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A sad lesson from the hammer-nail game: strength is better than dexterity
Hammer-Nail game:
Part 1: Show your strength in the hammer-nail game: a Nim game with incomplete
information” Working paper Beta 2023 n°5, January 2023
Part 2: A sad lesson from the hammer-nail game: strength is better than dexterity
Gisèle Umbhauer
Beta - University of Strasbourg
August 2023
Abstract
In this second paper on the hammer-nail game, we confront strength with dexterity. The
hammer-nail game, a game played in the French TV show “Fort Boyard”, goes as follows: two
players are in front of a nail slightly driven into a wooden support. Both have a hammer and in
turn hit the nail. The winner is the first player able to fully drive the nail into the support. A
player is of strength f if he is able, with one swing of the hammer, to drive the nail at most f
millimeters into the support. A player is of non dexterity e if he is unable to hammer smoothly,
so that, with one swing of the hammer, he drives the nail at least e millimeters into the support,
with In a previous paper, we mainly studied the impact of strength, both players being
of high dexterity ( ), and we transformed the hammer-nail game into a Nim game with
incomplete information on strength. In this paper we study the impact of both strength and
dexterity. We confront two players of different strength and dexterity and namely show a sad
result: strength is more useful than dexterity to win the game. We also study the behavior in
front of incomplete information, either on strength or on dexterity.
Keywords: Nim game, crossed cycles, Fort Boyard, subgame perfect Nash equilibrium,
strength, dexterity, incomplete information, heuristics of behavior.
JEL Classification: C72
1. Introduction
As in an earlier paper (Umbhauer, 2023), we study a game played in the French TV show “Fort
Boyard”
1
, the hammer-nail game. The game goes as follows: two players, player 1 and player
2, are in front of a nail slightly driven into a wooden support (see figure 1). Both have a (same)
hammer and in turn hit the nail. At the beginning of the game, the head of the nail is at a distance
umbhauer@unistra.fr, BETA-University of Strasbourg, 61 Avenue de la Forêt Noire, 67085 Strasbourg Cedex,
France
I thank the third-year class students (year’s class 2020/2021 and year’s class 2022/2023) at the Faculté des
Sciences Economiques et de Gestion (Faculty of Economic and Management Sciences) of the University of
Strasbourg, who played traditional Nim games with stones. I especially thank one of these students, Bastien
Landwerlin, who mentioned the hammer-nail game, and thereby gave me the idea to work on it.
1
Adventure Live Productions/Banijay Group for France Télévision.
2
D from the support. In the French TV show Fort Boyard, player 1 is a candidate who wants to
win the game, in order to get more time to catch the “Boyards” (coins) and player 2 is a “Maître
du temps”, a person who belongs to the Fort Boyard TV team, who wants to impede the
candidate from winning.
In this paper we work with both strength and dexterity. The strength is measured in numbers of
millimeters. We say that a player is of strength 5 if he is able, with one swing of the hammer,
to drive the nail at most 5 millimeters into the support
2
. This means that a player of strength 5
is able, with one swing of the hammer, to drive the nail 1 millimeter (mm), 2mm, 3mm, 4mm
or 5mm into the support. The notion of dexterity we introduce in this paper is somewhat limited
in that a lack of dexterity just means an inability to hammer smoothly, i.e. the inability to just
drive the nail 1 millimeter into the support. So non dexterity -unskillfulness- is also measured
in millimeters: a player of unskillfulness e, with one swing of the hammer, drives the nail at
least e millimeters into the support. Hence a player of strength f but unskillfulness e is able,
with one swing of the hammer, to drive the nail from e to f millimeters into the support, with
. A player of high dexterity is characterized by and we say that a player is
more skilled when he is of higher dexterity than his opponent.
We require that each player, at each turn of play, at least drives the nail e mm into the support,
which means that he cannot simulate hammering the nail. As a matter of fact, once the head of
the nail is close to the support, but too far from it to be fully driven into with one swing of the
hammer, a player may be incited to not drive the nail further into the support, in order to impede
the opponent from winning the game at his next turn of play
3
.
So the game becomes a special Nim game (see Bouton, 1901). TV shows sometimes work with
Nim games (see for example the sticks game also played in the Fort Boyard TV show
(Umbhauer, 2016) or the Game of 21, played in an American TV show (Dufwenberg et al.
2010, Gneezy et al., 2010)). In this special Nim game, the (pure) strategy set of a player of
strength f and unskillfulness e contains (pure) strategies
4
. Throughout the paper, we
will note f and e, respectively F and E, the strength and the unskillfulness of one player,
respectively of the other player.
In Umbhauer (2023) we mainly worked with two players of high dexterity but of
different strengths and we introduced incomplete information on strength. Each player knew
2
We work with millimeters but we could work with any (smaller) measure of distance.
3
This requirement does not exist in the TV Fort Boyard game, but it is mathematically necessary to avoid that
the game never stops.
4
We further only say strategies instead of “pure” strategies, given that we only work with pure strategies.
Head of the nail
Distance D
Wooden support
Figure 1 : the nail at the beginning of the play
3
his strength but had incomplete information on the strength of the opponent. We built a perfect
Bayesian equilibrium without making use of the prior probability distributions on the unknown
strengths, which is not usual when dealing with games with incomplete information. In this
paper we confront strength with dexterity. We show that strength is more useful than dexterity
to win the game. To put it more precisely, when both players have the same number of strategies
but are of different strength and dexterity, the stronger player wins more often than the more
skilled one. When the two players have not the same number of strategies, then the player with
the larger strategy set wins the game when D is sufficiently large, but strength keeps an
advantage on dexterity.
In this paper we also draw attention to crossed cycles, to the link between incomplete
information and crossed cycles and to heuristics of behavior, namely in contexts with
incomplete information, either on strength or on dexterity.
In section 2, we briefly recall some results out of Umbhauer (2023) we use in this paper and we
introduce the notion of crossed cycle. Section 3 shows the impact of dexterity on the optimal
strategies when both players are of same strength and same dexterity. Section 4 studies the
optimal behavior of two players of same strength but different dexterity and provides a first
result about the impact of incomplete information on dexterity. Section 5 is the core section of
the paper. We show that at the subgame perfect Nash equilibrium, if two players have the same
number of strategies but one player is more skilled and the opponent is
stronger ( but ), then starting the game being stronger leads to more often winning
the game than starting the game being more skilled. Section 6 gives the optimal behavior when
the more skilled player has either a smaller or a larger strategy set than the stronger player. We
show that the player with the largest strategy set always wins the game when D is sufficiently
large, but that strength keeps an advantage on dexterity. We conclude the paper in section 7 by
giving heuristics of behavior in a context of partial incomplete information, either on strength
or on dexterity.
2. Some results on strength
We here recall some results on strength (out of Umbhauer, 2023) when both players are of high
dexterity ( ).
Throughout the paper, we simply say that a player plays s to say that he drives the nail s
millimeters into the support.
Proposition 1 (out of Umbhauer, 2023)
Consider two players of same strength f and of same high dexterity e=1. We call r the remainder
of the division of D by The set of Subgame Perfect Nash Equilibria (SPNE) is given
by:
Player 1 : At her first turn of play (first round), she plays r if , and she plays any integer
from 1 to f if . At her other turns of play, she plays in such a way that the remaining
distance after her turn of play is a multiple of . If this is not possible (because the distance
she observes at her turn of play is a multiple of ), she plays any integer from 1 to f.
4
Player 2: At each potential turn of play, he plays in such a way that the remaining distance
after his turn of play is a multiple of . If this is not possible (because the distance he
observes at his turn of play is a multiple of ), he plays any integer from 1 to f.
5
We now suppose that one of the players is stronger than the other: one player is of strength f,
and the other player is of strength F, with . In that case the strongest player always
wins the game, except if and the weakest player begins the game; moreover, the strongest
player can win in many ways (see Umbhauer, 2023).
This leads us to drawing attention to the notion of crossed cycle, which is central to any Nim
game. Given that both payers play in turn, a crossed cycle is necessarily the sum of the
millimeters played by both players in two successive rounds. To put it more precisely, a cycle
is the number of millimeters that at least one of both players is able to complete regardless of
what is played by the opponent at the previous round. In each game, there are two crossed
cycles. So, when both players are of same high dexterity and same strength f, the two
crossed cycles are , given that, whatever is played by one player, the other can complete
the number of millimeters to at the next round. This is illustrated in figure 2a. When both
players are of high dexterity and of different strengths f and F (with ), the two
cycles are and given that, whatever the number of millimeters played by the weaker
player, the stronger player can complete it to or . In contrast, the weaker player
cannot complete the crossed cycles. These cycles are illustrated in figure 2b.
Figure 2a Figure 2b Figure 2c: the cycles in dashed lines are unknown by player i.
The strongest player, in front of any remaining distance , can bring the weakest player in
front of a multiple of , simply by playing the remainder r of the division of d by if
r is different from 0, or by playing if . Yet in proposition 2 (Umbhauer, 2023), we
exploited the crossed cycle . We showed that, even in front of a remaining distance that
is a multiple of , say , k , the strongest player wins the game by
behaving as follows: she chooses 1, so that the opponent, who can only play from 1 to f, leads
her to a remaining distance that is not a multiple of , because this distance goes from
to .
So she plays an integer from to so that the remaining distance is a multiple of
, which leads her to winning the game. This argument is illustrated in figure 3.
Proposition 2 (out of Umbhauer, 2023)
We note e=1 the dexterity of both players, f the weakest strength, F the strongest one, with
. A possible SPNE is characterized by:
5
Given that 0 is a multiple of any number, the proposition includes that in front of a remaining distance lower
than or equal to the strength, a player simply plays the remaining distance, i.e. fully drives the nail into its
support.
f
1
f
1
+
F
1
f
1
+
fi
1
?
?
?
1
+
5
The weaker player: if, at his turn of play, the remaining distance is lower than or equal to f, he
plays the remaining distance. If not, he can play from 1 to f.
The strongest player: if, at his turn of play, the remaining distance is lower than or equal to F,
she plays the remaining distance. Otherwise, at any turn of play, she plays so that the weaker
player faces a remaining distance that is a multiple of . If this is not possible (because
she faces a remaining distance that is a multiple of ) she can play any integer from 1 to
.
The stronger player always wins the game in any SPNE, except if the weaker player starts the
game, with D lower than or equal to f.
Legend of figure 3: the rectangle represents the distances from k(F+1) to (k+1)(F+1). The horizontal
lines are the distances at which the stronger player and the weaker player are playing. The blue and
red lines represent the number of millimeters respectively played by the stronger and the weaker player
when called on to play.
Proposition 2 allows the stronger player, in front of a remaining distance that is a multiple of
, to play 1, but also to play any integer from 1 to . Yet by playing 1, the player of
strength F wins in front of any weaker player, so playing 1 is the good way to play even if the
stronger player ignores the strength f of the weaker player. This fact is exploited in proposition
3 which gives the optimal behavior when both players ignore the strength of the opponent. We
suppose that the strength of player i , , belongs to a set of integers going from 1 to an
upper bound
with fi ≥ 2. Given that both players are of high dexterity ( ), each player I
only knows one crossed cycle of the game,
. This is illustrated in figure 2c for player i.
The Perfect Bayesian Equilibrium given in proposition 3 exploits this knowledge given that
each player only works with the crossed cycle he knows.
Proposition 3 (out of Umbhauer, 2023)
Consider two players of high dexterity (e=1) that ignore the strength of the opponent. A (Perfect
Bayesian) Nash equilibrium of this game with high dexterity and incomplete information on
strength goes as follows:
At each turn of play, a player of strength f, regardless of the value of f, plays as follows:
- If the distance d he is confronted to is lower than or equal to f, he plays d.
- If , he plays so as to put the opponent in front of a remaining distance that is a
multiple of . If this is not possible (because d is a multiple of ), then he plays
1.
This result is robust, in that it holds even if the players have no information at all on the strength
of the opponent. It does not need a distribution of prior probabilities on the types of the
-f -1
k(F+1)+F-f k(F+1)+F-1
k(F+1) (k+1)(F+1)
k(F+1)+F
-(F-f) -(F-1) -1
Figure 3
stronger player
weaker player
6
opponent. This explains that we do not introduce the player’s beliefs at each turn of play, given
that they play no role in the result. And this explains why we put (Perfect Bayesian) in brackets,
given that we make no use of the bayesian rule to get the optimal way of behavior.
It follows from proposition 3 that a player behaves as if he were the strongest player, by playing
1 if the remainder r of the division of d by ( ) is 0, and r when r is different from 0.
3. Dexterity, enlargement of the crossed cycle and set of losing distances
We now allow players to not be of high dexterity. We recall that non dexterity – unskillfulness-
just means the inability to hammer smoothly, so that a player of unskillfulness e, with one swing
of the hammer, drives the nail at least e millimeters into the support. Dexterity matters only if
the remaining distance d the player is confronted to is larger than e. If , then lack of
dexterity is not a problem, because hammering like a beast for sure fully drives the nail into the
support. The main impact of unskillfulness e is to reduce the strategy set from below: the
strategy set of a player of strength f and unskillfulness e, with becomes .
In this section we suppose that both players are of same strength and of same dexterity. A first
consequence of unskillfulness is the enlargement of the size of the crossed cycles: when the two
players are of strength f and of unskillfulness e, the size of the two crossed cycles becomes
(see figure 4a), because each player is able to complete to the number of millimeters
played by the opponent at the previous round (as the opponent plays at least e millimeters and
at most f millimeters). A second consequence is that unskillfulness reduces the probability to
win the game for the player who starts the game. As a matter of facts, in the subgame perfect
Nash equilibrium, a player, when starting the game at distance , with k an integer
and r an integer from 1 to , wins the game if r goes from 1 to f, and loses the game if r
goes from to . So unskillfulness reduces the number of winning distances by
, even if one starts the game (we recall that for , the starting player loses the game
only if the starting distance is a multiple of ).
Proposition 4
Consider two players of strength f and unskillfulness e, with f >e ≥1. Each player, in front of
the distance d= k(f+e)+r, with k an integer and r an integer from 1 to f+e, wins the game for r
from 1 to f, and loses the game for r from f+1 to f+e. More precisely, when both players are of
same strength f and same unskillfulness e, with f>e>1, then, at the SPNE, at any distance d he
is called on to play, each player plays as follows:
- He plays max(e,d) if d≤f
6
and if f<d≤f+e, he can play any integer from e to f.
- If d>f and d=k(f+e)+ r, with k an integer larger than or equal to 1, r an integer from 1
to f+e:
if r goes from 1 to e-1, he plays e
if r goes from e to f, he plays r
if r goes from f+1 to f+e, he can play any integer from e to f.
Example: For and , a player, if starting the game, wins the game only if D is a
multiple of 8 plus 1, 2, 3, 4 or 5. He loses the game, despite he starts the game, when D is a
multiple of 8, or a multiple of 8 plus 6 or 7, i.e. in front of 37.5% of all possible distances.
6
We recall that if d<e, playing e drives for sure the nail into its support.
7
Proposition 4 is illustrated in figure 4b.
Legend of figure 4b: The two horizontal lines represent the distances at which player A and player B
are playing. L and W mean that the player loses and wins the game at the corresponding distance. Each
rectangle is a crossed cycle (f+e). The horizontal blue -red- segments represent distances where player
A -player B- wins the game. The blue -red- lines going from one horizontal line to the other are the
number of millimeters played by player A -player B-. The dashed lines are optimal ways of playing but
they lead to losing the game.
Proof of proposition 4
A player’s behavior is optimal in front of a distance (she wins the game) and it is also
optimal in front of a distance d, with , given that she loses the game whatever
she plays. So let us suppose that the players’ behavior is optimal in cycle k, i.e. in front of all
distances with k an integer larger than 0 and r an integer from 1 to
, and that this behavior leads the player to winning the game for r from 1 to f, and to losing
it for r from to . This is true for (obvious)
So we just have to show that the players’ behavior is optimal in cycle k+1, i.e. in front of the
distances , with r from 1 to .
If a player is at distance , with r from 1 to , then playing e is optimal because
the remaining distance becomes , with r from 1 to , and so the
opponent loses the game by assumption.
If a player is at distance , with r from e to f, then playing r is optimal because the
remaining distance becomes and so the opponent loses the game by assumption.
Finally, if a player is at a distance , with r from to , playing any integer
from e to f is optimal, because the distance of the opponent becomes a number going from
to , and so the opponent wins the game anyway, regardless of the
chosen integer.
4. Different dexterity, same strength
We now suppose that both players are of same strength f, but of different unskillfulness, e and
E, with .
The two crossed cycles of the game become and (see figure 5a). The more skilled
player can complete both crossed cycles. As a matter of facts, if the less skilled player plays
from E to f, she can play from to e in order to complete the crossed cycle and
f
e
f
e
+
(k-1)(f+e)+1 (k-1)(f+e)+f k(f+e)+1 k(f+e)+e k(f+e)+f
(k-1)(f+e)+f+1 k(f+e)+f+1
(k-1)(f+e)+f+1 k(f+e)+f+1
(k-1)(f+e)+1 (k-1)(f+e)+f k(f+e)+1 k(f+e)+e k(f+e)+f
L W W L L W W W L L
L W W L L W W W L L
-e -e -f -f -f -e
-e -e -f -f -f -e
(k-1)(f+e) k(f+e) (k+1)(f+e)
Figure 4a Figure 4b : same strength f, same dexterity e, f>e>1
Player A
Player B
8
from f to E to complete the crossed cycle . Inversely, the less skilled player can complete
none of both crossed cycles. If the more skilled player plays f, he cannot complete the crossed
cycle , and if the more skilled player plays e, he cannot complete the crossed cycle ,
given that . So each time the more skilled player can lead the less skilled one to
a distance that is a multiple of ( ) or ( ) she wins the game. Proposition 5 works with
the smallest crossed cycle, i.e. the crossed cycle composed of the common strength (the strength
of the opponent) and the unskillfulness of the more skilled player (e)
7
.
In facts, the more the initial distance D grows, the less numerous are the situations where the
more skilled player loses the game. Yet there are a certain number of distances where she will
not win, even if she starts the game. As a matter of facts, in the first cycle (i.e. in front of
distances from 1 to ), the more skilled player loses the game when she is in front of the
distances . It follows that the less skilled player can win the game in front
of the distances from to . And, if , there are additional distances in the
second cycle (distances from to ), namely the distances from
to , in front of which the more skilled player again loses the game, given that she can
only lead the less skilled player to a distance where he is winning (because
and ). Yet there are less distances ( ) where she loses
the game in the second cycle than in the first one (e distances), given that . So there are
less additional distances where the less skilled player can win the game. And so on: given that
the number of losing distances of the more skilled player is decreasing in each cycle (e in the
first cycle, in the second, in the third, more generally in cycle
k) the number of losing distances generally quickly goes to 0. And if , then the more
skilled player wins in front of all distances except if she initially faces the distances from
to .
Proposition 5
Consider two players of same strength f but different dexterities, e and E, with f>E>e. The
subgame perfect Nash Equilibria lead the more skilled player to always winning the game,
provided the initial distance she is confronted to is larger than or equal to K(f+e), with K the
integer part of (E-1)/(E-e). For lower distances, the more skilled player wins the game at all
distances except the distances from tf+(t-1)E+1 to t(f+e), with t from 1 to K.
A SPNE way of playing for the more skilled player consists, at distance d, in playing f when a
simple swing is not sufficient to go to a multiple of f+e (lower than d), to play e when a simple
swing leads below the closest multiple of f+e (lower than d), and to play the remainder of the
division of d by f+e, if the remainder goes from e to f.
Proof of proposition 5
The main steps of the proof have been given in introduction to the proposition. It remains to
prove the recurrence step.
We suppose that in cycle k, (distances , with r from 1 to ( )), the
more skilled player loses the game at the distances from to and
wins the game at the distances from to . And we suppose
also that the less skilled player loses the game at the distances from to
– we go a little beyond cycle k- and wins the game at the distances from
to . These facts are true for .
7
We show in the conclusion that working with the largest crossed cycle leads to the same optimal behavior.
9
We now turn to cycle (distances , with r from 1 to ( )).
Given that the more skilled player loses the game at the distances from to
, the less skilled player wins the game at the distances from to
, because, by playing from E to f, he brings the more skilled player in front of a
distance going from to .
We now consider the more skilled player. Playing e at a distance from to
is optimal because the remaining distance goes from to
where the opponent loses the game. Playing r, for r from e to f, at the distance
, is optimal because it leads the less skilled player to the distance where he
loses the game. Playing f at a distance from to is optimal
because the less skilled player will be at a distance from to , where he
loses the game. And when the more skilled player is at a distance from to
, then she loses the game whatever she plays because the remaining distance
necessarily goes from to , so the opponent wins the game. So
playing f is an optimal action among others.
We now turn to the less skilled player. We have already shown that he wins the game at the
distances from to . We now show that he loses the game in front
of all distances from to . This follows from the fact that he
can play at least E and at most f, which leads the more skilled player to a distance from
to where she wins the game.
The number of losing distances for the more skilled player in cycle k is
. To make sense, this number has to be larger than or equal to 1 so we need
.
Therefore, if the more skilled player is at a distance larger than with K the integer
part of
, she wins the game.
The proof is partly illustrated in figure 5b.
Legend of figure 5b: similar to the legend of figure 4b
Let us comment proposition 5, by first giving two examples that illustrate the advantage of the
more skilled player. For , we get , so the more skilled player, if
starting the game, only loses the game if the initial distance she is in front to is 7,8,9, 17,18 or
27. For , we get and the more skilled player wins the game at all
distances, except for 6, 7 and 14.
f
e
f
E
+
Figure 5a Figure 5b :same strength f, different dexterity, f>E>e
(k-1)(f+e)+1 kf+(k-1)E k(f+e)+1 (k+1)f+kE
kf+(k-1)E+1 (k+1)f+kE+1
(k-1)(f+E)+1 kf+(k-1)e k(f+E) (k+1)f+ke (k+1)(f+E)
kf+(k-1)e +1 k(f+E)+1 (k+1)f+ke+1
(k-1)(f+e) k(f+e) (k+1)(f+e)
W W L L W W L L
W W L L W W L L
-E -f -f -E
-e -f -f -f -e
more skilled player
less skilled player
10
Yet this doesn’t mean that the less skilled player wins seldom the game, especially when the
initial distance D is not large. Assume . For , , if the more skilled
player starts the game, she wins at 22 among 25 possible distances. But the less skilled player,
when he starts the game, wins when D goes from 1 to 5, from 9 to 12, and from 17 to 19. So he
wins the game at 12 among 25 possible distances, i.e. almost half of time. And the possibilities
to win the game can even be much larger. For example, if the less skilled player starts the game,
if , , and , then he wins at of all possible initial
distances. But for larger initial distances these percentages fall and they also fall when the
difference between e and E rises.
In other terms, when both players are of same strength, more dexterity gives a strong advantage
when the initial distance is large (the more skilled player always wins the game if the first
distance she is confronted to is larger than ). But if the initial distance is not very large,
if f is much larger than E and E is not much larger than e, then the most important fact becomes
again the position in the game (first or second player). So more dexterity needs a sufficiently
large initial distance to counterbalance the position in the game (first or second player),
something we did not observe when only different strengths (with the same high dexterity) were
at work (proposition 2).
An immediate consequence of proposition 5 is that, if one player is of high dexterity, i.e.
, the other player being of unskillfulness , and if both players are of same strength f,
with , then the player with high dexterity always wins the game except at distance
f+1 (because in this special case). So the less skilled player always loses the game, except
if he starts the game at the distances from 1 to f and from f+1+E to 2f+1.
Corollary of proposition 5 (out of Umbhauer, 2023)
When both players are of same strength f, one player being of high dexterity e=1, the other
being less skilled, with f>E>1, a SPNE way of playing is given by:
For the player with high dexterity at distance d:
If d ≤ f, she plays d. If , she plays in order to bring the opponent in front of a remaining
distance that is a multiple of . If this is not possible (because d is a multiple of ),
then she plays f.
For the player with unskillfulness E>1 at distance d: If d ≤ f, he plays max (E,d). If
, he plays from f to E so that the player with high dexterity is in front of distance
. If and , he can play any integer from E to f.
Observe that in front of a distance equal to , with , the player with high
dexterity, as in proposition 5, simply plays f. By so doing she exploits the unskillfulness of the
opponent, who is not able to play less than E, and therefore is unable to bring her in front of a
multiple of (namely ). As a matter of facts, given that the opponent is unable to
play less than E and given that he cannot play more than f, the player with high dexterity will
be in front of a distance going from to , and
therefore she can lead the opponent to the multiple and win the game. This way
of playing is illustrated in figure 6.
This behavior is in sharp contrast with the behavior in propositions 2 and 3 which confront two
players of different strengths (both of high dexterity ), where the stronger player plays 1
to win when she is in front of a multiple of (her strength) . By playing 1, she exploits the
fact that the weaker player is unable to play F, hence to bring her back to a multiple of .
11
In other terms, in propositions 2 and 3, the more gifted (stronger) player exploits the fact that
the opponent cannot play large numbers, whereas in proposition 5 and its corollary, the more
gifted (with higher dexterity) player exploits the fact that the opponent cannot play low
numbers.
The corollary can be exploited in a context with incomplete information. So assume that it is
common knowledge of both players that they are of same strength f, but that both players ignore
the dexterity of the other player. This amounts to saying that each player knows his own
dexterity but that he only knows that the opponent’s unskillfulness e goes from 1 to .
Then, if one player is of high dexterity, in a Perfect Bayesian Equilibrium, she can play as in
the corollary of proposition 5.
Proposition 6
Suppose that it is common knowledge of both players that they are of same strength f, but that
each player ignores the dexterity of the opponent, so that he only knows that the opponent’s
unskillfulness goes from 1 to f-1. If so, in any Perfect Bayesian Nash Equilibrium of the game,
a player with high dexterity (e=1) plays optimally when she adopts the behavior given in the
corollary of proposition 5.
Proof of proposition 6
The proof immediately follows from the fact that the strategy of the player with high dexterity
in the corollary of proposition 5 is optimal regardless of the unskillfulness E of the opponent,
when . So the only thing we have to show is that the strategy remains optimal when the
opponent is also of high dexterity ( ). This is obvious. In front of an opponent of strength
f and high dexterity , playing f in front of a distance that is a multiple of is optimal
because any behavior leads to losing the game (given that the opponent can always lead the
player back to a multiple of ). And playing the remainder of the division of the distance
by f+1 is of course optimal when the remainder is different from 0.
Observe that we get this optimal behavior without any information on the unknown value E of
the opponent, because the player with high dexterity builds her strategy on the crossed cycle
she is informed on (f=strength of the opponent +1=her own dexterity).
5. Strength is better than dexterity when the strategy sets are of same dimension
Let us now suppose that one of the player is stronger and that the other is of higher dexterity,
but that both players have the same number of strategies, , and . So
f
1
f
E
+
(k-1)(f+1)+2 (k-1)(f+1)+f+2-E
(k-1)(f+1) k(f+1) (k+1)(f+1)
k(f+1)+1
-2 -f+E-2 -f
-f -E
player with high dexterity (e=1)
player with lower dexterity E
Figure 6 : same strength f, different dexterity, e=1 and f>E>1
12
the two crossed cycles become and and both players can complete them (figure
7a). We should conjecture the existence of a subgame perfect Nash equilibrium that gives the
same opportunities to win the game to both players. Yet this conjecture is wrong.
Proposition 7
Consider two players, one player with strength f and unskillfulness e, the other with strength F
and unskillfulness E, with F>f, E>e and F+e=f+E. At any SPNE, when the more skilled player
plays at distance d, she wins the game if the remainder of the division of d by f+E is 1 up to f,
and loses the game if else. When the stronger player plays at distance d, he wins the game if
the remainder of the division of d by f+E is 1 up to F, and loses the game if else.
The subgame perfect Nash equilibria of the game are given by:
For the more skilled player, in front of a distance k(f+E)+r, with k an integer and r an integer
from 1 to f+E: for r from 1 to e-1, she plays e, for r from e to f, she plays r, for r from f+1 to
f+E, she can play any integer from e to f.
For the stronger player, in front of a distance k(f+E)+r, with k an integer and r an integer from
1 to f+E: for r from 1 to E-1, he plays E, for r from E to F, he plays r, for r from F+1 to f+E,
he can play any integer from E to F.
Proposition 7 amounts to saying that the stronger player, when starting the game, wins the game
in situations whereas the player with better dexterity, when
starting the game, only wins the game in situations. So a better
dexterity is always to the advantage of the opponent, given that it lowers .
If two players are very different (a very strong –non skilled- player and a high skilled -but weak-
player) this may lead to very asymmetric probabilities of winning, which are always to the
advantage of the stronger player. For example, for , , and , the stronger
player wins the game if he starts the game in front of any distance different from a multiple of
7. By contrast, the more skilled player, if she starts the game, wins the game only in front of a
distance that is a multiple of 7 plus 1, 2 or 3. This amounts to saying that in front of a random
distance, the stronger player wins the game with probability 0.86, whereas the high dexterity
player only wins the game with probability 0.43 . So even if the more skilled player starts the
game, she loses more than half of time, when the starting distance is randomly chosen.
Things are less extreme if , in which case is larger than 0.5, so that
the more skilled player, if starting the game, wins the game in front of more than half of the
distances. Yet the stronger player, when starting the game, always wins more often. For
example, in the more equilibrated context where , , and , the more
skilled player loses the game in front of a distance that is a multiple of 8, or a multiple of 8 plus
r, with or 7. By contrast, the stronger player only loses the game when he is in front of a
multiple of 8. So, when starting the game, the high dexterity player loses the game in front of
37.5% possible distances whereas the stronger player, if starting the game, only loses in front
of 12.5% possible distances.
Proof of proposition 7
We proceed by recurrence. We call A the more skilled player (she) and B the stronger player
(he). Let us consider cycle k, i.e. the distances , with r from 1 to .
We suppose that A, in front of a distance d, wins the game if d goes from
to , loses the game if d goes from to
13
and that B wins the game if d goes from to and loses
the game if d goes from to .
This is true for when A and B play as in proposition 7 and these actions are optimal
(obvious).
We now consider cycle (distances with r from 1 to ).
First, for A, playing e is optimal if she faces a distance from to ,
because this leads B to a distance that goes from to where
he loses the game by assumption. And playing r if she faces a distance k(f+E)+r with r from
to f is optimal because it brings B in front of the distance where he loses the
game.
In a symmetric way, for B, playing E is optimal if he faces a distance from to
, because this leads A to a distance that goes from to
where she loses the game by assumption. And playing r if he faces a distance
with r from to F is optimal because it leads A to the distance where she
loses the game.
Second, A, at any distance from to ,
loses the game whatever she plays because she can only bring B in front of a distance that goes
from to where B wins the game. So any action is optimal given
that all actions lead to losing the game.
In a symmetric way, B, at any distance from to , loses the
game whatever he plays because he can only bring A in front of a distance that goes from
to , where A wins the game. So any action is optimal given that all actions
lead to losing the game.
So what was is true for cycle k is also true for cycle and we proved that the strategies in
proposition 7 are optimal.
The proof is partly illustrated in figure 7b.
Legend of figure 7b: similar to the legend of figure 4b.
6. Strength and dexterity when the strategy sets are of different sizes
In section 2 and section 4 we established that when two players are of same (high) dexterity but
different strength or of same strength but different dexterity, then the player with the larger
strategy set always wins the game, provided the initial distance he is confronted to is sufficiently
f
e
F
E
+
(k-1)(f+E)+1 (k-1)(f+E)+f k(f+E)+1 k(f+E)+f
(k-1)(f+E)+f+1 k(f+E)+e k(f+E)+f+1
(k-1)(f+E)+1 (k-1)(f+E)+F k(f+E)+1 k(f+E)+E k(f+E)+F
(k-1)(f+E)+F+1 k(f+E)+F+1
(k-1)(f+E) =(k-1)(F+e) k(f+E)=k(F+e) (k+1)(f+E)=(k+1)(F+e)
L W W L L W W W L L
L W W L L W W W L L
-e -e -f -f -e
-E -E -F -F -E
more skilled player
stronger player
Figure 7a Figure 7b : different strength and dexterity, same number of strategies, F>f, E>e, f-e=F-E
14
large. These two contexts gave rise to two crossed cycles of different sizes that only the player
with the largest strategy set was able to complete.
This leads us naturally to investigating the case where , with and ,
so that the more skilled player has a larger or a smaller strategy set than the stronger one
(respectively and ). This context gives rise to two crossed cycles
f+E and F+e of different sizes, and it is reasonable to conjecture that the player with the larger
strategy set will win the game in front of more distances than the other player. This will indeed
be the case, but strength keeps an advantage on dexterity.
To show this, we first study the case where the stronger player has a larger strategy set than the
more skilled one, so , which implies that only the stronger player is able to
complete the two crossed cycles and , with . So each time the
stronger player can lead the more skilled one to a distance that is a multiple of ( ) or (
) he wins the game. Proposition 8 works with the smallest crossed cycle .
Proposition 8
Consider two players, one player with strength f and unskillfulness e, the other with strength F
and unskillfulness E, with F>f, E>e and F+e>f+E.
If F>f+E, in the SPNE, the stronger player wins the game in front of any distance if he has the
opportunity to play. The more skilled player only wins the game if she starts the game at a
distance lower than or equal to f.
If F<f+E, then, in any SPNE, if the stronger player is confronted to any distance larger than
(f+E)K, with K the integer part of (e-1)/(F+e-(f+E)), he always wins the game. In each cycle k
(i.e. at distances (k-1)(f+E)+r, with r an integer from 1 to f+E and k an integer such that 1≤ k
≤ K), the stronger player only loses the game at the distances from kF+(k-1)e+1 to k(f+E); this
number of distances is decreasing in k .
The more skilled player, if confronted to a distance larger than (f+E)(1+K) always loses the
game. In each cycle k, with 1≤ k ≤ 1+K, she only wins the game at the distances from (k-
1)(F+e)+1 to kf+(k-1)E; this number of distances is decreasing in k .
Proof of proposition 8
If , then the stronger player (B, he) obviously wins the game at all distances of the
first cycle (distances from 1 to ). He ends – hence wins- the game at all distances from
to F. Then, at the distances from to , he can play E, so that the more skilled
player (A, she) can only lose the game. And at the distances from to , he
plays such that the more skilled player is in front of the distance , where she loses the
game. So B wins the game at all distances from 1 to and therefore also wins the game
at larger distances, because A can never lead him to a distance where he loses the game. It
follows from this fact that A can only win the game if she starts the game at a distance lower
than or equal to f.
If , we suppose that, in cycle k (distances , with r from
1 to ), the stronger player wins the game at the distances from ( to
and loses the game at the distances from to . We
also assume that the more skilled player wins the game at the distances from
to and loses the game at the distances from to
(we go a little beyond cycle k).
These facts are true for (obvious).
15
We now study cycle k+1(distances , with r from 1 to ).
The facts in cycle k imply that the stronger player B wins the game at the distances from
to , because by playing adequately from E to F, he leads A in front of a
distance going from to , where A loses the game. And these facts
imply that the more skilled player A wins the game at the distances from to
, because by playing adequately from e to f, she leads B to a distance going from
to where B loses the game.
In turn these new facts imply that the stronger player B loses the game at the distances from
to , because, whatever he plays, from E to F, he brings A
to a distance from to , where A wins the game. And these new
facts also imply that the more skilled player A loses the game at the distances from
to , because, whatever she plays, from e to f, she leads B to a distance
from to , where B wins the game.
So the facts are true for all cycles k, but only make sense if the distances from
to exist, hence for i.e.
. Observe that and are
decreasing in k given that , so the number of B’s losing distances, respectively
A’s winning distances, diminish from one cycle to the next, up to disappear.
The proof is illustrated in figure 8b.
Legend of figure 8b: similar to the legend of figure 4b
We comment this proposition with two examples, , , , and ,
, , .
In the first example, we get and, in the first cycle (distances from 1 to 8), the stronger
player B loses the game when starting at the distances from to , i.e. only at distance
8. At any other distance, he wins the game. And the more skilled player A, when starting the
game, only wins the game in the first cycle at the distances from 1 to , and in the second
cycle at the distances from to . At any other distance she loses the game.
In the second example, we get . In the first cycle (distances from 1 to 11), the stronger
player B loses the game when starting the game at the distances from to .
In the second, respectively in the 3rd cycle, he loses the game at the distances from
to , respectively at the distance . In
front of all other distances he wins the game.
f
e
F
E
+
Figure 8a Figure 8b : different strength and dexterity, different number of strategies, F>f, E>e, F-E>f-e
more skilled player
stronger player
(k-1)(f+E) k(f+E) (k+1)(f+E)
(k-1)(F+e) kf+(k-1)E k(F+e) (k+1)f+kE (k+1)(F+e)
(k-1)(F+e)+1 kf+(k-1)E+1 k(F+e)+1 (k+1)f+kE+1
(k-1)(f+E)+1 kF+(k-1)e k(f+E)+1 (k+1)F+ke
kF+(k-1)e+1 (k+1)F+ke+1
L W W L L W W L L
-e -f -f -e
-E -F -F -F -E
L W W L L W W L L
16
The more skilled player wins in front of the distances from 1 to 6 in the first cycle. In the second
cycle, respectively in the 3rd and in the 4th cycle, she wins at the distances from
to , respectively at the distances from to , and
at the distances from to . In front of all other distances she
loses the game.
So, when we focus on the three first cycles, i.e. at the distances from 1 to 33, where the stronger
player may lose, we see that the stronger player can only lose the game in front of 6 distances,
hence that he wins the game in front of 27 distances. By contrast, the more skilled player, when
starting the game, can only win the game in front of 15 distances.
Things are different when the more skilled player has a larger strategy set, i.e. when
, with and . This time, only the more skilled player is able to complete the
two crossed cycles and (see figure 9a), so that by bringing the stronger player to
a multiple of ( ) or ( ) she wins the game. But the winning process is slower than in
the previous case. Proposition 9 again works with the shortest crossed cycle, .
Proposition 9
Consider two players, one player with strength f and unskillfulness e, the other with strength F
and unskillfulness E, with F>f, E>e and f+E>F+e. In any SPNE, when the more skilled player
is confronted to any distance larger than (F+e)K, with K the integer part of (E-1)/(f+E-(F+e)),
she always wins the game. In each cycle k (distances (k-1)(F+e)+r, with r from 1 to (F+e) and
k an integer from 1 to K), she only loses the game at the distances from kf+(k-1)E+1 to k(F+e);
this number of distances is decreasing in k .
The stronger player, if confronted to a distance larger than (F+e)(1+K) always loses the game.
In each cycle k, with 1≤ k ≤ 1+K, he only wins the game at the distances from (k-1)(f+E)+1 to
kF+(k-1)e; this number of distances is decreasing in k .
We observe that, for a same value the threshold cycle K (largest cycle
where the player with the largest strategy set can lose the game) in proposition 8 is lower than
in proposition 9 (because ). We also see that, in a cycle k (lower than or equal to the
threshold cycle), when the more skilled player has the largest strategy set, she wins the game in
front of distances, whereas the stronger player, in the same
cycle, when he has the largest strategy set, wins at more distances given that he wins the game
at distances. By constrast, when the stronger player has the
largest strategy set, the more skilled player loses in front of
distances, whereas the stronger player, when the more skilled player has the largest strategy
set, only loses in front of distances. So clearly strength keeps an
advantage on dexterity.
Proof of proposition 9
We suppose that in cycle k (distances , with r from 1 to ) the more
skilled player (A, she) wins the game at the distances from to
and loses the game at the distances from to . We suppose also
that the stronger player (B, he) wins the game at the distances from to
and loses the game at the distances from to (we go a
little beyond cycle k).
These facts are true for (obvious).
We now study cycle k+1 (distances , with r from 1 to .
17
The facts in cycle k imply that the more skilled player A wins the game at the distances from
to because, by playing from e to f, she brings the stronger player
in front of a distance going from to where he loses the game.
Symmetrically, the stronger player B wins the game at the distances from to
, because by playing from E to F, he brings the more skilled player in front of a
distance going from to where she loses the game. These new
facts imply that A loses the game at the distances from to ,
because playing any action from e to f can only lead the stronger player to distances going from
to where he wins the game. And B loses the game at the distances
from to because playing any action from E to F can only
lead the more skilled player to distances going from to where
she wins the game.
So the facts are true for all cycles k, but only make sense if the distances from
to exist, i.e
The proof of proposition 9 is partly illustrated in figure 9b.
Legend of figure 9b: similar to the legend of figure 4b
We compare proposition 9 to proposition 8 thanks to the example: , , and
. This example is close to the second example we studied for proposition 8. For ,
, and , we have again two crossed cycles of length 11 and 12 but this time
. We have also two strategy sets with 4 and 3 strategies, but this time
the more skilled player has the largest strategy set.
According to proposition 9, the threshold cycle K checks , instead of in the
example for proposition 8, which clearly shows that the stronger player has more possibilities
to win the game than the more skilled player in the symmetric configuration.
To put it more precisely, respectively in the 1st, 2nd, 3rd, 4th and 5th cycle, the more skilled player
A loses the game respectively in front of 5 distances (from 7 to11), 4 distances (from 19 to 22),
3 distances (from 31 to 33), 2 distances (from 43 to 44) and 1 distance (55). She wins the game
in front of all distances larger than 55. The stronger player B, respectively in the 1st, 2nd, 3rd,
4th, 5th and 6th cycle, wins the game in front of respectively 8 distances (from 1 to 8), 7 distances
(from 13 to 19), 6 distances (from 25 to 30), 5 distances (from 37 to 41), 4 distances (from 49
to 52) and 3 distances (from 61 to 63). He loses the game at all distances larger than 63.
f
e
F
E
+
(k-1)(F+e)+1 kf+(k-1)E k(F+e)+1 (k+1)f+kE
kf+(k-1)E+1 (k+1)f+kE+1
(k-1)(f+E) kF+(k-1)e k(f+E) (k+1)F+ke (k+1)(f+E)
(k-1)(f+E)+1 kF+(k-1)e+1 k(f+E)+1 (k+1)F+ke+1
(k-1)(F+e) k(F+e) (k+1)(F+e)
-E -F -F -E
-e -f -f -f -e
L W W L L W W L L
L W W L L W W L L
more skilled player
stronger player
Figure 9a Figure 9b: different strength and dexterity, different number of strategies, F>f, E>e, F-E<f-e
18
So when we focus on the three first cycles (distances from 1 to 33), we see that, when starting
the game, both the more skilled player and the stronger player win the game in front of 21
distances, despite the more skilled player has a larger strategy set. This contrasts with the
previous example for proposition 8, where the stronger player, when starting the game, won the
game in front of 27 distances among 33, whereas the more skilled player, when starting the
game, only won the game in front of 15 distances. So strength has a strong advantage on
dexterity, when the starting distance D is not too large.
7. Conclusion, incomplete information and heuristics of behavior
Well, several facts emerge from our study.
First of all, we see that strength has a significant advantage on dexterity. This is obvious when
the more skilled player and the stronger player have the same number of strategies (section 5),
but keeps true when the size of both strategy sets is different. In the latter situation, the player
with the larger strategy set always wins the game when the starting distance D is larger than a
given threshold value. Yet, on the one hand, this threshold value is larger when the more skilled
player has a larger strategy set than in the opposite situation. And, on the other hand, in front
of distances lower than this threshold value, the winning distances of the stronger player (when
the more skilled one has a larger strategy set) are more numerous than the winning distances of
the more skilled player (when the stronger one has a larger strategy set). For the hammer-mail
game, this is clearly not good news, but can be tempered by the observation that the dexterity
we introduce is a quite limited notion of dexterity. We can reasonably expect that a more
sophisticated notion of dexterity would help us nuancing this result! Yet more generally, the
results in the paper establish that, in a Nim game, reducing the strategy set from below, i.e.
eliminating small strategies -which amounts to decreasing dexterity-, is not as damaging as
reducing the strategy set from above, i.e. eliminating large strategies – which amounts to
decreasing the strength.
Another fact is that we chose in the proofs and the propositions to only work with one crossed
cycle, most often the smallest one (from proposition 4 to proposition 9). This is of no
importance as regards the winning and losing distances for the players. Yet it is interesting to
do a study with the largest crossed cycle in that this will help us to give some hints of behavior
in presence of incomplete information on strength or dexterity (with or
.
So, without loss of generality, we suppose that , hence , and call
player A(she), respectively player B(he), the player with strength f and unskillfulness e,
respectively the player with strength F and unskillfulness E. From now on, F may be larger or
smaller than f and e may be smaller or larger than E , but we suppose that , to exclude
the case where player A wins the game in front of any distance. Player A can complete both
crossed cycles and and we now focus on the largest crossed cycle .
We suppose that in cycle k (distances , with r from 1 to ) player A
loses the game at the distances from to and wins the game at the
distances from to (so we go a little beyond cycle k).We also
suppose that player B wins the game in front of the distances from to
and loses the game at the distances from to . These facts
are true for . For A also wins the game at the distances from 1 to f.
19
We now consider cycle (distances , with r from 1 to ).
The facts in cycle k imply that player B wins the game at the distances from to
, because, by playing from E to F, he can bring A to distances going from
to where A loses the game. By contrast, B loses the game at the
distances from to , because any action from E to F leads A
to distances going from to where A wins the game. In turn, it
follows that player A loses the game at the distances from to
, because any of her action from e to f leads B to distances going from to
where B wins the game. By contrast, A wins the game at the distances from
to , given that, by playing from e to f, she can lead B to
distances going from to where he loses the game.
So the reasoning is true for all cycles k, as long as hence
.
This way of reasoning is illustrated in figure 10.
Legend of figure 10: similar to the legend of figure 4b
It follows from the reasoning, as expected, that figures 9b and 10 are identical except for the
size of the chosen cycle. Yet figure 9b and figure 10 will be useful in different contexts of
partial incomplete information.
As a matter of facts, it derives from our study that to build a strategy, each player has to know
at least one crossed cycle, so has to know either the strength of the opponent or his dexterity. It
is not possible to propose a behavior when a player both ignores the strength and the dexterity
of the opponent. Sometimes, like in proposition 3 and in proposition 6, only knowing the
dexterity or the strength is enough to build a Perfect Bayesian Equilibrium. But most often, this
is not sufficient. Yet, when a player knows one crossed cycle, we can propose a heuristic of
behavior that fits well if luckily the player has the largest strategy set: anyhow, if he has the
smallest strategy set, he has few chances to win the game, at least if the other player follows
the heuristic. In other terms, as long a player knows the lowest strategy of the opponent or his
largest one, we can propose a way to play the game.
So let us again consider player A (with strength f and dexterity e) and player B (with strength F
and dexterity E). F may be larger or smaller than f and e may be smaller or larger than E. For
player A, there are only two possible configurations of partial incomplete information given in
figure 11a and figure 11b. In configuration 1, player A knows the opponent’s strength F but not
Figure 10: different strength and dexterity, different number of strategies, f-e>F-E
kf+(k-1)E k(F+e) (k+1)f+kE (k+1)(F+e) (k+2)f+(k+1)E
kf+(k-1)E+1 k(F+e)+1 (k+1)f+kE+1 (k+1)(F+e)+1
(k-1)(f+E)+1 kF+(k-1)e k(f+E)+1 (k+1)F+ke
kF+(k-1)e+1 (k+1)F+ke+1
(k-1)(f+E) k(f+E) (k+1)(f+E)
-E -F -F -E
-f -e -e -e -f
W W L L W W L L W W W
L W W L L W W L L
Player A
Player B
20
his dexterity E, in configuration 2, player A knows the opponent’s dexterity E but not his
strength F.
We first study configuration 1. In this configuration, if player A expects to be the player with
the largest strategy set, hence , the crossed cycle she knows, , is the
smallest one, and we can build a strategy (a heuristic of behavior) that uses this crossed cycle.
The heuristic is: at any distance d, complete to a multiple of F+e if possible; if this is not
possible, play e when the remainder r of the division of d by (F+e) goes from 1 to e-1, and play
f for r=0 and r going from f+1 to F+e-1.
Observe that this is what we proposed to a player of high dexterity ( ) when she meets a
player of same strength but unknown dexterity (proposition 6). This heuristic immediately
follows from the behavior in figure 9b. In figure 9b, which illustrates the optimal behavior
behind proposition 9, there is a degree of liberty: when being at the distances from (
to , player A loses the game whatever she plays, so she can play f. By
contrast, when she is at distance , she has to play f to make player B lose the
game, and at all distances , with r from to ,
playing f is a good way of playing.
So by playing f at all distances from to , player A is sure to play
f at all the distances she has to do so (namely at the distance ) even if she does
not know E. So she plays well, even in a context of incomplete information on E, when
. Moreover, by so playing, she leads B faster to lower distances. So if E, luckily, is
large, she brings B faster to distances where he loses the game (the green segment in figure 12a
is growing in E). So, given that player B, when , cannot win more often than at
his winning distances in figure 9b, even if he knows player A’s strategy, player A wins at least
as often as in a context of complete information. These facts are illustrated in figure 12a.
We now study configuration 2. In this configuration, if player A expects to be the player with
the largest strategy set, hence , the crossed cycle she knows, , is the largest
f F
+
e ?
f ?
+
e E
Figure 11a Configuration 1
Figure 11b Configuration 2
k(F+e)+1 k(F+e)+e (k+1)f+kE
k(F+e)+f (k+1)f+kE+1
k(f+E) (k+1)F+ke
k(f+E)+1 (k+1)F+ke+1
k(F+e) (k+1)(F+e)
-f -e -e -f -f -f -f -e
L W W L L W
L L L W W L L
Figure 12a : player A
does not know E
Player B
Player A
21
one, and we can build a strategy (heuristic of behavior) that uses this crossed cycle (so that
works with figure 10).
The heuristic is: at any distance d complete to a multiple of if possible; if this is not
possible, play e.
Observe that this is what we proposed to a player of high dexterity who faces a player
of same dexterity but unknown strength (proposition 3). Here, we exploit optimally the degree
of liberty possible in the context of complete information studied in figure 10. In this context,
player A (the player with the largest strategy set) loses at the distances from
to whatever she plays, so she can play e. But when she is at distance
she has to play e to make player B lose the game. And playing e at the distances
from to and from to leads player
B to losing the game. So, by playing e each time she cannot bring player B to a multiple of
, player A, on the one hand, is sure to play e when it is necessary, namely at distance
, even if she does not know F. So she plays well even in a context of incomplete
information on F. Moreover, on the other hand, by playing e, when F, luckily, is low, player A
brings player B faster to the green segment (see figure 12b) where B is losing the game. So,
given that player B, when , cannot win more often than at his winning distances
in figure 10, even if he knows player A’s strategy, player A wins at least as often as in a context
of complete information. These facts are illustrated in figure 12b.
In some way, these two heuristics of behavior, that exploit e and f, may partly explain why so
many players in classic Nim games choose to either play f or e when they are unable to discover
the optimal strategy.
References
Bouton, C.L., 1901. Nim, a game with a complete mathematical theory. Annals of Mathematics,
3 (1/4), 35–39.
Dufwenberg,M., Sundaram, R. Butler, D.J., 2010. Epiphany in the game of 21, Journal of
Economic Behaviour and Organization, 75, 132-143.
Gneezy, U., Rustichini, A., Vostroknutov, A., 2010. Experience and insight in the race game,
Journal of Economic Behaviour and Organization, 75, 144-155.
Umbhauer, G., 2016. Game theory and exercises, Routledge Editors.
Umbhauer, G., 2023. Show your strength in the hammer-nail game: a Nim game with
incomplete information, Working Paper Beta 2023 n°05.
Figure 12b : player A
does not know F
(k+1)f+kE (k+1)(F+e)
(k+1)f+kE+1 (k+1)(F+e)+1
k(f+E)+1 (k+1)F+ke
(k+1)F+ke+1
k(f+E) (k+1)(f+E)
-e -e -f -e -e -e -e
W k(f+E)+e W L L W W
L L W W L L
L
Player B
Player A
