Honor Thy Father and Thy Mother

Many of us are familiar with the Fifth Commandment, given by the eponymous title of this post. But what does this mean in practice? In the Jewish tradition, the Rabbis interpreted one’s obligations under this commandment as the requirement to feed and clothe one’s parents, along with other similar duties. In other words, basically, one must take care of one’s parents in their old age.

We would like it if the kids had reason to actually fulfill these duties. For, as Immanuel Kant put it, “Nevertheless, in the practical problem of pure reason, i.e., the necessary pursuit of the summum bonum, such a connection is postulated as necessary: we ought to endeavour to promote the summum bonum, which, therefore, must be possible.”[1] In other words (abusing Kant a little bit), we would like the kids to actually take care of their parents in their old age in a Nash equilibrium solution.

Here’s the problem. The kids may well be rotten, and tell their parents, “So long, and thanks for all the fish!” Why should they waste their precious time and resources taking care of them when there’s nothing in it for them?

The standard reply is that, since you were raised by your parents from infancy, you ought to take care of them. Just as, when you were incapable of maintaining yourself, they clothed and fed you, you ought to do the same when they are incapable of maintaining themselves in their old age. If they would anticipate that you’d be so ungrateful, they wouldn’t have raised you in the first place! But, while the kid is happy that his parents raised him, this is still utterly absurd from a game theory perspective. Here’s the game tree:

Image

As per the story in the following paragraph,  T>D>N, and K>C>0.

We see from the game tree that, if we reach the second stage (where the kid has to make his decision, and the parents have already decided to raise the kid), then the optimal thing for the kid to do is to betray his parents. Since there’s nothing they can do about it, they are sort of stuck if that happens. Knowing this, the parents should expect that, if they raise the kid, they will receive a payoff of N. So, they should stick with the dog.

This solution seems paradoxical. After all, we don’t see parents choosing not to have children because of this. One might say that really, they’d still rather have the kid anyway, i.e. N>D; but this is not universally true, and was certainly more frequently not the case before the modern era. It used to be that children were considered a financial asset, as they could work the fields, bring in extra cash from factory work (in the 19th century), and/or support them in their old age. Without these incentives, the children would not have been thought worthwhile in the first place.

But in any case, we don’t see that all kids abandoning their parents in their old age. The stigma of not taking care of them no longer exists to a large extent, since we hear all the time of children estranged from their parents, sending them to nursing homes, etc. Even if N>D, game theory would seem to predict that they would do so, or else they are not acting rationally; as per the quote from Kant before, we’d like it to be in one’s rational interest to help one’s parents.

Kant would likely say that the game tree does not appear as it morally ought. That is, morally, one ought to have preferences where D>N, so that parents would choose to have children and children would choose to take care of their parents, as this leads to the socially optimal outcome. This falls in line with Kant’s categorical imperative: one ought to act in the way that one wishes were universal law. Since one would wish that it were universal law that children took care of their parents, their preferences and actions should fall into line.

But what if the preferences remain as above? Is there any way to save Kant, along with (more importantly) the incentive to honor one’s parents?

One (naïve) way to try to resolve this is by making the above game tree into a repeated game. If the kids are rotten, then this triggers the “dog strategy,” in which the parents expect that the kids will always be rotten going forward, and so always get the dog instead. Wary of this, the kids will be sure to toe the line. With discount factors sufficiently close to 1, this will be a subgame-perfect Nash equilibrium.

This, however, doesn’t quite work, since the kids are only kids once. By the time they have the ability to act rottenly, they have no concern for future stages of the game – the parents have no means by which to punish them for their deviance. So, the above proposed solution fails.

A more sophisticated method of enforcing taking care of one’s parents involves repeating this game with overlapping generations. After all, nobody lives forever: the kids will one day grow old and feeble themselves. So, we can have THEIR kids punish them for not taking care of their parents.

Here’s the idea: each generation lasts for two periods, each consisting of the two stages in the game tree above. In the first, they are the kids; in the second, they are the parents. If the kids (Generation B) decide not to take care of their parents (generation A), then their kids (Generation C) will not take care of them, either. Thus generation B will get a payoff of K+D, since they will know that their kids will not bother to take care of them to punish them for their own malfeasance.

Now, we might be worried that Generation C will not have incentive to punish their parents, out of fear of being punished themselves by their kids in turn (Generation D). We can resolve this by making an exception to the above rule, so that a generation (C) is not punished for not taking care of its parents (Generation B) if its parents (Generation B again), in turn, did not take care of their parents (here, Generation A). Thus Generation C would get the best of both worlds: free-riding from their parents (if they, Generation B, is dumb enough to have kids anyway), and support from their own children (Generation D). Moreover, the entire process resets once we get to generation D, so even if someone screws up and does the wrong thing, it doesn’t screw doom everyone forever.

Thus this proposed strategy profile is a subgame-perfect Nash equilibrium as long as K+D<(K-C)+T, so that all generations prefer to take care of their parents and be taken care of, over backstabbing their parents and being content with a dog. I think this is likely the case for many individuals, and so we can rest easy that our kids will likely not be so inconsiderate as to send us to a nursing home if they wouldn’t want that for themselves.

That being said, as per a fact known as the folk theorem,[2] multiple equilibria will exist in the repeated game framework, so everyone not taking care of their parents will also be an equilibrium. This can explain why some people fail to do their moral duty. Kant would not approve.


[1] Critique of Practical Reason, Book II, Chapter II, Section V. No, I’m not quite a Kantian, at least in this regard, but I like the quote.

[2] So called because it was among the “folklore” of game theory, sort-of known by everyone before it was rigorously proven.


Does White have an advantage in chess?

It seems likely, doesn’t it? After all, as White, you get to develop your pieces first, putting them in position to better attack Black, or at least defend against his/her attacks. And the statistics seems to bear this out (at least, according to Wikipedia, my source for all things true). Though it could turn out that Black has an advantage – it might be that any move by White fatally weakens his/her position, so that they are at a comparative disadvantage to Black.

Whatever the outcome might be, it turns out that if either side has an advantage, that advantage is necessarily complete. Put more formally: either (a) there exists a strategy profile for White that guarantees victory, or (b) a strategy profile for Black that guarantees victory, or (c) strategy profiles for both White and Black that guarantee a draw. It sounds somewhat trivial, but it’s not: for example, if (a) is true, this means that no matter what Black does, the outcome of the game is not in doubt – White will win.

Results such as these are common among many games without exogenous uncertainty, and many games have been solved, so we know which of the analogues to the above possibilities are true. For example, checkers was recently shown to have strategy profiles for both players which guarantee a draw. So that this holds true for chess as well should not come as a surprise.

To show that one of these three possibilities must hold, we can draw a game tree which contains all the possible move sequences in chess[1]. This is because chess ends in a finite number of moves: a draw is automatically declared if the same position is reached three times, or if fifty moves have gone by without a pawn move or a piece capture. Since there are only a finite number of possible pawn moves (given the size of the board) and piece captures (since there are only 32 pieces), the game is finite.

Next, we can use backward induction (as in my post on tic-tac-toe) from each possible ending of a game to determine the outcome from the beginning. At each node, the player involved (White or Black) deterministically selects the branch that leads to the best final outcome for him/her (using tie-breakers if necessary if several outcomes are equally good). We proceed in this manner all the way up to the initial node, corresponding to the starting position of the game. We can then go back down the tree, and since we have already determined the best response to any position, we can deterministically get to the best outcome for Black or White. This automatically yields a win for one of them, or a draw.

Unfortunately, while this works well in theory, in practice it is virtually impossible. Given the combinatorial explosion of positions in chess, the computing necessary to determine which possibility is correct is infeasible. I guess we’ll be stuck with just a good game of chess.


[1] That is, theoretically; the actual game tree is WAY too big to actually depict


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:

Image

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!


The Monty Hall Deception

When I was in middle school, I consumed a lot of typical nerd literature like Richard Feynman’s “Surely You’re Joking, Mr. Feynman” and anthologies of mathematics puzzles from the Scientific American by Martin Gardner. In the latter, I first encountered the Monty Hall Problem, and it goes something like this:

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

It turns out that, yes, it is always to your advantage to switch your choice. This is a solution that has been notoriously difficult for people to wrap their heads around. After all, when you picked a door, the probability of having picked the door with a car was still 1/3, and after a door was opened, there would still be a car and a goat behind the remaining two doors – it seems as through the probability of choosing the door with the car ought to be ½ regardless of the door chosen.

The Monty Hall Paradox is in fact not a paradox at all, but rather just some clever sleight of hand. The trick is that people are drawn to the fact that there are only two doors rather than three doors remaining, and assume that the host’s having opened a door is favorable to the player. People tend not to realize that the game has imperfect information – the player does not know where on the game tree he is, whereas the host does. Additionally, people assume that the host has no stake in the game (and this is not unreasonable, because the problem does not explicitly describe a parsimonious host! On the other hand, intuitively, we know that the host isn’t going to ruin the game by opening the door with the car.) So, if we assume that the host is profit maximizing and we model the problem as an extensive form game with imperfect information, then the conditional probabilities would be easy to see.

Now, just for fun, we’ll assign some utilities to the outcomes. What is a goat worth? According to a popular Passover song in Aramaic, a (small) goat is worth about 2 Zuz, and according to the traditional Jewish prenuptial document, a wife is worth about 200 Zuz. So, a goat is worth about 1/100th of a wife. I asked my roommate, Anna, how many cars she thought a wife was worth, and she determined that a wife was worth three cars. By transitivity, then, a car is worth about 33 goats. (I think goats have become quite a bit less valuable since that song was written, or maybe goats back then were a lot better than our goats.) So, if the player wins the game, he will walk away with a utility of 33, and the host will walk away with the 2 goats.

monty hall

In this game, the light gray branches are dominated because the host has no incentive to open the door that the player has already chosen, and the dark gray branches are dominated because, of the remaining two doors, the host would not open the door that has the car. We can tell that in the top branch, the host as 2 possible choices for doors to open, whereas in the lower two branches, the host is constrained to only one door (since, if the player has chosen a goat door, there is only one goat door left to open.)

So, since the player has no idea after the first stage what door has the car, we assume he picks door No.1 (as in the game). If he observes that the host opens door 3, he would know that there are two cases where the host opens door 2: in the world where the car is behind door 2, the host chooses door 3 100% of the time, and in the world where the car is behind door 1, the host chooses door 3 50% of the time. It’s actually twice as likely that we are on the 100% branch as that we are on the 50% branch – and that’s the branch where the car is hidden behind the other door.

What if we know that the host has opened a door, but we don’t know which one? Then, we can’t condition on a prior, because we don’t know what the prior is – we don’t get any new information by observing which door was opened, and switching doors would not help.


How to resolve a hostage crisis: Part I

Suppose an archvillain, with a few lackeys, comes into a bank lookin’ for loot. Knowing that he has the potential to get more out of the bank than merely what’s in the vaults, and knowing that it’s not so simple to just drive away, they decide to hold the innocent civilians in the bank hostage. Unfortunately, Batman is away in Shangri-La for now, so he can’t help you. Spiderman is busy trying to learn how to fly. And the other superheroes just suck. So your only option is to engage in a standoff, with an elite SWAT team standing outside, ready to storm the building if possible/necessary. Meanwhile, the criminals are inside, threatening to kill their hostages unless they get $100 million.

Suppose, for now, that the criminals are completely rational, and this is known to the SWAT team as well. Let’s start with the case where there is only one hostage, and the SWAT team cares about this dude so much that his life is worth $100 million to save. Nonetheless, if the SWAT team refuses to pay the ransom, what are the criminals to do? If they shoot the guy, then there’s nothing to prevent the SWAT team from storming the bank and shooting them dead, or at the very least capturing them, resulting in a long prison term for murder (on top of robbing the bank). This extra penalty makes it worse than not killing the hostage. Thus the SWAT team has no incentive to give in to their demands, as the threat of killing the hostage is not credible – the criminals are worse off if they do so.

Now let’s see what happens if there are more hostages. We can use induction to see that the hostage-takers face this predicament, no matter how many hostages are involved. Suppose that, once there are N hostages left, the SWAT team has no reason to give in. We then look at the case with N+1 hostages. If the hostage-takers kill the spare, they are left in a situation in which they automatically lose – they won’t get their ransom, and they’re stuck there until they are forced to surrender. Thus they won’t want to kill the N+1th hostage. We therefore end up with no dead hostages, no matter how menacing the threat.

I know what you’re thinking now: “Yeah right. Why don’t you try this strategy and see what happens? If you do, the hostages will end up DEAD.” And you’re probably right. So where have I gone wrong?

Uncertainty in the number of hostages doesn’t help much here, because the number of hostages will be bounded. We can therefore apply the same induction reasoning as before to the uncertain number of hostages, and once the criminals have claimed to kill the upper bound of the number of people that could possibly have been in the bank, they’re still dead meat. So we’ll have to find some other reason.

The key to my original argument was that the criminals are rational. Obviously, that’s not always going to be the case. After all, most rational, normal people do not go around robbing banks. So, chances are pretty good that the people robbing the bank aren’t quite rational, and will kill the hostages even though it ends up hurting them. In this case, it might be worth paying up to prevent the deaths of innocents. We’ll analyze this in part 2, which I will post in a week.


How to Win at “Shotgun” (or not)

I’ve known about this game for quite a while. I learned it in camp as a kid, and loved it at the time. In college, some of my friends rediscovered this game, and would play it at the dining hall table. So, I thought I would write about it in a post.

This game is a little bit more obscure than some of the others I’ve covered, so let me go through some of the details (which can be found here, anyway). Basically, you move simultaneously, by slapping your lap twice, then doing one of three moves: reload, shoot, or shield. Reloading gives your gun an additional bullet. Shooting means you’re trying to kill the other guy. Shielding blocks the (potential) bullet that the other guy is shooting at you. You win by shooting the other guy while he’s reloading; if you shoot each other simultaneously, the game ends in a draw.

Let’s assign the winner 1 point, the loser 0 points, and each player 1/2 a point in case of a draw. That seems pretty reasonable, right?

So, first thing, we can see that there’s a very simple subgame perfect Nash equilibrium (SPNE): each player always shields, no matter how many bullets he/she or the other player have. That this is a SPNE is pretty clear: nobody can do better by ever doing something else, since they’ll never get the opportunity to successfully shoot the other guy anyway, since his shield will always be up. Thus the game ends in a draw (or goes on forever, you pick). It’s pretty boring, but it works.

The question is, can we come up with something more interesting? Can anyone actually ever win in an SPNE? As we’ll see, the answer is NO.

Notice that in the SPNE, the expected number of points at any time cannot be less than 1/2 for each player: if it were, then they could do better by simply shielding forever, guaranteeing them at least 1/2 a point. Since the total number of points is 1, this means that each player can always expect exactly 1/2 a point.

This immediately implies that neither player will ever reload when the other guy has a bullet in the chamber of his gun (at least, not with a non-zero probability). If there ever were the slightest chance that he would, the other guy could shoot him, guaranteeing himself an expected number of points greater than 1/2. This is because, in the slight probability that he was reloading, the other guy would win (and get one point); if he ended up shielding while the other guy shot, the other guy could just shield forever. But this goes against the point in the previous paragraph: in any SPNE, the expected number of points for each player is exactly 1/2. Thus one will never reload, when there’s a possibility one could be shot, in any SPNE.

But this means that no one ever wins. Sounds pretty boring. I don’t think I’ll be playing this anymore.


Are you going to Hell?

One often hears Bible-thumpers declaring that anyone who does not adhere to (a rather narrow) religious tradition will be going to Hell when he or she dies. Often, their theology includes a God that is entirely good. Thus, when pressed why such a good God would institute such an awful thing as Hell, they claim that the purpose is to discourage people from deviating from the appropriate religious path. Since, if one does not adhere to the aforesaid religion, one will end up having very bad things happen, one will have incentive to actually do what is right.

We can write out a game tree to express this idea:

Here, H>L, and C,R>0. We illustrate God’s preferences of how the world should be through the “payoff” He gets; this is independent of the (much more controversial) thesis that God somehow enjoys certain states of the world in a hedonistic fashion.

The problem here with this reasoning for the existence of Hell is that it does not constitute a subgame-perfect Nash equilibrium. We’ve assumed that God only does something bad (sending people to Hell) to prevent worse things from happening (mass sinning). Consider the possibility that 100% of people adhere to the proper religion, and then God decides to send all those who acted righteously to heaven (100% of all people), and all those who sinned (egregiously?) to Hell (0% – no one). Then God is indeed doing what He prefers most by sending all sinners to Hell; since all sinners go to Hell, it is actually a Nash equilibrium for God to send all sinners to Hell.

The thing is, if someone actually did sin, God would have to consider what to do then and there. Since, once one is dead, sending you to Hell won’t (retroactively) prevent one from sinning, doing so would not serve its purported purpose. Moreover, Hell is unobservable to those who are still living, so it does not serve to deter the others who are alive from sinning – the conditional beliefs on those still alive are the same, regardless of what God actually does to the sinner. Thus, even if 100% of the world’s population would be sin-free, this scenario would not be a subgame-perfect Nash equilibrium, and thus the optimal thing for people to do would be to live it up. Given that the sinners would have already committed all of their sins, and there would be nothing possible to do about it after their deaths, the optimal thing for God to do would not be to send them to Hell. Indeed, since I don’t think there is any religion which doesn’t think somebody has existed who deserves to go to Hell, this isn’t a Nash equilibrium, either – the past sinners who should’ve gone to Hell under this worldview would not, since God would prefer to not send them to Hell (since He is good).

If one is then to posit the existence of Hell, one will have to address the issue that it cannot be based upon some purpose of deterrence where God acts completely rationally. It will probably have to be instead based on something else, such as a deontological framework, which may include a notion of just deserts.