# Battle of the sexes (game theory)

 Opera Football Opera 3,2 0,0 Football 0,0 2,3 Battle of the Sexes 1
 Opera Football Opera 3,2 1,1 Football 0,0 2,3 Battle of the Sexes 2

In game theory, battle of the sexes (BoS) is a two-player coordination game. Imagine a couple that agreed to meet this evening, but cannot recall if they will be attending the opera or a football match (and the fact that they forgot is common knowledge). The husband would prefer to go to the football game. The wife would rather go to the opera. Both would prefer to go to the same place rather than different ones. If they cannot communicate, where should they go?

The payoff matrix labeled "Battle of the Sexes (1)" is an example of Battle of the Sexes, where the wife chooses a row and the husband chooses a column. In each cell, the first number represents the payoff to the wife and the second number represents the payoff to the husband.

This representation does not account for the additional harm that might come from not only going to different locations, but going to the wrong one as well (e.g. he goes to the opera while she goes to the football game, satisfying neither). To account for this, the game is sometimes represented as in "Battle of the Sexes (2)".

Some authors refer to the game as Bach or Stravinsky and designate the players simply as Player 1 and Player 2, rather than assigning gender.[1]

## Equilibrium analysis

This game has two pure strategy Nash equilibria, one where both go to the opera and another where both go to the football game. There is also a mixed strategies Nash equilibrium in both games, where the players go to their preferred event more often than the other. For the payoffs listed in the first game, each player attends their preferred event with probability 3/5.

This presents an interesting case for game theory since each of the Nash equilibria is deficient in some way. The two pure strategy Nash equilibria are unfair; one player consistently does better than the other. The mixed strategy Nash equilibrium (when it exists) is inefficient. The players will miscoordinate with probability 13/25, leaving each player with an expected return of 6/5 (less than the return one would receive from constantly going to one's less favored event).

One possible resolution of the difficulty involves the use of a correlated equilibrium. In its simplest form, if the players of the game have access to a commonly observed randomizing device, then they might decide to correlate their strategies in the game based on the outcome of the device. For example, if the couple could flip a coin before choosing their strategies, they might agree to correlate their strategies based on the coin flip by, say, choosing football in the event of heads and opera in the event of tails. Notice that once the results of the coin flip are revealed neither the husband nor wife have any incentives to alter their proposed actions – that would result in miscoordination and a lower payoff than simply adhering to the agreed upon strategies. The result is that perfect coordination is always achieved and, prior to the coin flip, the expected payoffs for the players are exactly equal.

### Working out the above

The equilibria may be found by a generic Nash equilibrium solver such as the Lemke-Howson algorithm. But BoS solutions are also simple enough to be found though a few steps of simple algebra.

Let us calculate the four probabilities for the actions of the individuals (Man and Woman), which depend on their expectations of the behaviour of the other, and the relative payoff from each action. These four probabilities are:

• The Man goes to the Football (resp. Opera), denoted by MF (resp. MO).
• The Woman goes to the Football (resp. Opera), denoted by WF (resp. WO).

The Probability that the Man goes to the Football game, MF, equals the payoff if he does (whether or not the woman does), divided by the same payoff plus the payoff if he goes to the opera instead:

${\displaystyle M_{F}={\frac {W_{O}+3W_{F}}{W_{O}+3W_{F}+2W_{O}+0W_{F}}}={\frac {W_{O}+3W_{F}}{3W_{O}+3W_{F}}}}$

We know that she either goes to one or the other, so ${\displaystyle W_{O}+W_{F}=1}$, so:

${\displaystyle M_{F}={\tfrac {1}{3}}(W_{O}+3W_{F})}$

Similarly:

{\displaystyle {\begin{aligned}M_{O}&={\tfrac {2}{3}}W_{O}\\W_{F}&={\tfrac {2}{3}}M_{F}\\W_{O}&={\tfrac {1}{3}}(3M_{O}+M_{F})\end{aligned}}}

This forms a set of simultaneous equations. We can solve these, starting with ${\displaystyle M_{F}}$ for example, by substituting in the equations above:

{\displaystyle {\begin{aligned}M_{F}&={\tfrac {1}{3}}\left({\tfrac {1}{3}}(3M_{O}+M_{F})+3\left({\tfrac {2}{3}}M_{F}\right)\right)\\&={\tfrac {1}{3}}\left(M_{O}+{\tfrac {7}{3}}M_{F}\right)\\&={\tfrac {1}{3}}\left(1+{\tfrac {4}{3}}M_{F}\right)&&M_{O}+M_{F}=1\end{aligned}}}

Solving the last equation for ${\displaystyle M_{F}}$ yields:

${\displaystyle M_{F}={\tfrac {3}{5}}}$

Knowing that ${\displaystyle M_{F}+M_{O}=1}$, we deduce:

${\displaystyle M_{O}=1-{\tfrac {3}{5}}={\tfrac {2}{5}}}$
${\displaystyle W_{F}={\tfrac {2}{3}}M_{F}={\tfrac {2}{5}}}$
${\displaystyle W_{O}=1-{\tfrac {2}{5}}={\tfrac {3}{5}}}$

Then we can calculate the probability of coordination ${\displaystyle P_{c}}$ (that M and W do the same thing, independently), as:

${\displaystyle P_{c}=M_{F}W_{F}+M_{O}W_{O}={\tfrac {3}{5}}{\tfrac {2}{5}}+{\tfrac {2}{5}}{\tfrac {3}{5}}={\tfrac {12}{25}}}$

And the probability of miscoordination ${\displaystyle P_{m}}$ (that M and W do different things, independently):

${\displaystyle P_{m}=M_{F}W_{O}+M_{O}W_{F}={\tfrac {3}{5}}{\tfrac {3}{5}}+{\tfrac {2}{5}}{\tfrac {2}{5}}={\tfrac {13}{25}}}$

And just to check our probability working:

${\displaystyle P_{c}+P_{m}={\tfrac {12}{25}}+{\tfrac {13}{25}}={\tfrac {25}{25}}=1}$

So the probability of miscoordination is ${\displaystyle {\tfrac {13}{25}}}$ as stated above.

The expected payoff E for each individual (${\displaystyle E_{m}}$ and ${\displaystyle E_{w}}$) is the probability of each event multiplied by the payoff if it happens. For example, the Probability that the man goes to football and the Woman goes to football multiplied by the Expected payoff to the man if that happens (${\displaystyle E_{mmfwf}}$):

{\displaystyle {\begin{aligned}E_{m}&=M_{F}W_{F}E_{mmfwf}+M_{F}W_{O}E_{mmfwo}+M_{O}W_{O}E_{mmowo}+M_{O}W_{F}E_{mmowf}\\&={\tfrac {3}{5}}{\tfrac {2}{5}}3+{\tfrac {3}{5}}{\tfrac {3}{5}}1+{\tfrac {2}{5}}{\tfrac {3}{5}}2+{\tfrac {2}{5}}{\tfrac {2}{5}}0={\tfrac {39}{25}}\end{aligned}}}

Which is not the same as the ${\displaystyle {\tfrac {6}{5}}}$ stated above!

For comparison, let us assume that the man always goes to football and the woman, knowing this, chooses what to do based on revised probabilities and expected values to her:

${\displaystyle M_{F}=1}$
${\displaystyle M_{O}=0}$
${\displaystyle W_{F}={\tfrac {2}{3}}M_{F}={\tfrac {2}{3}}}$
${\displaystyle W_{O}=1-{\tfrac {2}{3}}={\tfrac {1}{3}}}$
{\displaystyle {\begin{aligned}Em&=M_{F}W_{F}E_{mmfwf}+M_{F}W_{O}E_{mmfwo}+M_{O}W_{O}E_{mmowo}+M_{O}W_{F}E_{mmowf}\\&=1{\tfrac {2}{3}}3+1{\tfrac {1}{3}}1+0{\tfrac {1}{3}}2+0{\tfrac {2}{3}}0={\tfrac {7}{3}}\end{aligned}}}

This is symmetric for ${\displaystyle E_{w}}$ if the woman always goes to the opera and the man chooses randomly with probabilities based on the expected outcome, due to the symmetry in the value table. But if both players always do the same thing (both have simple strategies), the payoff is just 1 for both, from the table above.

### Optimal decision

Given the above, how should players choose which of the three Nash equilibria to actually play in practice? One (contested) solution concept is that if they are both identical then they must arrive at the same rational answer, if one exists, and the only way to do this is to choose the mixed strategy equilibrium because it is symmetric. (citation needed).

## Burning money

 Opera Football Opera 4,1 0,0 Football 0,0 1,4 Unburned
 Opera Football Opera 2,1 -2,0 Football -2,0 -1,4 Burned

Interesting strategic changes can take place in this game if one allows one player the option of "burning money" – that is, allowing that player to destroy some of her utility. Consider the version of Battle of the Sexes pictured here (called Unburned). Before making the decision the row player can, in view of the column player, choose to set fire to 2 points making the game Burned pictured to the right. This results in a game with four strategies for each player. The row player can choose to burn or not burn the money and also choose to play Opera or Football. The column player observes whether or not the row player burns and then chooses either to play Opera or Football.

If one iteratively deletes weakly dominated strategies then one arrives at a unique solution where the row player does not burn the money and plays Opera and where the column player plays Opera. The odd thing about this result is that by simply having the opportunity to burn money (but not actually using it), the row player is able to secure her favored equilibrium. The reasoning that results in this conclusion is known as forward induction and is somewhat controversial.[2] In brief, by choosing not to burn money, the player is indicating she expects an outcome that is better than any of the outcomes available in the "burned" version, and this conveys information to the other party about which branch she will take.

## Battle of the Sexes Game with Ambiguity

Battle of the Sexes Game with Ambiguity
Player 2
Player 1
Left Middle Right
Top
0
0
100
300
x
50
Bottom
300
100
0
0
x
55

Decisions are said to be ambiguous if there are no objective probabilities given and it is difficult or impossible to assign subjective probabilities to events. Kelsey and le Roux (2015)[3] report an experimental test of the influence of ambiguity on behaviour in a Battle of Sexes game which has an added safe strategy, R, available for Player 2 (see Table). The paper studies the behaviour of subjects in the presence of ambiguity and attempts to determine whether subjects playing the Battle of Sexes game prefer to choose an ambiguity safe option.

The value of x, which is the safe option available to Player 2, varies in the range 60-260. For some values of x, the safe strategy (option R) is dominated by a mixed strategy of L and M, and thus would not be played in a Nash equilibrium. For some higher values of x the game is dominance solvable. The effect of ambiguity-aversion is to make R (the ambiguity-safe option) attractive for Player 2. R is never chosen in Nash equilibrium for the parameter values considered. However it may be chosen when there is ambiguity. Moreover for some values of x, the games are dominance solvable and R is not part of the equilibrium strategy. [4]

It was found that R is chosen quite frequently by subjects. While the Row Player randomises 50:50 between her strategies, the Column Player shows a marked preference for avoiding ambiguity and choosing his ambiguity-safe strategy. Thus, the results provide evidence that ambiguity influences behaviour in the games.

## References

• Luce, R.D. and Raiffa, H. (1957) Games and Decisions: An Introduction and Critical Survey, Wiley & Sons. (see Chapter 5, section 3).
• Fudenberg, D. and Tirole, J. (1991) Game theory, MIT Press. (see Chapter 1, section 2.4)
• Kelsey, D. and S. le Roux (2015): An Experimental Study on the Effect of Ambiguity in a Coordination Game, Theory and Decision.
1. ^ Osborne, Rubinstein (1994). A course in game theory. The MIT Press.
2. ^ For a detailed explanation, see [1] p8 Section 4.5.
3. ^ [2]
4. ^ For a detailed explanation, see [3].
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