1.5Y are the same than those lower than 15+1.5Y. So area A does not change when
players mainly play integers.
By contrast, area B increases when switching from 10 to 9.99 for three reasons. Area B, for 10,
is the set of couples , with , , and 1. Area B, for 9.99
is the set of couples , with , , and . So first
we add a whole range of couples, those where one of the opponent plays 30, and the other plays
a number Y with which may contain numerous couples, namely those where
one opponent plays 30 and the other plays 10. Second, for each Z from 30 to 100, we have to
add all the couples where one opponent plays 10 and the other opponent plays Z. Third, the
23
numbers lower than include the integer when Z is an integer divisible
by 3, which leads to add additional couples. So area B increases.
It follows from above that 9.99 has a lower success rate than 10. In facts, in the second
experiment with trained people, 9.99 leads to win with probability 0.7751, and so does worse
than 7, 8, 9 and 10.
What we explained for 10 is true for any played integer, namely 12 and 15, for the same reasons.
As long as one expects, in the beauty contest game with 3 players, that the two opponents play
integers, it is better to play an integer too. And this fact also holds for a different from 3/5
6
.
This is a rather unexpected result.
So what should we conclude? The 3-player beauty-contest game is a real guessing game and
there is no strategy that always wins the game. So we still have to guess what guides the others
in their choice and the kind of logic they may work with. The two classroom experiments
namely show that the students do not adopt the same heuristics of behavior in large N-player
guessing games and in 3-player guessing games. Yet the 3-player beauty contest game, in
comparison with the N-player game, has some mathematical properties that may help some
numbers to win with a large probability, even in a context where players follow different kinds
of logics. The 3-player game namely gives rise to win areas that are large for some numbers
and thin for others, and it brings dominance closer to standard level-k reasoning. So 10 becomes
a good number to play in a 3-player beauty contest game with parameter . In our
classroom experiments, whether the students are trained or not in game theory and guessing
games, whether they just think that the good way to proceed consists in guessing a number
between the two opponents’ numbers or whether they also know that low numbers can win the
game, 10 leads to win with a probability close to 0.82, which is a very nice performance. In
facts, for , 10 is a kind of nombre d’or that cumulates nice properties that help to win
with a large probability, with opponents sharing different logics. So let us play 10.
References
Breitmoser, Y. , 2012. Strategic reasoning in p-beauty contests, Games and Economic Behavior,
75, 555-569.
Costa-Gomes, M., Crawford, V.P., 2006. Cognition and behavior in two-person guessing
games: an experimental study, American Economic Review, 96, 1737–1768.
Grosskopf, B., Nagel R., 2008. The two-person beauty contest, Games and Economic Behavior
62, 93–99.
Ho, T.-H., Camerer C.,Weigelt, K., 1998. Iterated dominance and iterated best response in
experimental ‘p-beauty contests’, American Economic Review, 88(4), 947–969.
Ledoux, A., 1981. Concours résultats complets. Les victimes se sont plu à jouer le 14 d’atout.
Jeux et Stratégie 2 (10), 10–11.
6
For a different from 3/5, the facts justifying that areas A or B do not change are the same than for a=3/5, and
there always remains at least one of the reasons given for a=3/5 that justifies the enlargement of area A or B.
24
Lefebvre, M., Umbhauer, G. Traveler’s dilemma: how the value of the luggage influences
behavior, Cairn International, forthcoming.
Moulin, H., 1986. Game theory for social sciences, New York University Press.
Moulin, H, 1984. Comportement stratégique et communication conflictuelle : le cas non
coopératif. Revue Economique, 35,1, 109-146.
Nagel, R., 1995. Unraveling in guessing games: An experimental study, American Economic
Review, 85, 1313–1326.
Nagel, R., Bühren, C., Franck, B., 2017. Inspired and inspiring : Hervé Moulin and the
discovery of the beauty contest game, Mathematical Social Sciences, 90, 191-207.
