# Why do women (almost) never ask men on dates?

This is something I’ve asked a few of people about. It seems odd that in our modern, post-feminist age, it is almost always men who do the asking out. This is not so good for both men and women. For men, it puts a lot of pressure on them to make all of the moves. For women, I cite Roth and Sotomayor’s classic textbook on matching, which shows that, though the outcome from men always choosing partners is stable, it is the worst possible stable outcome for women. That is, women could get better guys to date if they made the moves.

I have a few hypotheses, but none of them seem particularly appealing:

1) Women aren’t as liberated as we think.

Pro: There doesn’t seem to be any point in history where this was any different, so this social practice may indeed be a holdover from the Stone Age (i.e. before 1960).

Con: If this is true, then it is a very bad social practice, and we should buck it! This is not a good reason to maintain it!

2) If a woman asks a man out, it reveals information about her. This could be a case of multiple equilibria. Suppose that a small percentage of “crazy types” of both men and women exists, and under no circumstances do you ever want to date one of them. The equilibrium in which we are is fully separating for women, where the “normal types” always wait for men to ask them out, while the “crazy types” ask men out. Since this is a perfect Bayesian equilibrium, men know that if they get asked out, the woman must be crazy, and so they reject. Knowing this, the “normal” women would never want to ask a man out, since it would involve the cost of effort/rejection with no chance of success.

Suppose the chance that someone is crazy is some very small $\epsilon > 0$. Consider the game tree:

Notice that the crazy women always want to ask the guy out, no matter what the beliefs of the guy are.

There are a few perfect Bayesian equilibria of this game, but I will highlight two. The first is that the normal women never ask guys out, and guys never accept. As $\epsilon \rightarrow 0$, this gives expected payoff to people of $(0,0)$. No one wants to deviate, because only crazy women ask guys out, and so a guy would never accept an offer, as that would give payoff $-10$ instead of $0$; knowing this, normal women will never ask men out, because that gives them payoff $-1$ instead of $0$.

Another equilibrium is that all women ask men out, and men always accept. As $\epsilon \rightarrow 0$, the expected payoff vector is $(2,2)$. Thus the former is a “bad” equilibrium, while the latter is a “good” one. In other words, we may be stuck in a bad equilibrium.

Pro: I think that there definitely some guys out there who think that women who would ask them out are “aggressive” or “desparate,” and so they wouldn’t go out with them.

Con: I don’t think the above sentiment is true in general, at least for guys worth dating! If a guy has that attitude, he’s probably an @#0!3 who’s not worth your time.

There may also be some elements of the problem with (1), but these would be harder to overcome, as the scenario here is an equilibrium.

Finally, while this might have some plausibility for people who don’t really know each other yet, I definitely don’t think this is true for people who know each other somewhat better, and therefore would already know whether the woman in question was crazy. That being said, I would expect it to be more likely that a woman who has known the man in question for longer to be proportionally more likely to ask him out (relative to the man), even if it is still less likely.

3) Women just aren’t as interested. If he’s willing to ask her out, then fine, she’ll go, but otherwise the cost outweighs the benefit.

Pro: It doesn’t have any glaring theoretical problems.

Con: I want you to look me in the eyes and tell me you think this is actually true.

4) They already do. At least, implicitly, that is. Women can signal interest by trying to spend significant amounts of time with men in whom they have interest, and eventually the guys will realize and ask them out.

Pro: This definitely happens.

Con: I’m not sure it’s sufficient to even out the scorecard. Also, this seems to beg the question: if they do that, why can’t they be explicit?

When I originally showed this to some friends, they liked most of these possibilities (especially (1) and (2)), but they had some additional suggestions:

5) Being asked out is self-validating. To quote my (female) friend who suggested this,

…many girls are insecure and being asked out is validation that you are pretty/interesting/generally awesome enough that someone is willing to go out on a limb and ask you out because they want you that badly. If, on the other hand, the girl makes the first move and the guy says yes it is much less clear to her how much the guy really likes her as opposed to is ambivalent or even pitying her.

ProThis is true of some women.

Con: Again to quote my friend, “There are lots of very secure, confident girls out there, so why aren’t they asking guys out?”

6) Utility from a relationship is correlated with interest, and women have a shorter window. This one is actually suggested by Marli:

If asking someone out is a signal of interest level $X > x$, and higher interest level is correlated with higher longterm/serious relationship probability, then women might be interested in only dating people with high interest level because they have less time in which to date.

Pro: It is true, women are often conceived to have a shorter “window,” in that they are of child-bearing age (for those for whom that matters) for a shorter period.

Con: This doesn’t seem very plausible. Going on a date doesn’t take very long, at least in terms of opportunity cost relative to the length of the “window.” As a friend put it in response,

Obviously one date doesn’t take up much time; the point of screening for interest $X > x$ is to prevent wasting a year or two with someone who wasn’t that into you after all. But then it would seem rational for (e.g.) her to ask him on one date, and then gauge his seriousness from how he acts after that. Other people’s liking of us is endogenous to our liking of them, it really seems silly to assume that “interest” is pre-determined and immutable.

So overall, it seems like there are reasons which explain how it happens, but no good reason why it should happen. I hope other people have better reasons in mind, with which they can enlighten me!

# How to resolve a hostage crisis: Part II

Last week, I showed that if the villains are perfectly rational, and this is “common knowledge,” then the villains will be best off by just surrendering immediately without killing any hostages. But the rationality assumption is a big one – let’s see what happens if we drop it.

Once there’s a good chance that the villains are “crazy,” even criminals who are perfectly sane will pretend to be crazy so they can extract money as ransom. We can then model this as a Bayesian game, with N periods corresponding to the $N$ hostages that are holed up in the bank. Each period will consist of two stages: in the first, the SWAT team decides whether to pay up or not; in the second, the villains choosing whether or not to kill a hostage; if they don’t, it’s tantamount to surrender. If they run out of hostages to kill, they are forced to surrender, and the surrender outcome gets progressively worse as they kill more hostages.

The payoffs are described as such:

Each dead hostage: gives $-H$ to the SWAT team, and $-C$ to the villains if they end up surrendering.

Surrender: gives $-S$ to the villains.

Pay up: gives $-P$ to the SWAT team, P to the villains, where $H > P$ (1)

Finally, we assume that there is a probability $k$ (initially) that the villains are nuts.

We’re going to construct a mixed strategy solution to this problem, which will yield a perfect Bayesian equilibrium. That is, in a given period $i$, the SWAT team pays up with probability $p_{i}$, and the villains execute a hostage with probability $k_{i}$. For those who are unfamiliar with mixed strategies, the basic idea is that both sides are indifferent between (at least) two options, so they may as well flip a coin as to which one to do. The trick is that the coin is weighted so as to make the other side indifferent as well. Thus they will flip their coins so as to make the first side indifferent. Hence this forms a Nash equilibrium – neither side can benefit by unilaterally changing their strategy.

The derivation is a little drawn-out, so I’m going to break it into steps:

(i) If the SWAT team always pays up in some period $i$, then the villains may as well execute their hostages until they reach that stage; after all, this guarantees them an automatic victory. But if they will do that, then the SWAT team should pay up immediately (i.e. in period 1): since they’ll lose anyway, they may as well save the lives of the hostages.

(ii) Conversely, if the villains will always execute a hostage at a certain stage, then the SWAT team should always pay up then. But then we run into the same issue as in (i), so the SWAT team will end up paying at the beginning. Combined, (i) and (ii) limit the types of equilibria we have to analyze.

(iii) If there is a period i at which it is known that the SWAT team will not pay up, no matter what, then the normal (not crazy) types, knowing this, will surrender in period $i-1$. But this possibility leads to an inconsistency. Knowing that the normal types will do this, the SWAT team will believe that anyone who hasn’t surrendered must be crazy. If so, it is a best response to pay up in order to avoid more casualties.

(iv) Thus, from (iii), there cannot be a perfect Bayesian equilibrium in which the SWAT team will maintain the siege to the bitter end – they must always cave in to the villains’ demands with some positive probability. Indeed, we see from (i) and (iii) that either they pay up immediately, or they pay up in period $N$ with some nonzero likelihood.

(v) There is no benefit to the villains in executing the $N^{th}$ hostage. Thus, if they are normal, they will not execute, and so this hostage will be killed if and only if the villains are crazy, which occurs with probability $k_{N}$.

(vi) In period $N$, by (iv), the SWAT team must still be indifferent between paying up and maintaining the siege. This implies that

$-P-(N-1)H=-(N-1)H(1-k_{N})-NHk_{N}$

$k_{N}=\frac{P}{H}$

(vii) In all periods $i$, the normal-type villains must be indifferent between surrendering and killing a hostage. Thus, if their expected payoff, if the game reaches the second stage of period $i+1$, is $\Pi_{villains}^{i+1}$, then we get

$-(i-1)C-S=p_{i+1}P+(1-p_{i+1})\Pi_{villains}^{i+1}$

But we know that since the villains will be indifferent as well in the next period (unless $i=N-1$), $\Pi_{villains}^{i+1}=-iC-S$. In any case, this formula still holds for $i=N-1$, since the normal types surrender in period $N$ anyway. Thus we get, after substitution,

$p_{i+1}=\frac{C}{P+iC+S}$

As expected, this probability goes down as more hostages are executed – after all, there is less of a risk of more future casualties, since there are fewer hostages left – a dark thought, indeed.

(viii) We now derive the probability that the villains execute a hostage in the periods other than the last. The SWAT team is indifferent between maintaining the siege and paying up. Hence if their expected payoff at the beginning of period $i+1$ is $\Pi_{SWAT}^{i+1}$, then

$-(i-1)H-P=-(1-k_{i})(i-1)H+k_{i}\Pi_{SWAT}^{i+1}$

But because the SWAT team will also be indifferent between paying up and maintaining the siege in the next period, $\Pi_{SWAT}^{i+1}=-iH-P$. Hence

$-P=-k_{i}H-k_{i}P$

$k_{i}=\frac{P}{P+H}$

This value is constant! Note that it is decreasing in $H$ – if the hostage is more valuable to the SWAT team, then the likelihood that they will kill another doesn’t have to be as high in order to deter the SWAT team from maintaining the siege. Conversely, if the hostage is less valuable, the villains will have to be more likely to kill the hostage to show they mean business.

(ix) Combining (vi) and (viii) gives us the the proportion of crazy types, at the beginning of any period $i$ (after the first) in the perfect Bayesian equilibrium will be $(\frac{P}{H+P})^{N-i}\frac{P}{H}$. Thus, if at period 1, the initial probability that the villains are crazy, $k$, is greater than this amount, then the risk that they are nuts (or wannabe-nuts) is too high, and the SWAT team should fold immediately. Otherwise, they should maintain the siege in the first period, while the villains execute the hostages with the exact probability so that the likelihood that they are nuts, as of the start of period 2, is exactly $(\frac{P}{H+P})^{N-2}\frac{P}{H}$. Afterwards, the villains follow the strategy given by (vi) and (viii), while the SWAT team follows the strategy given by (vii).

(x) As a final note, we see from (ix) that as $N\rightarrow\infty$, the threshhold for the initial likelihood of craziness necessary to enforce a SWAT team payout goes to 0. This makes sense – there’s a larger potential for more casualties as the number of hostages goes up.

(1) (footnote: though dropping this assumption only slightly changes the outcome. Also, we exclude the possibility of storming the bank, since it’s similar in concept to the other options).