I Don’t Think the Bus Driver Hates White People

I have an 8:00 meeting this morning, so I’m going to take the early (6:36) bus in. I almost never go in this early, so to be safe, I am already downstairs putting on my shoes at 6:29. Sometimes I joke that I live so close to the bus stop that if I hear the bus from the second floor window of my house, I still catch it as it goes by. That’s an exaggeration, but not by much. Now I hear a sound outside that could be the bus, but I know it’s not, it’s never this early. Still, I peek out the door.

It’s the bus, already on the corner. Crap.

I dash out. The bus stop is diagonally across from my house, and fortunately, the bus is waiting at a red light. This gives me a chance: I run down to the corner and across the street with the light. Now the bus and I are facing each other, I wave, cross the other way when the light turns as the bus waits an extra couple seconds, and get on. I am feeling good about the world and my place in it.

Still, 6:29, that’s super-early. Unless maybe this is the previous bus, running super-late? I think I should say something to the driver. “Excuse me, which bus on the schedule is this?” It’s the morning, everything is rushed, who knows if I actually said “Excuse me.”

The bus driver, not one I’m familiar with, is not pleased. He barks at me: “It’s whatever bus it is when I get here!” I look behind him, see the bus is mostly empty. I know from experience that if it were the previous bus, running late, it would be overcrowded, so this must really be the 6:36. I give the driver my ticket and try to explain: you’re running way too early, people who count on this bus will miss it, then the next buses are overcrowded and run late, etc. I ask him to wait a few minutes to get back on schedule. He remains very hot, says (yells) that I better stop talking to him, threatens to call the police if I don’t get away from him. The good thing about him yelling at me is that the bus is still not moving. A couple people, breathless from running down the street, get on the bus while we’re going back and forth. I count this as a small victory.

A large man walks up the aisle from the back of the bus. I am hoping for a little support from a fellow commuter. “You better fucking stop talking to him so he can go!” he yells at me. “You’re holding the bus up! He’s going to call the cops!” This guy looks like he’d like nothing better than to slug me. “He’s calling the cops!” The bus driver is not calling the cops. My fellow passenger, though he is louder than the driver, does not actually take a swing at me. I sit down in one of the many empty seats.

The bus doesn’t move.

I take my book out of my bag and try to disappear into it. It is eerily silent. In moments of suspense, time feels suspended. The bus still doesn’t move.

Finally, after what must have been only a short wait but didn’t feel that way, the bus huffs and trundles forward. I check the time. It is exactly 6:36. We have been waiting for just a few minutes.

I mull things over as the bus heads into the city. I feel good that I managed to stay calm through the whole episode. Maybe this will be a good story to tell the children — the importance of keeping your cool. It occurs to me that the bus driver ended up doing exactly what I asked him to — waiting to leave until he was scheduled to, to the minute. Or was that just a coincidence? Something makes me decide to talk to him again, though I can’t tell exactly how or why I make that decision.

The bus pulls in to Port Authority. I am going to wait for everyone else to get off and then try to approach the driver. As the people ahead of me are getting off, I hear the usual end-of-trip courtesies: passengers say thank you, driver says you’re welcome or have a good day. That’s encouraging: this isn’t one of those guys who’s so nasty that people give up on saying thanks at the end of the ride.

Once is everyone else is off, I get off too. The driver is standing to the side. “Thanks for waiting a few minutes at the bus stop,” I say to him. “I wasn’t trying to pick a fight, I just…”

The driver is looking me in the eye. “Yeah, I know,” he says. “I get what you were saying.” I talk to him about the schedule, he says he understands, sometimes they get off schedule and it’s hard to keep track. We are talking to each other like people now. He says he’s sorry, and he seems totally sincere. I tell him my name, because that seems like a talking-like-people thing to do, and he tells me his. I was hoping to make peace, but this is more than I expected; in under a minute, our conversation has turned comfortably fraternal. I am feeling good about the world again as I head inside the bus terminal.

Now another man slides over to me. He is fiftyish, tall, thin, with slightly graying hair, wearing jeans and a blazer. “You know,” he says,”I saw the bus driver yelling at you, I got a video on my phone. I’m going to call the bus company.”

I don’t actually want him to call the bus company. I tell him that I just had a good talk with the driver, and that he seemed very direct and sincere. My companion isn’t impressed. “He’s just afraid for his job,” he says. He tells me that the driver has done this (I think he means run ahead of schedule) once before, that people were running for the bus and he didn’t care. He looks at me a little conspiratorially. “I think this guy just hates white people.”

Well. The driver, like the majority of bus drivers I encounter, is black. I am white. The guy I’m talking to now is white. The passenger who was yelling at me earlier is white. Most people on the bus are white, with a few exceptions. What is the logic here, how do you decide that a black man driving a bus full of mostly white people hates them because they’re white? Does the white man who came up to yell at me before hate white people too? Why do we look differently at angry black people than at angry white people?

The man I’m talking to is going his way and I’m going mine, and there isn’t time to ask these questions. But I as I walk up 8th Avenue on a sunny end-of-summer morning, I realize that I’ve lost the tidy story I was going to tell my kids about the virtues of keeping calm. And perhaps a bit of my confidence in those virtues as well.

Cathy’s Book is Out!

Cathy O’Neil’s book Weapons of Math Destruction is out, and it’s already been shortlisted for a National Book Award! Here is a review of the book that I posted on Amazon.com:

So here you are on Amazon’s web page, reading about Cathy O’Neil’s new book, Weapons of Math Destruction. Amazon hopes you buy the book (and so do I, it’s great!). But Amazon also hopes it can sell you some other books while you’re here. That’s why, in a prominent place on the page, you see a section entitled:

Customers Who Bought This Item Also Bought

This section is Amazon’s way of using what it knows — which book you’re looking at, and sales data collected across all its customers — to recommend other books that you might be interested in. It’s a very simple, and successful, example of a predictive model: data goes in, some computation happens, a prediction comes out. What makes this a good model? Here are a few things:

  1. It uses relevant input data.The goal is to get people to buy books, and the input to the model is what books people buy. You can’t expect to get much more relevant than that.
  2. It’s transparent. You know exactly why the site is showing you these particular books, and if the system recommends a book you didn’t expect, you have a pretty good idea why. That means you can make an informed decision about whether or not to trust the recommendation.
  3. There’s a clear measure of success and an embedded feedback mechanism. Amazon wants to sell books. The model succeeds if people click on the books they’re shown, and, ultimately, if they buy more books, both of which are easy to measure. If clicks on  or sales of related items go down, Amazon will know, and can investigate and adjust the model accordingly.

Weapons of Math Destruction reviews, in an accessible, non-technical way, what makes models effective — or not. The emphasis, as you might guess from the title, is on models with problems. The book highlights many important ideas; here are just a few:

  1. Models are more than just math. Take a look at Amazon’s model above: while there are calculations (simple ones) embedded, it’s people who decide what data to use, how to use it, and how to measure success. Math is not a final arbiter, but a tool to express, in a scalable (i.e., computable) way, the values that people explicitly decide to emphasize. Cathy says that “models are opinions expressed in mathematics” (or computer code). She highlights that when we evaluate teachers based on students’ test scores, or assess someone’s insurability as a driver based on their credit record, we are expressing opinions: that a successful teacher should boost test scores, or that responsible bill-payers are more likely to be responsible drivers.
  2. Replacing what you really care about with what you can easily get your hands on can get you in trouble. In Amazon’s recommendation model, we want to predict book sales, and we can use book sales as inputs; that’s a good thing. But what if you can’t directly measure what you’re interested in? In the early 1980’s, the magazine US News wanted to report on college quality. Unable to measure quality directly, the magazine built a model based on proxies, primarily outward markers of success, like selectivity and alumni giving. Predictably, college administrators, eager to boost their ratings, focused on these markers rather than on education quality itself. For example, to boost selectivity, they encouraged more students, even unqualified ones, to apply. This is an example of gaming the model.
  3. Historical data is stuck in the past. Typically, predictive models use past history to predict future behavior. This can be problematic when part of the intention of the model is to break with the past. To take a very simple example, imagine that Cathy is about to publish a sequel to Weapons of Math Destruction. If Amazon uses only  purchase data, the Customers Who Bought This Also Bought list would completely miss the connection between the original and the sequel. This means that if we don’t want the future to look just like the past, our models need to use more than just history as inputs. A chapter about predictive models in hiring is largely devoted to this idea. A company may think that its past, subjective hiring system overlooks qualified candidates, but if it replaces the HR department with a model that sifts through resumes based only on the records of past hires, it may just be codifying (pun intended) past practice. A related idea is that, in this case, rather than adding objectivity, the model becomes a shield that hides discrimination. This takes us back to Models are more than just math and also leads to the next point:
  4. Transparency matters! If a book you didn’t expect shows up on The Customers Who Bought This Also Bought list, it’s pretty easy for Amazon to check if it really belongs there. The model is pretty easy to understand and audit, which builds confidence and also decreases the likelihood that it gets used to obfuscate. An example of a very different story is the value added model for teachers, which evaluates teachers through their students’ standardized test scores. Among its other drawbacks, this model is especially opaque in practice, both because of its complexity and because many implementations are built by outsiders. Models need to be openly assessed for effectiveness, and when teachers receive bad scores without knowing why, or when a single teacher’s score fluctuates dramatically from year to year without explanation, it’s hard to have any faith in the process.
  5. Models don’t just measure reality, but sometimes amplify it, or create their own. Put another way, models of human behavior create feedback loops, often becoming self-fulfilling prophecies. There are many examples of this in the book, especially focusing on how models can amplify economic inequality. To take one example, a company in the center of town might notice that workers with longer commutes tend to turn over more frequently, and adjust its hiring model to focus on job candidates who can afford to live in town. This makes it easier for wealthier candidates to find jobs than poorer ones, and perpetuates a cycle of inequality. There are many other examples: predictive policing, prison sentences based on recidivism, e-scores for credit. Cathy talks about a trade-off between efficiency and fairness, and, as you can again guess from the title, argues for fairness as an explicit value in modeling.

Weapons of Math Destruction is not a math book, and it is not investigative journalism. It is short — you can read it in an afternoon — and it doesn’t have time or space for either detailed data analysis (there are no formulas or graphs) or complete histories of the models she considers. Instead, Cathy sketches out the models quickly, perhaps with an individual anecdote or two thrown in, so she can get to the main point — getting people, especially non-technical people, used to questioning models. As more and more aspects of our lives fall under the purview of automated data analysis, that’s a hugely important undertaking.

 

Transcendence: The Bangles and Sleater-Kinney

You can measure how irresistible the Bangles are by the number of days (now two and counting since I saw them play live) that they’ve had me humming a song I had been pretty sure I hated.

Their show at Irving Plaza in New York on Saturday night really got cooking about half an hour in, with a lovely rendition of Big Star’s September Gurls. No matter what you think of the Bangles, you can’t deny the fact that they are the greatest cover band ever, and it’s not particularly close.

Next they played several tunes, at once primal and shiny, from Ladies and Gentlemen… The Bangles!, a just-released, and excellent, compilation of early recordings. “From when we were baby Bangles,” said one of the ladies, and my inability to remember which one was speaking only testifies to the band’s egalitarianism — everybody sings, everybody harmonizes, everybody talks. Then came Hazy Shade of Winter, a Simon and Garfunkel song once upon a time, but the Bangles took possession of it long ago, with blistering guitar lines and harmonies well beyond the original (!), and they’re not giving it back. Another highlight, kicking off the encore, was How is the Air Up There, which mixed garage rock with punk with a jangling and joyous charm that is entirely the Bangles’ own. When you hear the band on record, they sometimes suffer from 80’s-style overproduction, but all the songs they played at Irving Plaza were reduced to their essence, and the band got all them across. I am not a fan of recording shows on video, but I couldn’t help filming a bit of this one, just to have a keepsake.

The band closed the show, inevitably I suppose, with Eternal Flame, their biggest hit and the Freebird of their universe, more or less. I assumed this would be a letdown, especially after the vitality of the main set: despite Susanna Hoffs’s committed vocal, Eternal Flame always struck me as overproduced mush, pure pop cheese. But there’s a reason people yell and flick their lighters and refuse to go home till they hear Freebird: everybody likes a sing-along. After an hour and a half of speeding down Bangle freeway together, it was time for band and audience to meld into a single whole, and the ladies were smart and confident enough to make that happen by slowing down the pace and ending with a group hug of a ballad. So drummer Debbi Peterson came out from behind her kit and joined her bandmates on guitar, polka dots of light swirled through the audience, the cheesy lyrics became a call to harmony inside the club (Close your eyes! Give me your hand!), and there was no way you could keep yourself from singing along. It was corny, genuinely warm, and fully earned, a glorious way to send a thousand people jammed tightly in a small space out into the New York night. Maybe the song is treacle, but now I can’t stop humming it.

I was still feeling the thrill of the Bangles show two days later when I remembered how another band had recently made me feel almost exactly the same way. It was Sleater-Kinney, the Bangles’ spiritual successors in more ways than one, who came back last year after an almost ten year hiatus, playing just as fiercely as ever. When I went to see them, they too had ripped through an hour and a half of take-no-prisoners rock and roll… and then slowed things down and ended the show with their ironic ballad, Modern Girl. Only they played it as a sing-along, audience and band joined into one just like they would at the Bangles’ show, and the song completely transcended its irony. When we all sang the chorus together — My whole life is like a picture of a sunny day! — everyone knew the line as written was a deeply cutting takedown of consumer culture, but at the same time, in that room, there was no way you could help feeling it really was a sunny day, with this band back in your life. The moment was lovely, warm, musical, joyous, and more than a little Bangle-esque.

How I Learned to Stop Worrying and Love Pythagoras

 

Maybe the best thing about the Pythagorean theorem is how it puts math and non-math people on a pretty equal footing. We all know what it says (right triangle, squares of sides, hypotenuse), we all agree it’s Important Math with a capital M… and most of us don’t have much idea, if any, why it’s true. Seriously. Ask another math person if you don’t believe me. If you’re lucky, they might point you to a picture that looks more or less like this:

Pythagoras-proof-anim.svgThis is sometimes called a proof without words, but here are a few words to guide you, just in case. We’ve got the usual notation: a and b are sides of a right triangle, c is the hypotenuse. We build a super-square with side length a+b and break it up in two ways. On the right, we divide it into one (white) sub-square with side length a (area a2), another with side b (area b2), and four (colored) copies of the triangle. On the left, we rearrange the four triangles so that their complement is a new white sub-square with side c (area c2). White area on the left = white area on the right, so c2 = a2 + b2.

Lovely as this is, it feels like a nifty conjuring trick. The Pythagorean equation is the most direct thing in the world, a2 + b2 = c2, and the best we can do is to rearrange triangles inside a big square? Surely there must be a way to cut up c2 into a2 and b2 directly.

Well, there is! I first ran across what I’m about to show you a few weeks ago, loved it, and was surprised that (1) I hadn’t seen it before and (2) it doesn’t seem to be widely known, though the idea actually goes back to Euclid. Perhaps you’ll feel the same way once you see it. Here goes:

The set-up. What’s so special about right triangles? Well, one thing is that they have an amazing self similarity property.  Draw a line segment out from the vertex with the right angle toward the hypotenuse and perpendicular to it. It divides our original right triangle up into two smaller ones:

right-triangle

Let’s stick to the same notation we had before. Let c be the hypotenuse of our original right triangle, running along the bottom. Let a and b be the sides, a the one on the left, b the one on the right. Then a is the hypotenuse of the green right triangle on the left, and b is the hypotenuse of the blue right triangle on the right. Obviously (foreshadowing), the areas of the two smaller triangles add up to the area of the original.

The self similarity property is that the two smaller right triangles are both similar to the original one! In other words, all three triangles have the same three interior angles, which means that you can rotate and scale each one into any of the others. Put simply, all three triangles have the same shape. Can you see why? Let’s compare those interior angles: the big triangle has a right angle of 90 degrees, and two others, which we will call \theta (say the one on the left, in the green triangle) and \phi (on the right, in the blue triangle). The key point is that the angles of any triangle have to add up to 180 degrees, so two angles of a triangle always determine the third. The green triangle has an angle of \theta on the left that it inherits from the original big triangle, and a 90 degree angle in the middle, so its third angle, at the top, must be \phi. (Essentially: the green triangle and the big triangle have two angles in common, so they must have all three in common.) Similarly, the blue triangle has an angle of \phi on the right that it inherits from the original triangle, and a 90 degree angle in the middle (again, two angles in common), so its third angle must be \theta. Same angles, same shape.

The pay-off. Stare for a minute at those three similar triangles, with the areas of the two smaller ones adding up to the area of the bigger one. Wouldn’t it be great if the triangle with hypotenuse a had area a2, the one with hypotenuse b had area b2, and the one with hypotenuse c had area c2? It’s not true, of course. But it’s almost true! If you read my last post, you know that in fact the area of a right triangle with hypotenuse c and interior angle \theta is

\frac14 \cdot \sin(2 \theta) \cdot c^2.

The green and blue triangle have exactly the same interior angles as the big one. So their areas are given by the same exact formula, with c replaced by a and b, respectively. The areas have to add up, so we have:

\frac14 \cdot \sin(2 \theta) \cdot a^2 + \frac14 \cdot \sin(2 \theta) \cdot b^2 = \frac14 \cdot \sin(2 \theta) \cdot c^2.

Now just divide out the common factor of \frac14 \cdot \sin(2 \theta), and you’re left with

a^2 + b^2 = c^2.

What just happened? The way Pythagoras’s equation fell out of equating areas may seem like a bit of a magic trick too, but it’s actually based on the very fundamental idea of scale invariance. To recap: we (1) wrote a completely explicit formula for the area of a right triangle, (2) equated formulas corresponding to equal areas, and (3) found that the bulk of the formulas, everything except the part corresponding to the Pythagorean theorem, went away. The key to understanding all that is this picture:

square

It shows the area of a right triangle embedded in the area of the corresponding square, with the hypotenuse matching one side of the square. The precise formula for the ratio of the areas, \frac14 \cdot \sin(2 \theta), doesn’t matter so much — what matters is that when we blow this picture up or down, the ratio of the areas doesn’t change. That’s scale invariance. If the two smaller triangles add up to the big triangle, then the squares corresponding to the smaller triangles have to add up to the square corresponding to the big triangle. And that’s exactly the Pythagorean theorem.

I like to think of a2b2, and c2 as units of area corresponding to each triangle. In a nutshell, the Pythagorean theorem decomposes a right triangle into two smaller, similar ones, and says that if the triangles add up, the units of area have to add up too. It’s deep, it’s direct, and I’ll never forget it. How about you?

For Pi Day 3/14/16: Triangle to Circles, and Back Again

Around this time last year, we heard a lot about 3/14/15 being a once-in-a-century Pi Day, because the date, including the year, matches the first few digits of pi (3.1415…). But the next digit after 3.1415 in the decimal expansion is 9, giving us 3.14159… So pi is a lot closer to 3.1416 than it is to 3.1415, which is why I want to wish you a Happy once-in-a-century Pi Day today, 3/14/16!

Pi fascinates us because it appears in so many unexpected places. Since pi is ultimately about circles (it’s defined, more or less, as the circumference of a circle of diameter 1), what this really means is that circles appear in many unexpected places. I want to show you one.

You need something with a corner (I used a Rubik’s cube), a cable or string of some kind, and two fixed, firm endpoints (I used the ends of a towel rod). Stretch the cable between the two endpoints across the corner. Now move the corner from one side to the other between the two endpoints, keeping the cable stretched across all three, like this:

Question: what’s the shape that the corner traces out? A triangle, a circle, a parabola, an ellipse, something else?

I told you there was a circle coming, so, yes, it turns out to be a circle. (Well, a semicircle, tracing out 180 degrees, or pi!) I find this a little surprising, because this set-up is quite different from a compass, which is based on spinning a fixed length around a center. Here, the length of cable stretched across the corner extends and then contracts as you move the corner from one side to the other. One good reason to try this out physically is that you can really feel the extension and contraction!

Here is an animation that might start to convince you:

Animated_illustration_of_thales_theorem

The mathematical fact afoot here is Thales’s Theorem, which says that if you put two points on two opposite ends of a circle (so the line between them is a diameter), then the line segments connecting those points with any third point on the circle will meet at a right angle. To give you an idea of why this is true, and play with pi some more, here’s a picture:

2000px-Thales'_Theorem.svg

The idea here is that if O is the center of the circle, then the line segments OA, OB, and OC all have the same length, because they’re all radii of the circle. This makes the two triangles AOB and BOC isosceles: each triangle has two equal sides, hence two equal angles, as illustrated in the picture. With this notation, the sum of the angles of triangle ABC is 2 \alpha + 2 \beta. But we know that the sum of angles of a triangle is 180 degrees (pi again!). Dividing by 2, we find that \alpha + \beta must be 90 degrees, which is Thales’s theorem.

One last thing: this picture also leads to a formula for the area of the right triangle in terms of the length of the hypotenuse. Let d be this length, i.e., the diameter extending from A to C. Look at triangle BOC again. The sum of its angles must also be 180 degrees, and since we know from above that 2 \alpha + 2 \beta = 180, we find that the missing angle, between segments OB and OC, must be 2 \alpha. This means the height of our triangle is \sin (2 \alpha) times the radius of the circle, or \sin (2 \alpha) \cdot d/2. Since the area of a triangle is half the base times the height, we find

{\rm{Area}} = \frac14 \cdot \sin(2 \alpha) \cdot d^2.

This last formula is actually connected in a very interesting way to the Pythagorean theorem. But that’s another blog post.

 

 

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…

CHOOSE YOUR OWN ADVENTURE STORY!

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.

YOU: Tell me about your kids.

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.