# Probability For Us Dummies 2: More Boy-Girl Variants

I want to disturb you with probability and its (apparent) paradoxes some more.

Let’s warm up with something that looks really straightforward. (Uh-oh!) You are shown a red door and a blue door, and told there is a prize behind one of them. What’s the likelihood that the prize is behind red as opposed to blue?

You probably said 1-in-2 for each door, and you may well be right. Or maybe you were playing Let’s Make a Deal, you originally picked the red door from among three (say red, green, and blue), the host showed you the prize wasn’t behind the green door, so now you think it’s 1-in-3 for red and 2-in-3 for blue.

But, depending on the back story, the probabilities could be literally anything. I mean it. Pick any fraction a/b, where a and b are integers and a < b. For concreteness, say a = 13, b = 63, so your fraction is 13/63. Now let’s play Let’s Make a Deal with 63 doors. It works like this: there’s a car behind one door, and you get to pick 13 doors out of 63. I’ll open 12 of the 13 doors that you picked, and 49 of the 50 doors you didn’t pick. I’ll make sure the car is not behind any of the doors I open.

Of the two remaining doors, say the one you picked is red and the one you didn’t pick is blue. Then the probability that the car is behind the red door is 13/63 (or a/b). We can play the same game for any fraction you give me.

So when you see just two doors to pick between, remember that you might be missing a whole back story that can impact your choice. In fancier language: to make inferences based on observations, you need to know the process by which those observations were arrived at.

In my last post, I talked about this in the context of the so-called Boy-Girl Puzzle. This puzzle has a number of interesting variants, and I want to reinforce the importance of back story by touching on a couple of them.

The set-up: you’re at a school reunion, chatting at a reception with a friend who you find out has two kids. Some of your classmates’ kids are in the room as well, although you haven’t paid any attention to them because you’ve been focusing on the grown-ups. Now suppose that while you’re talking, your friend points toward a cluster of kids (you can’t tell which kid he’s pointing to) and says, “There’s my daughter.” The Boy-Girl puzzle asks: what’s the likelihood that your friend has two kids of the same gender? Meaning, in this case, that their other kid is also a girl.

Imagine two scenarios:

Scenario 1. Since everyone brought kids to the reunion, the organizers have kindly put together some events to keep the kids entertained. Today there’s a basketball camp. But there isn’t enough gym space for all the kids, so this has been scheduled in two phases: a girls’ session from 4 to 5, and then a boys’ session from 5 to 6. Assume all the kids go to their respective session. It is now 5:15, the girls’ session let out a little while ago, and now all the girls are in the reception room while the boys are playing basketball.

Scenario 2. Same as Scenario 1, only the school has a really big gym, so the basketball camp is at the same time for everybody. It’s 5:15 again, and all the kids, girls and boys alike, are in the reception room.

Analyzing Scenario 1, let’s imagine that there are 400 two-kid families in your class, 100 each for each of the four possible gender combinations: (Girl, Girl), (Girl, Boy), (Boy, Girl), and (Boy, Boy). (For now I’m ordering the kids by age, though any unambiguous ordering will do.) When the girls’ camp lets out, 400 girls make their way to the reception room. Now 300 of the 400 parents of a two-kid family have a daughter in the room. You just found out that your friend is one of them. Of those 300 parents, 200 have a girl and a boy, and 100 have two girls. So the probability that your friend’s other kid is a girl is 1-in-3.

In Scenario 2, there are 400 girls and 400 boys from two-kid families in the reception room. To keep things simple, let’s assume that each of your classmates will eventually see both of their kids. Assume also that a parent with both a girl and a boy is just as likely to see the girl first as the boy. This means that if we list the kids in each family in the order that their parents see them, each of the four gender combinations ((Girl, Girl), etc.) is equally likely. For 200 of the 400 2-kid families, the parent will see a daughter first: 100 with (Girl, Girl), and 100 with (Girl, Boy). One of these parents is your friend, so the probability that their other kid is a girl is 1-in-2.

Even if you believe me that back story is important, the different conclusions here might still seem a little weird. Aren’t you finding out the same information in each case? Why does it matter if you’re in a room full of girls vs. a room with girls and boys?

I find it helpful to think about this in terms of the possibilities that are being ruled out. You start out with a 50-50 chance that your friend has two kids of the same gender. When you’re in a room full of girls only, two things could happen. Your friend could eventually tell you they see their daughter, or, if they have two sons, they wouldn’t see any of their kids at all. So when you hear, “I see my daughter,” you’ve ruled out some (half) of the possible two-kids-of-the-same-gender scenarios. So of course the same-gender probability goes down.

Whereas, if both girls and boys are in the room, you will eventually hear either “I see my daughter” or “I see my son.” (The story ends as soon as your friend sees one of their kids.) “I see my daughter” rules out half the same-gender scenarios, but it also rules out half the different-gender scenarios (the ones where your friend sees their son before their daughter). So the same-gender probability remains the same, 50-50.

If having the same-gender probability bounce between 1-in-2 and 1-in-3 isn’t weird enough for you, you can actually try to make it land somewhere in between!  (In the spirit of the Let’s Make a Deal example, you can imagine a basketball camp with different numbers of girls and boys.) But I want to take things in a slightly different direction. Let’s work off of Scenario 1, so we have a room full of girls, your friend with two kids tells you they see their daughter (1-in-3 chance at this point that your friend has two girls), but now let’s say your friend keeps talking. Suppose you hear your friend say…

Scenario 1A. “That girl’s my oldest kid.”

Scenario 1B. “That girl was born on a Tuesday.”

Now suppose I tell you that in one of these scenarios, the 1-in-3 probabilities of a same-gender pair changes, and in the other it doesn’t. Care to guess which is which? Be careful, the answer might not be what you think!

To analyze Scenario 1A, let’s go back to writing gender combinations in birth order. When we heard “I see my daughter” in Scenario 1, we ruled out all the (Boy, Boy) pairs. In Scenario 1A, we can rule out all 100 (Boy, Girl) pairs (oldest kid is a boy), and none of the (Girl, Boy) pairs. What about the (Girl, Girl) pairs? Well, if we assume that your friend will see their oldest daughter first half the time, and their youngest daughter first the other half, we can rule out half, or 50, of the (Girl, Girl) pairs. So we have 150 possible pairs: 100 (Girl, Boy) and 50 (Girl, Girl). (Symmetrically, if your friend had said “That girl’s my youngest kid,” you would have gotten the other 150 pairs with a girl: 100 (Boy, Girl), and the other 50 (Girl, Girl).) Of the 150 pairs, 100 are opposite-gender and 50 are same-gender. So the same-gender probability is still 1-in-3.

We already saw a version of this analysis in my last post. It basically comes down to this: the same-gender-pair probability depends on what the alternative to “My older kid is a girl” is. If the alternative is “My older kid is a boy,” the probability is 1-in-2. If, as here, the alternative is “My younger kid is a girl,” the probability is 1-in-3.

Now let’s move on to Scenario 1B. To make the arithmetic we’re about to do easier, let’s throw out 8 of our 400 families (two for each gender combination), so now we have 98 pairs of kids under each of (Girl, Girl), (Girl, Boy), and (Boy, Girl). (There are also 98 under (Boy, Boy), but they don’t count since one of your friend’s kids is a girl.) Now you ask: what difference could it make that the girl you saw was born on a Tuesday? Well, let’s see which of the above pairs you can rule out once you know this.

We assume that kids are as likely to be born on any day as any other. So 1/7 of all kids are born on Monday, 1/7 on Tuesday, etc. Of the 98 (Girl, Boy) pairs, the girl was born on Tuesday 1/7 of the time. That makes 14 (Tuesday Girl, Boy) pairs, and 84 (non-Tuesday Girl, Boy) pairs.

Similarly, of the 98 (Boy, Girl) pairs, there are 14 (Boy, Tuesday Girl) pairs and 84 (Boy, non-Tuesday Girl) pairs.

The (Girl, Girl) pairs are different, though. In 14 of them, the older girl was born on Tuesday, and in 14 of them, the younger girl was born on Tuesday. Does that make 28 pairs with a Tuesday girl? Not quite, because there’s overlap. How much? Well, in 1/7 of the 14 (Tuesday Girl, Girl) pairs, the younger girl was born on Tuesday too. 1/7 of 14 is 2. So there are 2 (Tuesday Girl, Tuesday Girl) pairs, 12 (Tuesday Girl, non-Tuesday Girl) pairs, and 12 more (non-Tuesday Girl, Tuesday Girl) pairs — 26 in all. Taking 26 away from 98, we have 72 (non-Tuesday Girl, non-Tuesday Girl) pairs.

The important thing here is that unlike Scenario 1A, in which we got to rule out half the opposite-gender pairs and half the same-gender pairs, in Scenario 1B we ruled out 6/7 of the opposite-gender pairs but only about 5/7 of the same-gender pairs. If we pare down the opposite-gender pairs more than the same-gender pairs, the same-gender probability should go up. We have 14+14+26 = 54 pairs with a Tuesday Girl, and 26 of them are (Girl-Girl) pairs, so the same-gender probability is now 26/54 = 13/27 (almost a half!).

What drove things here was that in Scenario 1B, the additional condition (born on a Tuesday) is non-exclusive: either or both kids in the pair could be born on a Tuesday. Whereas, in Scenario 1A, only one kid in the pair could be the older kid. The non-exclusivity means that the additional information (kid you see was born on a Tuesday) is more restrictive in the opposite-gender pairs, when you know which kid it applies to (the girl in the pair) than in the same-gender pairs, when you know it applies to the girl you see, but that girl could be either of the girls in the pair.

Still, I don’t blame you if you find Scenario 1B kind of a head-scratcher. Or, if you don’t, and you have an alternate explanation that you like, please write it down in the comments!

# Probability For Dummies (And We’re All Dummies)

Sometimes it feels like probability was made up just to trip you up. My undergraduate advisor Persi Diaconis, who started out as a magician and often works on card shuffling and other problems related to randomness, used to say that our brains weren’t wired right for doing probability. Now that I (supposedly!) know a little more about probability than I did as a student, Persi’s statement rings even truer.

I spent a little time this weekend thinking lately about why probability confuses us so easily. I don’t have all the answers, but I did end up making up a story that I found pretty illuminating. At least, I learned a few things from thinking it through. It’s based on what looks like a very simple example, first popularized by Martin Gardner, but it can still blow your mind a little bit. I actually meant to have a fancier example, but my basic one ended up being more than enough for what I wanted to get across. (Some of these ideas, and the Gardner connection, are explored in a complementary way in this paper by Tanya Khovanova.) Here goes.

Prologue. Say you go to a school reunion, and you find yourself at a dimly-lit late evening reception, talking to your old friend Robin. You haven’t seen each other for years, you’re catching up on family, and you hear that Robin has two children. Maybe the reunion has you thinking back to the math classes you took, or maybe you’ve just been drinking too much, but for some reason, you start wondering whether Robin’s children have the same gender (two boys or two girls) or different genders (one of each). Side note: if you’ve managed to stay sober, this may be the point at which you realize that you’ve not only wandered into a reunion you’re barely interested in, you’ve wandered into a math problem you’re barely… um, well, anyway, let’s keep going.

The gender question is pretty easy to answer, at least in terms of what’s more and less likely. Assuming that any one child is as likely to be a girl as a boy (not quite, but let’s ignore that), and assuming that having one kid be a girl or boy doesn’t change the likelihood of having your other kid be a girl or boy (again, probably not exactly true, but whatever), we find there are four equally likely scenarios (I’m listing the oldest kid first):

(Girl, girl)      (Girl, boy)     (Boy, girl)     (Boy, Boy)

Each of these scenarios has probability 25%. There are two scenarios with two kids of the same sex (total probability 50%), and two scenarios with two kids of opposite sexes (total probability also 50%). Easy peasy.

But things won’t stay simple for long, because you’ve not only wandered into a school reunion and a math problem, you’ve also wandered into a…

Really. So you’re at the reunion, still talking to Robin, only you might be sober, or you might be drunk. Which is it?

Sober Version: You and Robin continue your nice lucid conversation, and Robin says: “My older kid is a girl.” Does the additional information change the gender probabilities (two of the same vs. opposites) at all?

This one looks easy too, especially given that you’re sober. Now that we know the older kid is a girl, things come down to the gender of the younger kid. We know that having a girl and having a boy are equally likely, so two of the same vs. opposite genders should still be 50-50. In terms of the scenarios above, we’ve ruled out the last two scenarios and have a 50-50 choice between the first two.

But now let’s turn the page to the…

Drunk Version: You and Robin have both had more than a little wine, haven’t you? Maybe Robin’s starting to mumble a bit, or maybe you’re not catching every word Robin says any more, but in any case, in this version what you heard Robin say was, “My umuhmuuh kid is a girl.” So Robin might have said older or younger, but in the drunk version, you don’t know which. What are the probabilities now? Are they different from the sober version?

Argument for No: Robin might have said, “My older kid is a girl,”in which case you rule out the last two scenarios as above and conclude the probabilities are still 50-50. Or Robin might have said, “My younger kid is a girl,” in which case you would rule out the second and fourth scenarios but the probabilities would again be 50-50. So it’s 50-50 no matter what Robin said. It doesn’t make a difference that you didn’t actually hear what it was.

Argument for Yes: Look at the four possible scenarios above. All we know now is that one of the kids is a girl, i.e., we’ve only ruled out (Boy, Boy). The other three are still possible, and still equally likely. But now we have two scenarios where the kids have opposite genders, and only one where they have the same gender. So now it’s not 50-50 anymore; it’s 2/3-1/3 in favor of opposite genders.

Both arguments seem pretty compelling, don’t they? Maybe you’re a little confused? Head spinning a little bit? Well, I did tell you this was the drunk version!

To try to sort things out, let’s step back a little bit. Drink a little ice water and take a look around the room. Let’s say you see 400 people at the reunion that have exactly two kids. I won’t count spouses, and I’ll assume that none of your classmates got together to have kids. That keeps things simple: 400 classmates with a pair of kids means 400 pairs of kids. On average, there’ll be 100 classmates for each of the four kid gender combinations. One of these classmates is your friend Robin.

Now imagine that each of your classmates is drunkenly telling a friend about which of their kids are girls. What will they say?

• The 100 in the (Boy, Boy) square would certainly never say, “My umuhmuuh kid is a girl.” We can forget about them.
• The 100 in the (Girl, Boy) square would always say, “My older kid is a girl.”
• The 100 in the (Boy, Girl) square would always say, “My younger kid is a girl.”
• The 100 in the (Girl, Girl) square could say either. There’s no reason to prefer one or the other, especially since everyone is drunk. So on average, 50 of them will say “My older kid is a girl,” and the other 50 will say, “My younger kid is a girl.”

All together, there should be 150 classmates who say their older kid is a girl, 150 who say their younger kid is a girl, and 100 who don’t say anything because they have no girl kids.

In the drunk version, where we don’t know what Robin said, Robin could be any of the 150 classmates who would say “My older kid is a girl.” In that case, 100 times out of 150, Robin’s two kids have opposite genders. Or Robin could be any of the 150 classmates who would say, “My younger kid is a girl,” and in that case again, 100 times out of 150, Robin’s two kids have opposite genders.

This analysis is consistent with the Argument for Yes, and leads to the same conclusion: there is a 2-in-3 chance (200 times out of 300) that Robin’s kids have opposite genders. But, it seems to agree with the spirit of the Argument for No as well! It looks like knowing Robin was talking about the older kid actually didn’t add any new information: that 2-in-3 chance would already hold if Robin had soberly said “My older kid is a girl” OR if Robin had just as soberly said “My younger kid is a girl.”

But now something seems really off. Because now it’s starting to look like our analysis of the sober version, apparently the simplest thing in the world, was actually incorrect. In other words, now it seems like we’re saying that finding out Robin’s older kid was a girl actually didn’t leave the gender probabilities at 50-50 like we thought. Which is just… totally… nuts. (And not at all sober.) Isn’t it?

Not necessarily.

Here’s the rub. In the sober version, the conversation could actually have gone a couple different ways:

Sober Version 1.

ROBIN: I’ve got two. My older kid is a junior in high school, plays guitar, does math team, runs track, and swims.

YOU: That’s great. Girls’ or boys’ track? The girls’ track team at my kids’ school is really competitive.

ROBIN: Girls’ track. My older kid is a girl.

Sober Version 2.

YOU: I teach math and science, and I’m really interested in helping girls succeed.

ROBIN: That’s great! Actually, if you’re interested in girls doing math, you might be interested in something that happened to one of my kids. My older kid is a girl, and…

Comparing Versions. In both versions, it looks like you ended up with the same information (Robin’s older kid is a girl). But the conclusions you get to draw are totally different!

Let’s view things in terms of your 400 classmates in the room. In Sober Version 1, the focus is on your classmate’s older kid. The key point is that, in this version of the conversation, in the 100 scenarios in which both of your classmate’s kids are girls, you would hear “my older kid is a girl” in all of them. Of course in the 100 (Girl, Boy) scenarios, you would hear “my older kid is a girl” as well. That makes for 200 “my older kid is a girl” scenarios, 100 of which are same-gender scenarios. The likelihood that both kids are girls is 50-50.

Whereas in Sober Version 2, the focus is on girls. In the 100 scenarios in which both of your classmate’s kids are girls, you should expect to hear a story about the older daughter about half the time, and the younger daughter the other half. (Perhaps not exactly, because the older kid has had more time to have experiences that become the subject of stories, but I’m ignoring this.) Combining this with the 100 (Girl, Boy) scenarios, we get 150 total “my older kid is a girl” scenarios. Only 50 of them are same-gender scenarios, and the likelihood that both kids are girls is only 1-in-3.

Why Probability Makes Us All Dummies. Probability is about comparing what happened with what might have happened. Math people have a fancy name for what might have happened: they call it the state space. What we see in this example is that when you talk about everyday situations in everyday language, it can be very tricky to pin down the state space. It’s hard to keep ambiguities out.

Even the Sober Version, which sounds very simple at first, turns out to have an ambiguity that we didn’t consider. And when we passed from the Sober Version to the Drunk Version, we got confused because we implicitly took the Sober Version to be Version 1, with a 200-person state space, while we took the Drunk Version to be like Version 2, with a 150-person state space. In other words, in interpreting “My older kid is a girl” vs. “One of my kids is a girl,” we fell into different assumptions about the background. I think this is what it means that our brains aren’t wired right to do probability: it’s incredibly easy for them to miss what the background assumptions are. And when we change the state space without realizing it by changing those background assumptions, we get paradoxes.

Note: while I framed what I’ve been calling the Drunk Version (one of my kids is a girl) in a way that makes Version 2 the natural interpretation, it can also be reframed to sound more like Version 1. In that case, the Argument for No in the Drunk Version is fully correct, and the probabilities are 50-50. From a quick online survey, I’ve found this in a few places, including Wikipedia and the paper I linked at the start. I haven’t seen anyone else note that what I’ve been calling the Sober Version (my oldest kid is a girl) can be also framed in multiple ways. Just more proof that it’s really easy to miss background assumptions!

Another point of view on this is in terms of information. The Sober vs. Drunk versions confused us because it looked like we had equivalent information – one of the kids is a girl – but ended up with different outcomes. But in fact we didn’t have equivalent information; in fact in the Sober version, there was an essential ambiguity in what information we had! The point here is that just knowing the answer to a question (my oldest kid is a girl) usually isn’t the full story when it comes to probability problems. We need to know the question (Is your oldest kid a girl vs. Is one of your kids a girl) as well. The relevant information is a combination of a question and a statement that answers it, not a statement (or set of statements) floating on its own.

# Monty Hall and Loss Aversion

After my last post, I started thinking about other ways to extend the Monty Hall problem to more doors. If you want to take your intuition in a very different direction, try this: there’s a car behind one of 100 doors. You, the player, get to pick 98 doors. Then Monty opens one of the two remaining doors, showing you the car isn’t there. He asks you if you want to stand pat or trade one of the doors you originally picked for the remaining unopened door.

The same general logic still applies. Your original win probability is 98/100. If you switch, you lose only if the car was behind the door you gave up (probability 1/100), so your win probability after switching is 99/100. Now I don’t know about you, but in this version, I find it really hard to make my brain grasp intuitively that there’s anything to be gained by switching. Which might be a little more evidence for a psychological basis for loss aversion, or the endowment effect, or whatever you want to call that phenomenon whereby people are reluctant to give up, or risk, what they already have. Or maybe it shows our inability (or unwillingness) to distinguish between numbers once they get really big.