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?Well. Is it?
The answer, which seems obvious to me, but which most people seem not to mention, is that there isn't enough information in the question for you to be able to answer it. This is discussed in detail at the wikipedia article I linked to above, but I thought I'd write about it myself, as I was reminded of it recently by the furore over the Tuesday Boy problem. This question, which was mentioned at a recent Gathering for Gardner is:
I have two children. One is a boy born on a Tuesday. What is the probability that I have two boys?Well. What is it?
I'm pretty sure the answer is, once more, that you haven't been given enough information. But I'm getting ahead of myself.
In the Monty Hall problem, most people will tell you something like "if you decide to stick with your original choice, you can only win if the door you chose first time had the car behind it, which happens with probability 1/3, so switching must win with probability 2/3". But it ain't necessarily so.
Imagine it's Monty's last day, and he has decided to make every effort to make sure you win a car. Imagine also that Monty gets to choose whether he offers you a switch or not. Now what are the possibilities? You could have picked the car first time, in which case Monty doesn't offer you a switch, or you could have picked a goat, in which case the only remaining door has the car behind it. If Monty's feeling generous (and he let's you know this beforehand) then you should definitely switch! You win the car every time!
Or imagine Monty is in a really bad mood - maybe one of the goats bit him when he was putting them behind the doors. Now, he will only offer you a switch if you've already chosen the car (this version is sometimes referred to as Monty Hell). If you choose to switch, you never win the car.
Or imagine Monty just likes his gameshows to be fair, so he wants "switch" and "stick" to both be equally good options. Now he offers you a switch exactly 1/2 the time when you've picked a goat, and exactly none of the time otherwise. It doesn't matter if you switch or stick, you win the car 1/2 of the time (1/3 of the time you picked a goat first time, and he offered you a switch, 1/3 of the time you picked a car first time and he offered you a switch, 1/3 of the time you picked a goat first time and he didn't offer you a switch, but you know you're not in this case, because you're being given the option).
In other words - the solution to the Monty Hall Problem as usually posed is "well, I don't know - which side of bed did Monty Hall get out of this morning?" If the hidden assumptions in the question are more clearly stated, the answer can be anything from "I stick" to "I switch" to "well it doesn't matter, does it? Why don't I just toss a coin?"
Now, let's have a look at the Tuesday Boy Problem. The standard answer goes something like this: there are 14 possible combinations of gender and birthday for each of the children. So there are 196 possible gender/birthday combinations for 2 children. Of these, 27 feature boys born on Tuesdays (14 + 14, minus the case where *both* are boys born on a Tuesday). Of these, 13 feature two boys (with the obvious notation (and no, I have no idea why I chose capital letters for boys and lower case letters for girls) MT TT WT ThT FT ST SuT TM TW TTh TF TS TSu) and 14 feature a boy and a girl (mT tT wT thT fT sT suT Tm Tt Tw Tth Tf Ts Tsu). So the answer is 13/27.
This is unambiguously the correct answer to some question related to the Tuesday Boy Problem. In fact, it's unambiguously the correct answer to the question Steve Landsburg posed today when he wrote about this problem:
We gather all those women in the world who have exactly two children, tell each of them to “go home unless you have a boy born on a Tuesday”, and select a woman randomly from those who remain. Assume that births are equally likely to occur on any day of the week, and that on any given day, boys and girls are equally likely.However, I don't think it's unambiguously the correct answer to the version I stole from Alex Bellos. I think this for the same reason as I don't think "always switch" is the correct answer to the Monty Hall Problem. We don't have any idea why the woman chose to come up to us and say such a bizarre thing in the first place.
Imagine she used the following procedure: pick my oldest child, note their gender G and their birthday D, now walk up to the mathematicians and tell them "I have two children, one of who is of gender G and was born on day D". Now the probability that there are two children of gender G (two boys, in the case we happened to find ourselves in earlier) is clearly 1/2 - they are both the same iff the youngest child is the same gender as the oldest, which happens 1/2 the time (note that, for some reason, I'm happy to keep the independence assumption, while not happy to keep the others - if you like go through and replace all the children being born with coin flips).
Or she did the following: if at least one of my children is a boy, pick one of the boys at random, note his day of birth, D and go up to the mathematicians and say "I have two children, one of whom is a boy born on day D", otherwise, say nothing. Now, all the information we actually have is that at least one of the children is a boy... and we all know that this means the probability of them both being boys is 1/3.
The "standard" assumption is equivalent to the woman generating potential statements at random, and walking up to us and saying them if they are true. With this assumption, it is indeed true that 13/27 of the women who say "I have two children, one is a boy born on a Tuesday" have two boys. Personally though, I don't think this is a very sensible model for how people make conversation (even at maths conferences).
Finally, Peter Cameron originally posted the Tuesday Boy problem because he found the 13/27 answer "counterintuitive". I personally think this is because it *is* counter-intuitive. Intuitively, we have a model for how people generate statements, and in this model the "born on a Tuesday" part of the puzzle statement is indeed irrelevant. However, if you want to get an intuition for why extra information about one of the boys should move the probability of both being boys from 1/3 towards 1/2 in the standard model, consider the following version:
I have two children. One is a boy born at 21.03 on Monday 21 June 1982. What is the probability that I have two boys?Well?