What is the (finite) maximum number of Nash equilibria in a 2x2 game?
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I think the answer should be 3: 2 pure and 1 mixed. But that is just because I fail to construct more for at least 3 hours. I don't know how to prove this, there are just too many variables. Am I right? If yes, how would I go about proving it?
game-theory nash-equilibrium
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up vote
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favorite
I think the answer should be 3: 2 pure and 1 mixed. But that is just because I fail to construct more for at least 3 hours. I don't know how to prove this, there are just too many variables. Am I right? If yes, how would I go about proving it?
game-theory nash-equilibrium
Your answer is correct if you assume no ties in a player's payoffs.
– mlc
Nov 17 at 22:06
What do you mean by that? I want to consider all 2x2 games.
– SlowerPhoton
Nov 17 at 23:46
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I think the answer should be 3: 2 pure and 1 mixed. But that is just because I fail to construct more for at least 3 hours. I don't know how to prove this, there are just too many variables. Am I right? If yes, how would I go about proving it?
game-theory nash-equilibrium
I think the answer should be 3: 2 pure and 1 mixed. But that is just because I fail to construct more for at least 3 hours. I don't know how to prove this, there are just too many variables. Am I right? If yes, how would I go about proving it?
game-theory nash-equilibrium
game-theory nash-equilibrium
asked Nov 17 at 18:53
SlowerPhoton
371111
371111
Your answer is correct if you assume no ties in a player's payoffs.
– mlc
Nov 17 at 22:06
What do you mean by that? I want to consider all 2x2 games.
– SlowerPhoton
Nov 17 at 23:46
add a comment |
Your answer is correct if you assume no ties in a player's payoffs.
– mlc
Nov 17 at 22:06
What do you mean by that? I want to consider all 2x2 games.
– SlowerPhoton
Nov 17 at 23:46
Your answer is correct if you assume no ties in a player's payoffs.
– mlc
Nov 17 at 22:06
Your answer is correct if you assume no ties in a player's payoffs.
– mlc
Nov 17 at 22:06
What do you mean by that? I want to consider all 2x2 games.
– SlowerPhoton
Nov 17 at 23:46
What do you mean by that? I want to consider all 2x2 games.
– SlowerPhoton
Nov 17 at 23:46
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Consider the $2 times 2$ game
begin{array}{c|cc|}
& L & R \
hline
T & a_1,b_1 & a_2,b_3 \
B & a_3,b_2 & a_4,b_4 \
hline
end{array}
Assume no ties: $a_1 ne a_3$ and $a_2 ne a_4$. (For simplicity, I only discuss the Row player. Replace $a$ with $b$ to cover the Column player.)
There are four possible cases:
1.1) if $a_1>a_3$ and $a_2>a_4$, then the strategy $T$ is strictly dominant and thus is the only best reply for Row;
1.2) if $a_1 < a_3$ and $a_2<a_4$, then the strategy $B$ is strictly dominant and thus is the only best reply for Row;
1.3) if $a_1>a_3$ and $a_2<a_4$, then both T and B may be best replies for Row;
1.4) if $a_1<a_3$ and $a_2>a_4$, then both T and B may be best replies for Row.
Make the analogous list for Column (replace $a$ with $b$ and number items 2.x instead of 1.x), assuming she has no ties either. Combining the two lists, you get a total of 16 cases that can be grouped in three categories:
a) a unique equilibrium in pure strategies; f.i., 1.1 and 2.1 yield (T,L) as the unique equilibrium in pure strategies;
b) a unique equilibrium in mixed strategies; f.i., 1.3 and 2.4 yield (aunique equilibrium in mixed strategies;
c) two equilibria in pure strategies and one in mixed strategies; f.i., 1.3 and 2.3 yield (T,L) and (B,R) as equilibria in pure strategies and there is also an equilibrium in mixed strategies.
The above may be summarised as follows: generically (=assuming no ties), the maximum number of Nash equilibria in a $2 times 2$ game is three.
If you allow for ties (as per your comment), the picture gets messy. You can have games with only two equilibria in pure strategies, such as:
begin{array}{c|cc|}
& L & R \
hline
T & 1,1 & 0,0 \
B & 0,0 & 0,0 \
hline
end{array}
where only $(T,L)$ and $(B,R)$ are equilibria.
More importantly for your question, you can have games with an infinite number equilibria such as
begin{array}{c|cc|}
& L & R \
hline
T & 0,1 & 0,1 \
B & 0,1 & 0,1 \
hline
end{array}
where any strategic profile (pure or mixed) is a Nash equilibrium. If you feel "cheated" by this example because every player is indifferent over every outcome, you might appreciate how particular (=non-generic) is the assumption that ties are possible.
Simple, concise, clear and interesting. Thank you for this!
– SlowerPhoton
Nov 19 at 19:01
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Consider the $2 times 2$ game
begin{array}{c|cc|}
& L & R \
hline
T & a_1,b_1 & a_2,b_3 \
B & a_3,b_2 & a_4,b_4 \
hline
end{array}
Assume no ties: $a_1 ne a_3$ and $a_2 ne a_4$. (For simplicity, I only discuss the Row player. Replace $a$ with $b$ to cover the Column player.)
There are four possible cases:
1.1) if $a_1>a_3$ and $a_2>a_4$, then the strategy $T$ is strictly dominant and thus is the only best reply for Row;
1.2) if $a_1 < a_3$ and $a_2<a_4$, then the strategy $B$ is strictly dominant and thus is the only best reply for Row;
1.3) if $a_1>a_3$ and $a_2<a_4$, then both T and B may be best replies for Row;
1.4) if $a_1<a_3$ and $a_2>a_4$, then both T and B may be best replies for Row.
Make the analogous list for Column (replace $a$ with $b$ and number items 2.x instead of 1.x), assuming she has no ties either. Combining the two lists, you get a total of 16 cases that can be grouped in three categories:
a) a unique equilibrium in pure strategies; f.i., 1.1 and 2.1 yield (T,L) as the unique equilibrium in pure strategies;
b) a unique equilibrium in mixed strategies; f.i., 1.3 and 2.4 yield (aunique equilibrium in mixed strategies;
c) two equilibria in pure strategies and one in mixed strategies; f.i., 1.3 and 2.3 yield (T,L) and (B,R) as equilibria in pure strategies and there is also an equilibrium in mixed strategies.
The above may be summarised as follows: generically (=assuming no ties), the maximum number of Nash equilibria in a $2 times 2$ game is three.
If you allow for ties (as per your comment), the picture gets messy. You can have games with only two equilibria in pure strategies, such as:
begin{array}{c|cc|}
& L & R \
hline
T & 1,1 & 0,0 \
B & 0,0 & 0,0 \
hline
end{array}
where only $(T,L)$ and $(B,R)$ are equilibria.
More importantly for your question, you can have games with an infinite number equilibria such as
begin{array}{c|cc|}
& L & R \
hline
T & 0,1 & 0,1 \
B & 0,1 & 0,1 \
hline
end{array}
where any strategic profile (pure or mixed) is a Nash equilibrium. If you feel "cheated" by this example because every player is indifferent over every outcome, you might appreciate how particular (=non-generic) is the assumption that ties are possible.
Simple, concise, clear and interesting. Thank you for this!
– SlowerPhoton
Nov 19 at 19:01
add a comment |
up vote
2
down vote
accepted
Consider the $2 times 2$ game
begin{array}{c|cc|}
& L & R \
hline
T & a_1,b_1 & a_2,b_3 \
B & a_3,b_2 & a_4,b_4 \
hline
end{array}
Assume no ties: $a_1 ne a_3$ and $a_2 ne a_4$. (For simplicity, I only discuss the Row player. Replace $a$ with $b$ to cover the Column player.)
There are four possible cases:
1.1) if $a_1>a_3$ and $a_2>a_4$, then the strategy $T$ is strictly dominant and thus is the only best reply for Row;
1.2) if $a_1 < a_3$ and $a_2<a_4$, then the strategy $B$ is strictly dominant and thus is the only best reply for Row;
1.3) if $a_1>a_3$ and $a_2<a_4$, then both T and B may be best replies for Row;
1.4) if $a_1<a_3$ and $a_2>a_4$, then both T and B may be best replies for Row.
Make the analogous list for Column (replace $a$ with $b$ and number items 2.x instead of 1.x), assuming she has no ties either. Combining the two lists, you get a total of 16 cases that can be grouped in three categories:
a) a unique equilibrium in pure strategies; f.i., 1.1 and 2.1 yield (T,L) as the unique equilibrium in pure strategies;
b) a unique equilibrium in mixed strategies; f.i., 1.3 and 2.4 yield (aunique equilibrium in mixed strategies;
c) two equilibria in pure strategies and one in mixed strategies; f.i., 1.3 and 2.3 yield (T,L) and (B,R) as equilibria in pure strategies and there is also an equilibrium in mixed strategies.
The above may be summarised as follows: generically (=assuming no ties), the maximum number of Nash equilibria in a $2 times 2$ game is three.
If you allow for ties (as per your comment), the picture gets messy. You can have games with only two equilibria in pure strategies, such as:
begin{array}{c|cc|}
& L & R \
hline
T & 1,1 & 0,0 \
B & 0,0 & 0,0 \
hline
end{array}
where only $(T,L)$ and $(B,R)$ are equilibria.
More importantly for your question, you can have games with an infinite number equilibria such as
begin{array}{c|cc|}
& L & R \
hline
T & 0,1 & 0,1 \
B & 0,1 & 0,1 \
hline
end{array}
where any strategic profile (pure or mixed) is a Nash equilibrium. If you feel "cheated" by this example because every player is indifferent over every outcome, you might appreciate how particular (=non-generic) is the assumption that ties are possible.
Simple, concise, clear and interesting. Thank you for this!
– SlowerPhoton
Nov 19 at 19:01
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Consider the $2 times 2$ game
begin{array}{c|cc|}
& L & R \
hline
T & a_1,b_1 & a_2,b_3 \
B & a_3,b_2 & a_4,b_4 \
hline
end{array}
Assume no ties: $a_1 ne a_3$ and $a_2 ne a_4$. (For simplicity, I only discuss the Row player. Replace $a$ with $b$ to cover the Column player.)
There are four possible cases:
1.1) if $a_1>a_3$ and $a_2>a_4$, then the strategy $T$ is strictly dominant and thus is the only best reply for Row;
1.2) if $a_1 < a_3$ and $a_2<a_4$, then the strategy $B$ is strictly dominant and thus is the only best reply for Row;
1.3) if $a_1>a_3$ and $a_2<a_4$, then both T and B may be best replies for Row;
1.4) if $a_1<a_3$ and $a_2>a_4$, then both T and B may be best replies for Row.
Make the analogous list for Column (replace $a$ with $b$ and number items 2.x instead of 1.x), assuming she has no ties either. Combining the two lists, you get a total of 16 cases that can be grouped in three categories:
a) a unique equilibrium in pure strategies; f.i., 1.1 and 2.1 yield (T,L) as the unique equilibrium in pure strategies;
b) a unique equilibrium in mixed strategies; f.i., 1.3 and 2.4 yield (aunique equilibrium in mixed strategies;
c) two equilibria in pure strategies and one in mixed strategies; f.i., 1.3 and 2.3 yield (T,L) and (B,R) as equilibria in pure strategies and there is also an equilibrium in mixed strategies.
The above may be summarised as follows: generically (=assuming no ties), the maximum number of Nash equilibria in a $2 times 2$ game is three.
If you allow for ties (as per your comment), the picture gets messy. You can have games with only two equilibria in pure strategies, such as:
begin{array}{c|cc|}
& L & R \
hline
T & 1,1 & 0,0 \
B & 0,0 & 0,0 \
hline
end{array}
where only $(T,L)$ and $(B,R)$ are equilibria.
More importantly for your question, you can have games with an infinite number equilibria such as
begin{array}{c|cc|}
& L & R \
hline
T & 0,1 & 0,1 \
B & 0,1 & 0,1 \
hline
end{array}
where any strategic profile (pure or mixed) is a Nash equilibrium. If you feel "cheated" by this example because every player is indifferent over every outcome, you might appreciate how particular (=non-generic) is the assumption that ties are possible.
Consider the $2 times 2$ game
begin{array}{c|cc|}
& L & R \
hline
T & a_1,b_1 & a_2,b_3 \
B & a_3,b_2 & a_4,b_4 \
hline
end{array}
Assume no ties: $a_1 ne a_3$ and $a_2 ne a_4$. (For simplicity, I only discuss the Row player. Replace $a$ with $b$ to cover the Column player.)
There are four possible cases:
1.1) if $a_1>a_3$ and $a_2>a_4$, then the strategy $T$ is strictly dominant and thus is the only best reply for Row;
1.2) if $a_1 < a_3$ and $a_2<a_4$, then the strategy $B$ is strictly dominant and thus is the only best reply for Row;
1.3) if $a_1>a_3$ and $a_2<a_4$, then both T and B may be best replies for Row;
1.4) if $a_1<a_3$ and $a_2>a_4$, then both T and B may be best replies for Row.
Make the analogous list for Column (replace $a$ with $b$ and number items 2.x instead of 1.x), assuming she has no ties either. Combining the two lists, you get a total of 16 cases that can be grouped in three categories:
a) a unique equilibrium in pure strategies; f.i., 1.1 and 2.1 yield (T,L) as the unique equilibrium in pure strategies;
b) a unique equilibrium in mixed strategies; f.i., 1.3 and 2.4 yield (aunique equilibrium in mixed strategies;
c) two equilibria in pure strategies and one in mixed strategies; f.i., 1.3 and 2.3 yield (T,L) and (B,R) as equilibria in pure strategies and there is also an equilibrium in mixed strategies.
The above may be summarised as follows: generically (=assuming no ties), the maximum number of Nash equilibria in a $2 times 2$ game is three.
If you allow for ties (as per your comment), the picture gets messy. You can have games with only two equilibria in pure strategies, such as:
begin{array}{c|cc|}
& L & R \
hline
T & 1,1 & 0,0 \
B & 0,0 & 0,0 \
hline
end{array}
where only $(T,L)$ and $(B,R)$ are equilibria.
More importantly for your question, you can have games with an infinite number equilibria such as
begin{array}{c|cc|}
& L & R \
hline
T & 0,1 & 0,1 \
B & 0,1 & 0,1 \
hline
end{array}
where any strategic profile (pure or mixed) is a Nash equilibrium. If you feel "cheated" by this example because every player is indifferent over every outcome, you might appreciate how particular (=non-generic) is the assumption that ties are possible.
answered Nov 18 at 11:37
mlc
4,83431332
4,83431332
Simple, concise, clear and interesting. Thank you for this!
– SlowerPhoton
Nov 19 at 19:01
add a comment |
Simple, concise, clear and interesting. Thank you for this!
– SlowerPhoton
Nov 19 at 19:01
Simple, concise, clear and interesting. Thank you for this!
– SlowerPhoton
Nov 19 at 19:01
Simple, concise, clear and interesting. Thank you for this!
– SlowerPhoton
Nov 19 at 19:01
add a comment |
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Your answer is correct if you assume no ties in a player's payoffs.
– mlc
Nov 17 at 22:06
What do you mean by that? I want to consider all 2x2 games.
– SlowerPhoton
Nov 17 at 23:46