## At Least One Daughter Exists

In the class I’m teaching we’ve entered probability. This is a fun subject. It’s one of the bits of mathematics which people encounter most often, about as much as the elements of geometry enter ordinary life. It seems like everyone has some instinctive understanding of probability, at least given how people will hear a probability puzzle and give a solution with confidence. You don’t get that with pure algebra problems. Ask someone “the neighbor’s two children were born three years apart and twice the sum of their ages is 42; how old are they?” and you get an assurance of how mathematics was always their weakest subject and they never could do it. Ask someone “one of the neighbor’s children just walked in, and was a girl; what is the probability the other child is also a girl?” and you’ll get an answer.

But it’s getting a correct answer that is really interesting, and unfortunately, while everyone has some instinctive understanding and will give an answer as above, there’s little guarantee it’ll be the right one. Sometimes, and I say this looking over the exam papers, it seems our instinctive understanding of probability is designed to be the wrong one. I’m happy that people aren’t afraid of doing probability questions, not the way they are afraid of algebra or geometry or calculus or the more exotic realms, though, and feel like it’s my role to find the most straightforward ways to understanding which start from that willingness to try.

Some of the rotten track record people have in probability puzzles probably derives from how so many probability puzzles start as recreational puzzles, that is, things which are meant to look easy and turn out to be subtly complicated. I suspect the daughters-question comes from recreational puzzles, since there’s the follow-up question that “the elder child enters, and is a girl; what is the probability the younger is a girl?” There’s some soundness in presenting the two as a learning path, since they present what looks like the same question twice, and get different answers, and learning why there are different answers teaches something about how to do probability questions. But it still feels to me like the goal is that pleasant confusion a trick offers.

But I don’t think escaped-recreational-puzzles accounts for all the confusion, except possibly by bulk (since the Monty Hall Problem is exactly one of these). Here’s another factor: the straightforward method of calculating the probability of an event occuring is to identify all the possible outcomes, and then to identify all the possible outcomes which count as the event occurring. The ratio of event-occurring outcomes to the total-possible outcomes is the probability of the event occurring. Multiply this by 100 if you like your probabilities expressed as percent chances. Do something I never quite get straight if you prefer probabilities expressed as something-to-one shots.

(There are textbooks worth of qualifiers here. For the most prominent one, this approach seems to be assuming that all the possible different outcomes are equally likely. This implies we know what we mean by different outcomes being “equally likely” when we’re still in the process of defining the probability of an event, which is some collection of one or more outcomes. If you’ll let me pretend we can do that, we can save a great deal of work. It’s obvious that we know what we mean by a couple outcomes being equally likely. The mathematical folklore I wish to invoke at that point is that the two kinds of things one must never attempt to prove are the false and the obvious.)

Characterizing all the possible outcomes of something, though, seems to be a real difficulty. The simple method is just to list all the possibilities, but except for the simplest possible problems there’s just no doing that before one loses interest in the problem. Rolling two dice and looking for their sum gives us 36 possible outcomes, and if that isn’t near the limit of what can be done by hand, it’s close. Three dice have 216 possible outcomes and I don’t know anyone who’d write them all out, at least not now that it’s easier to write a Perl script to list them all. And three dice are easy; imagine trying to check all the ways ten items can be arranged with some property satisfied. Ten items, if we can tell them apart, can be arranged in 3,628,800 different ways. Nobody’s listing that, and even the Perl script might not bother to keep them all in memory.

But even when a complete listing is possible it might not be used. One problem with highly variable results asked about the 17th-century belief that a 9 and a 10 were equally likely to turn up in throwing of three dice. That there are 216 possible outcomes is easy to calculate; but the number of ways to make a 9 or a 10 have to be worked out. This isn’t too difficult to do: there are 25 ways to make a 9, and 27 ways to make a 10, on three dice, and if you use any kind of system they can be written out in a few minutes.

But it’s learning to make that system which is the hard part, apparently. I started with 6-2-1 on the three dice, to make 9, and then 6-1-2; then 5-3-1, 5-2-2, 5-1-3; then on to 4 on the first die; and, for 10, starting with 6-3-1 and carrying on about like that. It’s easy to understand how someone might have no idea where to even begin on this problem; it’s less easy, at least for me, to quite remember what it’s like starting out a nice little list of all these combinations and then overlooking two or three.

I suppose that’s what I’m learning, though, from my class and from little essays like this: what the things were which I had to learn so I could build a system like I did above.