Eponine Lupo
May 3, 2011
MAT 5900
Seminar paper
Game Theory and the Nash Equilibrium
Game Theory is a mathematical theory that deals with models of conflict and
cooperation. It can be applied to many social sciences, evolutionary biology, and has many
applications in economics. In economics Game Theory can be applied to oligopolies, market
equilibriums, choosing the location of firms, bargaining, public goods and more. (Nash 164). It is
a precise and logical description of a strategic setting. It considers how one should make
decisions and, to an extent, how one makes them. Game Theory is often used in more complex
situations and is also used as a decision-making tool in circumstances where chance and a
player’s choice are not the only factors that are contributing to the outcome. (Davis 4). A major
aspect of Game Theory is the Nash Equilibrium. A set of strategies is a Nash Equilibrium if no
player can do better by unilaterally changing his strategy. This paper looks to briefly explain the
basics of Game Theory and the Nash Equilibrium and then to look at various games and the
Nash Equilibrium in the statistical software package R.
In Game Theory, a “game” is a situation where the outcome is determined by the strategy
of each person involved. In a game, there are players and each player has a strategy to make a
decision. The result of each game is a payoff. A “player” is a participant in the game, and is not
necessarily a single person. It can be a corporation, a country, or a team with all members having
the same goals and strategies. (Davis 6). Each game can have two or more players. It can never
have only one player. If a “game” only has one player, it is not a game because there is no
strategy involved. A “strategy” is a complete contingent plan outlining all the actions a player
will do under all possible circumstances. (Watson 24). All decisions are merged into one to form
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a single strategy for a game. A “game” cannot be a true game if there is no strategy. For instance,
if there is only one decision, to turn right or to turn right, there is no strategy, the player turns
right; this is not a game.
In Game Theory, there are a few key assumptions each player undertakes. Each player
knows how many players are in the game, has complete information, knows the goals of the
other players are to maximize their payoffs, and assumes each player is a rational decision
maker. This last assumption implies that players are constantly in pursuit of their own objectives,
to maximize their payoffs. In addition, an assumption is made that players are rational because if
players are irrational, there would be no way to make sense of a game or to solve it with any
degree of certainty. There would be no point to Game Theory if players were irrational because
nothing would make sense; it would just be a jumble of one player after another player making a
decision without rhyme or reason. The assumption of complete information means that each
player knows the strategies of the other players and knows how they will act in order to
maximize their payoffs. Each player knows the other players’ payoffs for each strategy they can
take. In other words, in a two player game, player one knows everything there is to know about
player two and vice versa. When a game theorist analyzes a game, he says “a player in the game
is intelligent if he knows everything [the game theorist] knows about the game and can make any
inferences about the situation.” (Myerson 4).
A game can be modeled in two forms: a normal, or matrix, form or an extensive form.
The normal form is good when players move simultaneously. However, when one player acts
before another and his act is observed by the other players, the normal form is incapable of
completely conveying this strategically crucial piece of information. It is better to model this
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game in the extensive form. (Heap 45). An example of a two player game is the classical game of
prisoner’s dilemma, shown here.
Figure 1.
Prisoner’s dilemma is one of the most well-known classical games. The story goes as
follows. The authorities have captured two criminals whom they know are guilty of a certain
crime. They only have enough evidence to convict them of a minor offense. If neither criminal
confesses, then they will only be charged with this minor offense. The authorities separate the
criminals in the hope that one of them will turn on the other. If both criminals stay quiet and
collude (C) with each other, they will get a minimum sentence in jail. However, if one criminal
deviates (D) from the other criminal, the criminal who deviates will get a lower sentence (a
higher payoff) than if they both stayed silent. If they both deviate, they will both have a heavier
sentence (lower payoff) than if they had both remained silent. The best outcome for a criminal is
to deviate while the other player colludes; the next best outcome occurs when both criminals
colludes; the worst payoff for a criminal occurs when that criminal colludes and the other
criminal deviates. The extensive form of the prisoners’ dilemma is shown below. It is the exact
same game as above but depicted in a tree form. A dotted line joining two or more of a player’s
nodes indicates the players move simultaneously, player two does not know what decision player
one has made.
Figure 2.
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Two player games with more than two strategies or with an unequal number of strategies, for
example, player one has three strategies and player two has two strategies, are depicted in a
similar fashion as the prisoners’ dilemma above. Three or more player games are also depicted in
a similar manner. However, the extensive form is more conducive to portraying games with four
or more players because the normal form gets too complicated if too many players are involved.
A three player game in the normal form can look like this:
L
R
Figure 3.
3
Three concepts to solve a game include dominance, efficiency, and best response.
Dominance says S1 is dominated by S11 if S11 gives player one better payoffs than S1, no matter
what the other players do. S1 is a strategy for player one and S11 is another strategy. The
dominance concept compares two or more strategies for a single player. (Watson 52). This is
exhibited in the prisoners’ dilemma game. In the prisoners’ dilemma, strategy D, deviate,
dominates strategy C, collude, for player two because a payoff of 3 is greater than a payoff of 2
and a payoff of 1 is greater than a payoff of 0. The same is true for player one, 3>2 and 1>0.
Therefore, for both players, D dominates C and the solution to the game is (D,D). Efficiency
compares two strategy combinations involving all the players of a game. It says S is more
efficient than S1 if every player prefers S to S1 or is indifferent between the two. S is said to be
efficient if there is nothing that is more efficient than S. (Watson 55). Efficiency is clearly seen
in the classical game of pareto coordination.
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Figure 4.
Here, (A,A) is more efficient than all the other strategy combinations because both player one
and player two prefer (A,A) to (A,B), (B,A,) or (B,B). Therefore, (A,A) is efficient and the
solution to the game. The best response concept dictates that a player will select the strategy that
gives him the greatest expected payoff knowing what his opponants’ strategies are. In other
terms, S1 is a best response to S2 if S1 gives player one the highest payoff given player two is
playing S2. Looking at the above game of pareto coordination, if player one chooses to play B
and player two knows this, player two’s best response is to play B as well. Players in games have
pure strategies and, additionally, can have mixed strategies. In the prisoners’ dilemma, player
one is said to have two pure strategies, C and D. According to Heap, “If a player has N available
pure strategies (S1,S2,…,SN), a mixed stategy M is defined by the probabilities (p1,p2,…,pN) with
which each of his pure strategies will be selected.” All of the probabilities must be between zero
and one and must sum to one. A mixed strategy is a probability distribution over the pure
strategies of a player. There are an infinite number of mixed strategies for a player. A player
chooses a mixed strategy over a pure strategy in order to keep one’s opponents guessing. For
example, when the opponent can benefit from knowing the next move. A player also chooses a
mixed strategy if a game is unsolvable using pure strategies, whether or not there are dominant
or efficient strategies. In order to calculate the expected payoffs for each player, the mixed
strategies of each player must be known. If s1 = (p , (1-p)) and s2 = (q , (1-q)) in the game:
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Figure 5.
then the general payoff for player one would be u1(s1 , s2) = p*q*6 + p*(1-q)*2 + (1-p)*q*1 + (1p)*(1-q)*8 and the general payoff for player two would be u2(s1 , s2) = p*q*5 + p*(1-q)*2 + (1p)*q*3 + (1-p)*(1-q)*5.
In R, a game must be split up into a payoff matrix for each player. To keep matters
simplified, I only looked at games with two players with each player only having two strategies.
The following code was used to create the payoff matricies for prisoners’ dilemma:
n<-2
m<-2
##number of pure strategies for player 1
##number of pure strategies for player 2
C1<-c(2,0)
D1<-c(3,1)
C2<-c(2,0)
D2<-c(3,1)
##
##
##
##
row
row
col
col
1
2
1
2
for
for
for
for
P1
P1
P2
P2
P1<-t(matrix(cbind(C1,D1),n,m))
P2<-matrix(cbind(C2,D2),n,m)
The output was:
P1
[1,]
[2,]
[,1] [,2]
2
0
3
1
P2
[1,]
[2,]
[,1] [,2]
2
3
0
1
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In order to find the dominant strategy, a loop must be created for both players. For player one the
code to find the dominant strategy is:
for (i in 1:n){
if(P1[1,i]>P1[2,i]){print("P1[1,] dominates")}
else if(P1[1,i]<P1[2,i]){print("P1[2,] dominates")}}
If there is a dominant strategy, as is the case for prisoners’ dilemma, the output will look similar
to the following:
"P1[2,] dominates"
"P1[2,] dominates"
If "P1[2,] dominates" is not repeated in the output, and "P1[1,] dominates" is displayed with
"P1[2,] dominates", then there is no dominant strategy. The 2 in the brackets indicates that the
strategy corresponding to row two for player one dominantes row one. The code to find the
dominant strategy for player two is as follows:
for (j in 1:m){
if(P2[j,1]>P2[j,2]){print("P2[,1] dominates")}
else if(P2[j,1]<P2[j,2]){print("P2[,2] dominates")}}
The output is the same as player one’s except “P1” will be “P2” and the brackets will have the
first space blank and the number will respresent the column corresponding to the strategy for
player two.
In R, I was also able to create a code to find the best response for each player given the
other player’s strategy. For player one the code is:
if(P1[1,1]>P1[2,1]){print("The best response for player 1 when
player 2 chooses column 1 is row 1")}else {print("The best
response for player 1 when player 2 chooses column 1 is row 2")}
if(P1[1,2]>P1[2,2]){print("The best response for player 1 when
player 2 chooses column 2 is row 1")}else {print("The best
response for player 1 when player 2 chooses column 2 is row 2")}
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The code is similar for player two. This only works for games with 2x2 dimensions.
A major aspect of Game Theory is the Nash Equilibrium, named after the American
mathematician John Nash. A strategy profile is a Nash Equilibrium if and only if each player’s
prescribed strategy is a best response to the strategies of the others. No player can do better by
unilaterally changing his strategy. (Heap 42). The Nash Equilibrium is not necessarily the best
joint outcome. For instance, in the prisoners’ dilemma, if both players collude they will have a
higher payoff than if they both deviate. However, both players deviating is the Nash Equilibrium
because neither player can get a better payoff if he changes his strategy, taking into account that
his opponent is choosing the strategy to deviate. There are pure and mixed strategy Nash
Equilibriums. Some games do not have a pure strategy Nash Equilibrium, but one always exists
in a mixed form. All finite games have at least one Nash Equilibrium and, in an infinitely played
game, a Nash Equilibrium will be played in the last stage of the game. In a mixed strategy profile
is a Nash Equilibrium if no player can increase his payoff by switching to any other strategy,
given the other players’ strategies. (Watson 124).
In a two person game, each player chooses his mixed strategy such that his opponent has
no advantage in playing any pure strategy. In the game displayed in Figure 5, player one wants to
have a mixed strategy (p, 1-p) that forces player two to be indifferent between playing pure
strategy L and pure strategy R. To find p, the following calculation is performed: u2((p, 1-p),L) =
u2((p, 1-p),R)  5p+3(1-p) = 2p+5(1-p)  2p+3 = (-3)p+5 5p = 2  p=2/5. Thus, S1 = (2/5,
3/5). Likewise, player two wants a mixed strategy (q, 1-q) such that player one is indifferent
between playing pure strategy U and pure strategy D. The same calculation is performed:
u1(U,(q, 1-q)) = u1(D,(q, 1-q))6q+2(1-q) = 1q+8(1-q)  4q+2 = (-7)q+8 11q = 6 q=6/11.
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Thus, S2 = (6/11, 5/11) and the mixed strategies Nash Equilibrium for the game in Figure 5 is
((2/5, 3/5), (6/11, 5/11)). This mixed strategy ensures that neither player will be able to “exploit”
the other player. If player one attempts to change the value of p, his payoff will not be changed,
but player two’s payoff will be affected. Therefore, in order to not help out his opponent, player
one will not change his mixed strategy. The same is true for player two.
In R the pure strategy Nash Equilibrium for a 2x2 is found using the following code:
for (i in 1:n){
for (j in 1:m){
if (P1[i,j]==max(P1[,j])&&P2[i,j]==max(P2[i,])){print(c(i,j))}}}
The output will be the coordinates corresponding to the pure strategies of each player that
constitute a Nash Equilibrium. For example, in prisoners’ dilemma, the ouput will be “2 2”
indicating that the strategy for player one corresponding to the second row and the strategy for
player two corresponding to the second column yields a pure strategy Nash Equilibrium.
If found analytically, the mixed strategy Nash Equilibrium is stable, otherwise it is
difficult to find and unstable. For some games, if it does not randomly begin at the mixed
strategy Nash Equilibrium, it is impossible to stochastically find it. Theoretically, the game will
end up at a pure strategy of the form ((p, 1-p), (q, 1-q)) equalling one of the following: ((0,1),
(0,1)), ((0,1), (1,0)), ((1,0), (0,1)), or ((1,0), (1,0)). I theorized that if a random p and q were
chosen for players one and two, each player would take turns adjusting his probaility in response
to his opponent’s probability in order to maximize his payoff. This back and forth interaction
will continue until both players can no longer change their strategies to increase their payoffs.
Eventually, this would lead to a pure strategy Nash Equilibrium. I attempted to create a code in R
to find the mixed strategy Nash Equilibrium for 2x2 games. The code I wrote, with explanatory
comments is as follows. (The code is in red, the comments are in black.)
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nruns<-10000
##number of times the code is run
mixedNE1<-double(nruns) ##vector holding all the p values found
mixedNE2<-double(nruns) ##vector holding all the q values found
for(run in 1:nruns){
p<-runif(1,0,1) ##a random p value is initially generated
q<-runif(1,0,1) ##a random q value is initially generated
payoffold<-double(2)##vector holding the p and q values found
payoffnew<-double(2)##vector holding the new p and q values found
difference<-c(1,1) ##difference between new and old payoff vectors
PQ<-double(2)
##vector holding the p and q values
##the while loop below causes the players to change their values of p and
q as long as p and q are between 0 and 1 and the change in the payoffs is
minimal
while(p>0.01 && p<.99 && q>0.01 && q<.99 && (difference[1]>0.00000001 ||
difference[2]>0.00000001)){
payoff1<-((p*q*P1[1,1])+(p*(1-q)*P1[1,2])+((1-p)*q*P1[2,1])+((1-p)*(1q)*P1[2,2]))
##payoff1 holds the payoff for player 1 using the original p and q values
payoff1a<-(((p+.01)*q*P1[1,1])+((p+.01)*(1-q)*P1[1,2])+((1(p+.01))*q*P1[2,1])+((1-(p+.01))*(1-q)*P1[2,2]))
##payoff1a holds the payoff for player 1 using the original q value and
p+.01
payoff1b<-(((p-.01)*q*P1[1,1])+((p-.01)*(1-q)*P1[1,2])+((1-(p.01))*q*P1[2,1])+((1-(p-.01))*(1-q)*P1[2,2]))
##payoff1b holds the payoff for player 1 using the original q value and p.01
payoffnew[1]<-max(payoff1,payoff1a,payoff1b)
##payoffnew[1] holds the maximum payoff for player 1 just found above
if(payoff1>payoff1a && payoff1>payoff1b){PQ[1]=p}else if(payoff1a>payoff1
&& payoff1a>payoff1b){PQ[1]=(p+.01)}else{PQ[1]=(p-.01)}
##the above if statement determines whether p, p+.01, or p-.01 generated
the highest payoff for player 1 and, along with the line below, makes it
the new p value
p<-PQ[1]
payoff2<-((q*p*P2[1,1])+(q*(1-p)*P2[2,1])+((1-q)*p*P2[1,2])+((1-q)*(1p)*P2[2,2]))
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##payoff2 holds the payoff for player 2 using the original q and the new p
value
payoff2a<-(((q+.01)*p*P2[1,1])+((q+.01)*(1-p)*P2[2,1])+((1(q+.01))*p*P2[1,2])+((1-(q+.01))*(1-p)*P2[2,2]))
##payoff2a holds the payoff for player 2 using q+.01 and the new p value
payoff2b<-(((q-.01)*p*P2[1,1])+((q-.01)*(1-p)*P2[2,1])+((1-(q.01))*p*P2[1,2])+((1-(q-.01))*(1-p)*P2[2,2]))
##payoff2b holds the payoff for player 2 using q-,01 and the new p value
payoffnew[2]<-max(payoff2,payoff2a,payoff2b)
##payoffnew[2] holds the maximum payoff for player 2 just found above
if(payoff2>payoff2a && payoff2>payoff2b){PQ[2]=q}else if(payoff2a>payoff2
&& payoff2a>payoff2b){PQ[2]=(q+.01)}else{PQ[2]=(q-.01)}
##the above if statement determines whether q, q+.01, or q-.01 generated
the highest payoff for player 2 and, along with the line below “q<-PQ[2]”,
makes it the new q value
difference<-payoffnew-payoffold
##difference between the old and new payoffs
payoffold<-payoffnew
##the old payoff disappears and the new payoff becomes the old payoff
q<-PQ[2]}
##The following sequence of code is only
or 0 and q is still between 1 and 0. The
here. It is executed assuming that since
strategy to improve his payoff, player 2
his payoff is maximized.
executed if the value for p is 1
same procedure above happens
player 1 can no longer change his
will continue to alter his until
if(p>.99||p<.01){
while((q<.99&&q>.01)){
payoff2<-((q*p*P2[1,1])+(q*(1-p)*P2[2,1])+((1-q)*p*P2[1,2])+((1-q)*(1p)*P2[2,2]))
payoff2a<-(((q+.01)*p*P2[1,1])+((q+.01)*(1-p)*P2[2,1])+((1(q+.01))*p*P2[1,2])+((1-(q+.01))*(1-p)*P2[2,2]))
payoff2b<-(((q-.01)*p*P2[1,1])+((q-.01)*(1-p)*P2[2,1])+((1-(q.01))*p*P2[1,2])+((1-(q-.01))*(1-p)*P2[2,2]))
payoffnew[2]<-max(payoff2,payoff2a,payoff2b)
if(payoff2>payoff2a && payoff2>payoff2b){PQ[2]=q}else if(payoff2a>payoff2
&& payoff2a>payoff2b){PQ[2]=(q+.01)}else{PQ[2]=(q-.01)}
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payoffold[2]<-payoffnew[2]
q<-PQ[2]}}
q
##The following sequence of code is only
or 0 and p is still between 1 and 0. The
here. It is executed assuming that since
strategy to improve his payoff, player 1
his payoff is maximized.
executed if the value for q is 1
same procedure above happens
player 2 can no longer change his
will continue to alter his until
if(q>.99||q<.01){
while((p<.99&&p>.01)){
payoff1<-((p*q*P1[1,1])+(p*(1-q)*P1[1,2])+((1-p)*q*P1[2,1])+((1-p)*(1q)*P1[2,2]))
payoff1a<-(((p+.01)*q*P1[1,1])+((p+.01)*(1-q)*P1[1,2])+((1(p+.01))*q*P1[2,1])+((1-(p+.01))*(1-q)*P1[2,2]))
payoff1b<-(((p-.01)*q*P1[1,1])+((p-.01)*(1-q)*P1[1,2])+((1-(p.01))*q*P1[2,1])+((1-(p-.01))*(1-q)*P1[2,2]))
payoffnew[1]<-max(payoff1,payoff1a,payoff1b)
if(payoff1>payoff1a && payoff1>payoff1b){PQ[1]=p}else if(payoff1a>payoff1
&& payoff1a>payoff1b){PQ[1]=(p+.01)}else{PQ[1]=(p-.01)}
payoffold[1]<-payoffnew[1]
p<-PQ[1]}}
p
##below the 2 matrices that were set up to store the p and q values found
have the final values stored in them.
mixedNE1[run]<-p
mixedNE2[run]<-q}
I attempted to find the mixed strategy Nash Equilibrium, that did not work. Then I attempted to
show that if the probability distribution over the strategies was not orignially at the mixed
strategy Nash Equilibrium, the mixed strategies ended up being pure strategies, in one of the
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corners of a unit square ((p,q)=(0,0), (0,1), (1,0), or (1,1)). I created code to show how often the
strategies ended up in the corners and graphed the results.
ZeroZero<-0
ZeroOne<-0
OneZero<-0
OneOne<-0
##Each of the above hold the number of times (adding up to nruns) the game
ended up with a pure strategy (p,q)=(0,0), (0,1), (1,0), or (1,1)
for (i in 1:nruns){
if(mixedNE1[i]<.01 && mixedNE2[i]<.01){ZeroZero<-ZeroZero +1}
else if(mixedNE1[i]<.01 && mixedNE2[i]>.99){ZeroOne<-ZeroOne +1}
else if (mixedNE1[i]>.99 && mixedNE2[i]<.01){OneZero<-OneZero +1}
else if (mixedNE1[i]>.99 && mixedNE2[i]>.99){OneOne<-OneOne +1}}
##The above loop determines whether a pure strategy occurred and then
placed it in the appropriate holder
corners<-c(ZeroZero,ZeroOne,OneZero,OneOne)
##corners holds all the values together in one vector
barplot(corners,
ylab="Frequency",names.arg=c("ZeroZero","ZeroOne","OneZero","OneOne"),main
="Frequency of Corner Probabilities")
##the above is a barplot of the frequency of each pure strategy outcome
(sum(corners)/nruns)*100
##the above is the percentage of the total runs that the game ended with a
pure strategy
The results I found were confusing. When prisoners’ dilemma was entered into the
program, the following barplot was the outcome. Only 8.22% of the time did the game end at a
pure strategy.
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300
100
200
Frequency
400
500
600
Frequency of Corner Probabilities
0
Figure 6.
ZeroZero
ZeroOne
OneZero
OneOne
I also printed out a scatter plot of all the mixed strategies my R code generated for the prisoners’
0.6
0.4
0.2
mixedNE2
0.8
1.0
dilemma. The code for the scatter plot is: plot(mixedNE1,mixedNE2) It is pictured below.
0.0
Figure 7.
0.0
0.2
0.4
0.6
0.8
1.0
mixedNE1
The x-axis, mixedNE1, is the p value and the y-axis, mixedNE2, is the q value for every mixed
strategy combination found. Clearly there is a wide variety of mixed strategies found using this R
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code, including the four pure strategies. However, this should not happen because the prisoners’
dilemma only has one Nash Equilibrium, and that is to deviate all the time. With the above code
and graphs, that is represented by (0,0): S1=(0,1) and S2=(0,1). However, the Nash Equilibrium I
was expecting to find, I did not find.
Next I entered another classical game, battle of the sexes, into my code. The normal form
of this game is pictured here:
Figure 8.
The outcome I found was different than the outcome I found trying to analyze the prinsoners’
dilemma game. At first, when a random p and q value were chosen, 69.11% of the time the game
ended with a pure strategy and 69.05% of the time the game ended on a pure strategy Nash
Equilibrium with both p and q equalling one or both p and q equalling zero. The bar plot is
pictured here.
2000
1500
1000
500
Figure 9.
0
Frequency
2500
3000
Frequency of Corner Probabilities
ZeroZero
ZeroOne
OneZero
OneOne
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When compared to the analysis of prisoners’ dilemma, the results of battle of the sexes is better,
0.6
0.4
Mixed Strategy Nash Equilibrium
of ((2/3, 1/3), (1/3, 2/3))
0.0
0.2
mixedNE2
0.8
1.0
but it should still be much different. The scatter plot of all the mixed strategies is:
Figure 10.
0.0
0.2
0.4
0.6
0.8
1.0
mixedNE1
As stated earlier, roughly 69% of the final strategies ended at a pure strategy Nash Equilibrium,
on this scatter plot the points (0,0) and (1,1) correspond to those strategies. A few end at the
mixed strategy Nash Equilibrium of ((2/3, 1/3), (1/3, 2/3)) which is represented at the bottom
right corner of the triangle in the scatter plot. There is a high concentration of mixed strategies
along the x=y line and between that line and the marked mixed strategy Nash Equilibrium. This
concentration along the x=y line must be another kind of equilibrium that I did not expect to find,
but would have explored further if time had permitted. At the time of writing this paper, I do not
know how to explain this equilibrium or why it was found with the code I wrote for R.
A third game I looked at with my R code was the game depicted in Figure 5. This game
has two pure strategy Nash Equilibriums, ((1,0), (1,0)) and ((0,1) (0,1)), and the mixed strategy
Nash Equilibrium found earlier in this paper of ((2/5, 3/5), (6/11, 5/11)). When I analyzed it with
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my code I found that the game ended in one of the pure strategies 96.87% of the time, a
significant improvement from the prisoners’ dilemma and battle of the sexes games. A pure
strategy Nash Equilibrium was the result 96.86% of the time. The bar chart showing this result
is:
3000
2000
1000
Frequency
4000
5000
Frequency of Corner Probabilities
0
Figure 11.
ZeroZero
ZeroOne
OneZero
OneOne
The resulting scatter plot was even more astounding than the percentage of pure strategy
0.6
0.2
0.4
Mixed Strategy Nash Equilibrium!
0.0
mixedNE2
0.8
1.0
outcomes.
Figure 12.
0.0
0.2
0.4
0.6
mixedNE1
0.8
1.0
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The remaining 3.13% of the time the game ended on or around the mixed strategy Nash
Equilibrium of ((2/5, 3/5), (6/11, 5/11))!
In one respect, my code worked, but only for this game. I cannot explain why my code
works for this game, but not for the other games I explored. A truly efficient code would work
for any game, not just one, and the payoffs inside each game should have no baring on the
outcome. Unfortunately, after a few runs with other games, I discovered my code is sensitive to
the payoffs within each game. My code is, therefore, inefficient for stochastically finding the
Nash Equilibrium of a 2x2 game.
From this instability found through simulations, I have deduced several implications
about equilibriums in life. In life, everyone reacts to other people’s choices in order to increase
his own utility or happiness. For example, if a sibling is being irritating, a person can either
choose to ignore him or say something. Rationally speaking, the person would choose the
response that will give him the best payoff, make him the happiest or most content. Also, once
this person reacts, the irritating sibling will react to his reaction and an endless cycle will persist
until they decide to put an end to their “game.” It is also very rare that people are in a Nash
Equilibrium in any area of their lives. Even if two people have a stable equilibrium between
themselves, outside forces and circumstances can very quickly and easily cause the two people to
be out of equilibrium. In addition, there are almost always choices that can be made to better a
person’s current utility.
In conclusion, Game Theory is an important tool in life when making controlled, strategic
decsions and can be applied to many aspects in life. As a large part of Game Theory, the Nash
Equilibrium seeks to solve a game such that, given a strategy set, not one of the players can
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increase his payoff by unilaterally changing his strategy. A mixed strategy Nash Equilibrium can
be found rather quickly and easily using an analytic method, by solving a simple equation.
However, when attempting to find a mixed strategy Nash Equilibrium through simulation,
stochastically, it is less easy and less simple. In fact, I was unable to find the mixed strategy
Nash Equilibrium in R for the majority of the games I researched, the game in Figure 5 being the
exception. However, this project was not completely fruitless. I was able to enter a 2x2 game and
find the dominant strategy, if one existed, and find the pure strategy Nash Equilibrium, if one
existed. I was able to look at Game Theory from a different perspective and to further explore
some specific games. This project also showed the instability of life scenarios and the fact that
not very often are we in equilibrium in our lives. Even though I was unable to find a mixed
strategy Nash Equilibrium stochastically, the Nash Equilibrium is an interesting, important
feature of Game Theory that should be continued to be studied until no further discoveries can be
made.
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Works Cited
Davis, Morton D. Game Theory: A Nontechnical Introduction. New York: Basic Books, Inc.,
Publishers, 1983.
Heap, Shaun P. Hargreaves, and Yanis Varoutakis. Game Theory: A Critical Text. London:
Routledge, 2004.
Myerson, Roger B. Game Theory: Analysis of Conflict. Cambridge, Massachusetts: Harvard UP,
1991.
Nash, John F. “The Work of John Nash in Game Theory.” Economic Sciences (1994): 160-190.
Watson, Joel. Strategy: An Introduction to Game Theory. New York: WW Norton & Company,
2008.