Why your crazy girlfriend always gets what she wants: battle of the sexes game theory

You want to watch the football game. She wants to go out for a movie. Guess which one you end up doing.

It’s not exactly a scientific observation, but time and again, I have seen friends whipped by crazy girlfriends. (I would equally say there are a lot of crazy boyfriends, so don’t get caught up in the gender).

It turns out there is a game theory explanation for this phenomenon which is what I want to explore in this article.

You’ll see why crazy people get their way, and how you can use a similar strategy to fight back.

Battle of the sexes

To begin, consider the following model of conflict. This game is one of the simplest examples in game theory, often the first example presented in a game theory course.

Alice and Bob are planning to go out for an evening. Bob wants to go to the football game, but Alice wants to go to the opera. Still, both would prefer to be with the other person than going out to an event by themselves.

Let’s say each person gets 3 points to go to their favored event, 1 points to go to the other event but be with their significant other, and 0 points if the two do not go together.

The outcomes can be represented in the following matrix:

The question is: how will this game play out?

Solution 1: opera or football

This game is very easy to solve. We need to consider what each person’s best response is, given what the other person might be doing.

Here is how Bob thinks about the problem:

–If Alice goes to the football game, then I should also go to get 3 points rather than going to the opera for 0 points

–If Alice goes to the opera, I might not like the opera, but if I go I’ll at least get 1 point. If I watch football I would instead end up with 0 points.

Bob’s strategy is to guess what Alice wants to do and follow.

Alice’s reasoning is exactly the same: she wants to be with Bob, so she should pick the choice she expects him to do.

The outcomes that both go to football, or both go to opera are the solutions of the game.

These are the Nash equilibria of the game.

The pleasant finding is that both players end up choosing mutually beneficial outcomes.

But the annoying part is the game has two different solutions. I mean which one do they end up doing?? Do they go to football or opera? The concept of the Nash equilibrium falls short here in providing a specific prediction.

However, there are other ways to think about the game.

Solution 2: correlated equilibrium

If the game is repeated, as it would be in a relationship, there is a good compromise that Alice and Bob could agree to.

What they could do is flip a coin to decide whether they go to football or the opera. Over time the events will balance out, and they both guarantee that they are together. They will get an average payout of 2.

This seems like a very reasonable solution. It also makes a lot more sense they flip a single coin and coordinate their choices, rather than each player flipping a coin individually and hoping to end up at the same place (the mixed strategy Nash equilibrium doesn’t make much sense in this game: a good chunk of time the two end up in different places)

In a healthy relationship, there is compromise and people may choose the correlated equilibrium.

But a crazy girlfriend would not tolerate this. She wants to get her way, and she uses another strategy.

Solution 3: changing the game

Let’s suppose Alice really, really wants to go to the opera, and she wants to convince Bob that it’s in his best interest too.

Here is a drastic strategy that Alice could use to change the game.

Alice takes out a $20 bill from her purse, and then announces the following:

Look Bob, I really want to go to the opera. And if I think there is even a chance we are not going, that would stress me out. I’ll vent by burning this $20 bill.

I’m going to the room to think about what I’m going to do. Come knock on my door with your choice in one minute. But think about what’s best for both of us.

Alice has introduced another strategy to the game called burning money. In this game, Alice first chooses whether to burn her own $20 bill, and then both players choose where they would like to go.

Let’s say that burning a $20 bill will destroy 1 point of utility for Alice. How will this game play out?

The way to analyze this game is to write out a large matrix with all the choices and solve as one normally would.

Alice has two moves in her strategy: she can either burn the money or not, and she can either go to the opera or not. Let’s abbreviate Alice’s choice of “burn the money, go to opera” as BO, and similarly the rest of her choices as BF, NO, NF.

Bob also has two moves in his strategy. He needs to decide what he will do if Alice burns the money, and what he will do if Alice does not burn the money. Let’s abbreviate Bob’s choice of “if she burns the money I go to the opera, if she does not burn I go to football” as OF, and similarly the rest of his choices as OO, FO, FF.

The matrix of payouts is as follows:

We can now eliminate bad strategies by a concept of iteratively deleting weakly dominated strategies (see more in this game and this game).

To begin, Alice will realize that NO is weakly better than BF, so she will never play the strategy of burning the money and choosing football. Both players realize this and “cross out” that option in the matrix:

Looking at the matrix, Bob realizes that OO is weakly better than OF, and FO is weakly better than FF. So both Alice and Bob eliminate those options in their mind.

The process continues, and you can check the only strategies that remains are NO for Alice and OO for Bob.

The result is this: the equilibrium outcome is that Alice does not burn the money and goes to the opera, and Bob chooses to go to the opera whether Alice burns money or not.

That is, they end up going to the opera just like Alice wanted!

Just think about what happened: Alice got her way because she threatened to burn money. But she never actually has to burn the money: she gets her way because she threatens to torch her own utility.

This seems to capture an element of how spoiled brats in real life operate. They do not always throw tantrums. They only have to threaten to throw a tantrum and act unhappy to force everyone into their choice.

(While I find the solution interesting, I should mention there is controversy about the idea. It is odd that Alice can change the game by threatening to use bizarre behavior. This is an issue raised in this paper )

How to fight back

Bob has a couple of options for fighting back in this game. He can threaten to burn money pre-emptively too, which might get Alice to see his side and drop the pettiness.

Or he can play it safe and change the game once and for all. While crazy people do change and grow, it is a question of how fast and whether it is worth the effort.

Of course there are other ways to deal with crazy people, and I could go on and on. But in my opinion it is often not worth the time and effort. Sometimes breaking it off is the best move.



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  • guywholearnedthehardway

    Solution R (real world)
    Alice posses certain physical resources that, on occasion, she shares with Bob.  Though Bob also posses related physical resources, Alice’s desire for them are less than Bob’s desire for Alice’s.  

    Each player is fully aware of the strengths and weaknesses of the other player, though Bob is generally less astute until the inevitable winner of the game becomes clear.  Through the years Alice has learned that simply implying access to her resources will be completely eliminated for an indeterminate time causes Bob to fold his cards and comply.  Through numerous ‘frustrations’ in playing the game Bob has learned that his best strategy is to accept some short-term loss (going to the opera for instance) in order to more quickly access Alice’s resources.

  • http://twitter.com/preshtalwalkar Presh Talwalkar

    I forgot to mention a relevant topic: if the SOPA and PIPA blackout day protest works, we might view it as evidence of how the “burning money” strategy can work.

  • Norcross

    Not really, since SOPA was effectively killed in Congress well before the protest.  But ironically, from a game theory standpoint that makes the “protest” more likely!

    Let’s look at the payouts:
    Protest occurs, SOPA passes anyway: big loss for protesters, since they get laws they don’t want and look powerless
    Protest occurs, SOPA doesn’t pass:  big win for the protesters, since new law doesn’t get passed and they look powerful
    Protest does not occur, SOPA passes:  loss for the protesters, since law passes
    Protest does not occur, SOPA does not pass: minor win for the protesters, law remains status quo with protesters seen as having no influence

    So, if SOPA was likely to pass, the expected payoff for the protest would depend on how likely the protest was to change the passage (ie, whether the damage from a big loss exceeded the marginal gain from the chance of turning a loss into a big win).  Since SOPA is already dead, the only choices left are big win vs minor win, so the protest is much more likely to occur than it would be if there were still something to protest!

    Another surprising result from game theory…

  • William

    I think there’s an error in the payouts for the 4×4 grid.  Given your definition, “if she burns I go to the opera, if she does not burn I go to football” as OF, I think the payouts (or labels) for columns FO and OF are swapped.

    Also, NF seems to be an inconsistent option for Alice, because if she decided to go to football, she would be stressed and burn the money.

  • http://mindyourdecisions.com/blog/ Presh Talwalkar

    Thanks you’re right. I will correct this by changing the wording of the strategy to avoid having to modify all the tables.

  • http://mindyourdecisions.com/blog/ Presh Talwalkar

    Never mind, I’ll just change the tables.

  • Paul Smith

    I agree. I just tried to populate the grid myself, and then looked up and noticed it was swapped.

  • Anonymous

    Can someone please explain the ”
    Looking at the matrix, Bob realizes that OO is weakly better than OF, and FO is weakly better than FF. So both Alice and Bob eliminate those options in their mind. ” ?

    OO and OF adds up to the same total points, and so FO and FF.

  • http://mindyourdecisions.com/blog/ Presh Talwalkar

    You can’t look at the total points for each strategy: you have to look at the payouts of the strategy OO line by line versus what Alice can do. You’ll see that OO always has at least as high a point total as OF. That’s what weak dominance means.

  • John

    Your reply, while correct, doesn’t actually address Rajtantajtan’s question. If OO beats FO (it’s FO and not OF as OF beats OO for NF- you forgot to change your references when you “corrected” your definition) for each of Alices choices, then by necessity the total points will also be greater. The points being greater does not mean it dominates, but if it dominates, the points will be greater.

    The only way I can get the points to come out the same for OO and FO is to include BF (which has already been eliminated), and by counting both Bob’s points AND Alice’s points. Since OO and FO are possible strategies for Bob and not Alice, only Bob’s points are relevant here.

    It also seems to me that this is probably the least educational post of your’s I’ve come accross (for the most part I’ve been quite impressed which is why I check regularly). I’m not sure exactly what the game theory definitions are, but I gather that a weakly dominated strategy is dominated by at least one other, while a strongly dominated (or stricly dominated) strategy is one dominated by all others. So, BF is weakly dominated because it is dominated by NO (and it is also dominated by NF).

    If I understand definitions correctly, then this seems a demonstration of the weakness of eliminating weak strategies. It seems to make a lot more sense to me to first eliminate all squares on your grid that involve her going to the football game while he goes to the opera. If they couldn’t communicate, then removing weakly dominated strategies might make sense. But they can communicate.

    Other little problems are:

    you give the same points for going to either place if they go seperately (i.e. neither cares where they go if it is alone – so if she was off on business and he had to go alone, he would be just as happy going to the opera as the game?)

    She’s offering to burn her money? His response seems obvious.”Go ahead and burn your money. When you’ve got that childishness out of your system and you’re ready to discuss things as adults, come back and we’ll decide where to go.”

    If she REALLY wanted to go to the opera and was clever instead of just crazy (this type of game theory only works when all involved are “rational” – so if she is crazy, the explanation given has no bearing), she would “change the game” simply by saying “I’m going to the opera. Period. Are you coming?”

    The real reason “crazy” mates always get their way is that the only ones they will tolerate and who will tolerate them are the ones who will continually give in without getting upset enough to do anything about it.

  • Anonymous

    For Bob I count 1,0,1 for OO and 0,3,1 for OF when the last row is eliminated. 

    If ruling out NF because Alice said she would burn if going to the football, OO would be the best choice for Bob.

    Anyway, I was looking at the total points because of the rule Alice introduced that says “But think about what’s best for both of us.” 

  • Anonymous

    By saying that she would burn the money she effectively said that she’s going to the opera. 1) she ruled out NF 2) She made sure that she can’t get a positive score from BF. 

    Hence she will choose to go to the opera no matter what because that’s the only way she can get any points at all. And bob knows this when making his decision.

  • http://mindyourdecisions.com/blog/ Presh Talwalkar

    Hey John, a weakly dominant strategy is one that is at least as good as another strategy, whereas a strongly dominant strategy is better than another.

    I admit the story is a bit concocted, but this “burning hand” is one of the extensions of the Battle of the Sexes model, so I was trying to do justice to the textbook example:

    http://en.wikipedia.org/wiki/Battle_of_the_sexes_(game_theory)#Burning_money 

  • john

    I’m afraid I was a little too quick to comment last time. I made a couple of errors.
    First, I checked the definitions of weakly and strictly dominant and dominated strategies. Both refer to the comparison of only 2 strategies (when more than 2 possible strategies exist, the rest are ignored). If strategy A strictly dominates B then for all possible strategies of other participants, the outcome of A is strictly superior to the outcome of B. If strategy A weakly dominates B then there is at least one strategy of other participants for which the outcome of A is strictly superior to the outcome of B, and there are no strategies of other participants which give B a superior outcome. The value mentioned for the outcome is the value to the participant choosing between A and B.
    I also checked the definition of a Nash equilibrium (since I’m not a game theory expert). A Nash equilibrium is a combined strategy of all players (or participants) such that none can benefit from a unilateral change of strategy (If only one person changes strategy, that person’s outcome cannot be improved). If any unilateral change by any player always leads to a strict decrease to the value of that players outcome, then the equilibrium is a strict Nash equilibrium, otherwise it is a weak Nash equilibrium.
    Based on these definitions, the following seems evident:
    Iteratively removing strictly dominated strategies cannot remove any Nash equilibria. Iteratively removing weakly dominated strategies cannot remove any strict Nash equilibria, but can remove weak Nash equilibria. Which means, iteratively removing strictly dominated strategies (when possible) should lead to the Nash equilibrium, and iterative removal of weakly dominated strategies (when possible) should lead to a Nash equilibrium, but not necessarily the only one.
    So, removal of dominated strategies is a way of finding Nash equilibria. However it must be kept in mind that a Nash equilibrium, even a strict Nash equilibrium in a game with only one, is NOT necessarily the best combined strategy for all players. There are games which have a combined strategy that is better than the Nash equilibrium for each and every player. The prisoners’ dilemma is a perfect example. Both suspects confessing is the strict Nash equilibrium where each gets 5 years. This equilibrium can be found by removing strictly dominated strategies. Yet, if they cooperate, they could each end up with only 6 months (better by far for each of them).
    My second mistake was to confuse the situations described with the games described by the utility functions represented by the matrices (the 2×2 matrix for the original situation, and the 4×4 matrix for the situation created by Alice’s threat to burn her money). The 2×2 matrix describing the utility function for the prisoners isn’t likely to represent a true situation either. For one, the game is the same regardless of the guilt or innocence of the suspects (which is why it is reasonable to consider this tactic “unfair” of the police _ and those who think it is clever are either wrong, or have the mind set of the Spanish Inquisition, or the Salem witch hunt). So, any just and rational judge and jury would dismiss any confession obtained in this manner as being unreliable. Then there is the issue of the suspects trusting the police or worrying about retaliation by the partner at the end of the jail term.
    For the original situation here, it seems a little odd that the utility function gives the same value when Bob goes to the game and Alice goes to the opera as when Bob goes to the opera and Alice goes to the game. This can be justified by suggesting that each gets a warm fuzzy knowing that the other was willing to give up their event so that they could be together (but this really only applies if they can’t communicate).
    For the situation after Alice’s threat, it seems ludicrous to suggest that Bob can’t also use a “game changing” strategy. So there are a lot more strategies available than are represented in the matrix, making the matrix pretty meaningless. It also seems to me that Alice’s threat could change the utility function as well as introduce the extra strategies mentioned. If Bob is sufficiently disgusted by Alice’s selfish tactics, he might decide he’d prefer being alone. In that case, it might be better to assign him 3 points if he goes to the football game while she goes to the opera (or anywhere other than the game), 1 point if he goes to the opera while she is at the game, and no points if they go to either event together.
    Now that my opinion of these utility functions is (I hope) clear, I’ll ignore the real situations described, and look at the games represented by the matrices.
    The 2×2 matrix has 2 strict Nash equilibria. Each individual strategy includes a strict Nash equilibrium, so there are no strictly or weakly dominated strategies, and there can be no elimination of such strategies. Yet both combined strategies that weren’t Nash equilibria were immediately removed. This could be done because Bob and Alice were communicating. Without communication, each of the 4 combined strategies would be equally likely.
    As a repeated game, any combined coordinated mixed strategy is a strict Nash equilibrium. Initially, both Bob and Alice should easily agree to remove all of the non coordinated strategies. Presh suggests a 50/50 mix could or should be agreed to, and flipping a coin each time would determine where they should go. I agree with the 50/50 mix, but would suggest alternating (or switching after the first and alternating every 2 afterwards) rather than flipping a coin. When the best solution in a competition game (rather than a cooperation game) involves a mixed strategy, the mix must be randomized or the opponent might detect the pattern and use it to coordinate strategies and gain an edge. Here, both benefit from the same coordination of strategies, and randomization is unnecessary.
    Why the 50/50 mix? The simple answer is because that seems fair. The longer answer involves something very similar to one of the principal rules used by philosophers to determine ethical behavior: if everyone practicing the behavior is detrimental to all, then the behavior is unethical. To not both lose, Bob and Alice must agree on the same mix. If Alice insists on a mix skewed towards the opera, she should expect Bob to insist on a mix skewed towards football, and they both lose. The only way not to lose is for one to cave, or go with the 50/50 mix. If it is common knowledge that they are both rational, the 50/50 mix will be the result.
    The 4×4 matrix has 4 weak Nash equilibria. Three of the equilibria have them going to the opera, and the other has them at the game. For the one at the football game, Bob’s strategy is to go to the game regardless, and Alice does not burn her $20 (Alice=NF, Bob=FF). One of the opera equilibria is where Bob goes to the opera only if Alice burns her bill (A=BO, B=OF). And for the last 2, Alice does not burn any money (A=NO, B=OO and B=FO).
    I don’t know if this really pertains to the solution of the game, but the 4 equilibria have some interesting properties (at least I find them interesting).
    _ The 2 A=NO equilibria together are like a strict Nash equilibrium: for either equilibrium, both Bob and Alice have strict incentive to not leave the pair. I would call the pair a stable equilibrium.
    _ The equilibrium at the football game seems unstable as Bob appears not to have any reason not to switch to OF. But doing so would lead Alice to switch to BO resulting in the last equilibrium which would cause Bob to lose 2 points. Since he knows that if he switches to OF Alice will then switch to BO, Bob does have incentive to stay with FF. So If Bob has foresight, the equilibrium at the game is stable.
    _ The last equilibrium at the opera (A=BO) is unstable. Bob has no incentive not to switch from OF to OO, and once he does, Alice will switch to NO (back to one of the stable pair). This still has no effect on Bob, so he really has no incentive to stay with OF.
    So the equilibrium that involves burning money decays to the stable pair at the opera, and the one at the football game is stable.
    There are only 2 orders in which the weakly dominated strategies can be removed. After A=BF has been removed, B=FO and B=FF can be removed in either order. But the result is the same: they both go to the opera and no money is burned. Is that what would really happen if 2 rational people were playing the game? If it wasn’t repeated and there was no communication possible, then it might well be the best way to make coordination most likely (but I’m not sure even for this case). But it is my belief that if discussion is possible and/or the game is repeated, rational players will see the game as identical to the 2×2 game.
    There is only one solution if you iteratively remove dominated strategies individually. What do I mean by individually? Alice has 4 individual strategies, but one could instead compare the strategy to burn against the strategy to not burn (or going to the opera versus going to the game. What if we look at Bob’s choice to take Alice’s threat into account (H=heed) or not (I=ignore)? If he does, he ends up with the 4×4 matrix game which leaves him going to the opera. If he doesn’t, Alice’s burn strategies become strictly dominated and the original 2×2 matrix remains. H gives Bob 1 point, and I gives him 1 if they go to the opera and 3 if they go to the game. So strategy I weakly dominates H leaving the original 2×2 matrix.
    However, that is NOT why I think rational players view the 2 games identically. The argument in the previous paragraph is no better (and I’m sure some will say is weaker) than the one given by Presh. It just demonstrates that iterative removal of dominated strategies depends on perspective.
    Here is how I see it. There are only 2 outcomes of interest. Alice wants both to go to the opera, and Bob wants both to go to the game. Wherever they go, Bob does not care one bit if Alice burned her money or not , and Alice would rather not (based on the utility function he has no gain or loss from her burning money, and she has a loss of 1 point). If Bob did care, and didn’t want the money burned, nothing changes (for the moment I’m assuming both are rational and I’m assuming that rationality to be common knowledge). On the other hand, if Bob got a kick out of watching Alice burn her money, then he might well decide that watching her burn her money and going to the opera is a good compromise. But then, the utility function should show him getting more than 1 point for that option.
    The argument for the 50/50 mix in the game described by the 2X2 matrix depends only on the symmetry of the points for the two desired outcomes, and the symmetry of the penalties if they can’t agree. That means the argument is just as valid here. So, If it is common knowledge that both players are rational, a single game will be decided arbitrarily (by the flip of a coin possibly) and a repeated game will have a 50/50 mix.
    But in reality, many “crazy” girlfriends do win with such tactics, so what is going on? My take is that the answer is in the title to the article: SHE’s CRAZY! What does that mean? It simply means that she is not rational, and can be counted on to go to the opera even if he goes to the game. Actually, Alice doesn’t even have to be crazy, she only needs to convince Bob that she is. Offering to burn her own money seems like a good way to do that.
    This is one of those games where 2 rational people will share the winnings, 2 selfish irrational people will simply lose, but a rational person with an irrational one will accept the lesser prize rather than lose even though that gives the big prize to the irrational person. But it also means that a selfish rational person will want to appear irrational thus making the big prize attainable.

  • john

    Sorry, I was a little quick to comment last time. It seemed the utility function matrix wasn’t very appropriate for the situation. I’ve corrected my point of view in a reply to myself if you’re interested. But be warned: it is long.

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