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?
7 comments:
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.
But the problem doesn't say that she did say such a thing in the first place. It asks for the probability that she's got two boys, conditional on her having (at least) one Tuesday boy.
That, I think, is unambiguous. When I posed the problem on my blog, I anticipated objections like the one you've made here and therefore added the clarifying language. But I am actually not terribly sympathetic to your objection and I don't think the clarifying language was really necessary.
The Monty Hall problem is quite different, because it specifies that you got your information from Monty, making Monty's algorithm relevant. But this problem has no Monty-equivalent.
Steve - did you actually read the version of the problem I quoted? It is written in the first person, and so very definitely does leave open the question of the speaker's algorithm. To repeat:
"I have two children. One is a boy born on a Tuesday. What is the probability that I have two boys?"
The "I" that is saying this statement is your "Monty-equivalent".
I agree that the answer to the version of the problem that you posted is 13/27, even without the clarification, but I do think that the difference between your version of the problem and the version Alex Bellos posted is significant enough to be worth commenting on.
In fact, as I've already said, I think it's the fact that we conflate the two versions of the problem in our heads that leads our intuition to balk at the correct answer to your version of the problem.
And, just to clarify - although the link I gave above was actually to Peter Cameron's page, the version I quoted is actually the version that Alex Bellos used when he first wrote about the puzzle: http://alexbellos.com/?p=725
John: I stand corrected. I overlooked the first-person statement in the version you quoted. Mea culpa.
In the Monty Hall Dilemma, most people will not tell you to switch doors; they will tell you to stick with your original answer, partly on the grounds that they believe the odds are only 1/2 that it has a car behind it. But those are not the odds, and that is not the best answer. The best answer is to switch doors, because the odds increase to 2/3 once the host -- who we can assume will always open the door with a goat behind it, since there's no reason to vary this factor -- opens the door revealing the goat. The best answer (switch doors) is unintuitive while the "obvious" answers (don't switch or it doesn't matter) are wrong. This is what makes the dilemma interesting.
As for your objection that there are additional assumptions which can alter the answer, that's true but the assumptions are reasonable and most people accept them without even thinking but still believe that the odds are only 1/2 that the unopened door has a car behind it. So these assumptions don't alter the best answer, which is to switch. (It's best because it illustrates an unintuitive point; by contrast, saying that the dilemma doesn't explicitly state reasonable assumptions that most people already read into the problem anyway, isn't particularly interesting.)
Moreover, there's no good reason to "imagine" other scenario-changing assumptions into the question. For example, whether the host will offer a switch every time isn't a relevant consideration in the problem. Raising these kinds of irrelevant assumptions is pointless. You might as well say, "No, it's not to my 'advantage' to switch, because I happen to love goats more than cars! How dare you assume otherwise! A goat, after all, is a living creature with feelings!" It's easy, and, again, pointless and silly to manufacture these kinds of "hidden assumptions."
Similarly, notice that both of us have indeed "answered" the question. So you cannot be correct when you claim that "there isn't enough information in the question for you to be able to answer it." Reading these kinds of ambiguities into the problem is a mere parlor trick.
The Man: did you actually bother to read my post?
"The best answer is to switch doors, because the odds increase to 2/3 once the host -- who we can assume will always open the door with a goat behind it, since there's no reason to vary this factor -- opens the door revealing the goat."
My entire point is that we are given no reason to assume that the host always opens a door. Hosts on gameshows aren't always so obliging. Monty Hall certainly wasn't.
"whether the host will offer a switch every time isn't a relevant consideration in the problem."
Well, yes, it is, as demonstrated above - whether or not the host always offers you a switch changes the optimal course of action completely. I'm not imagining other scenarios, I'm pointing out that the scenario as given isn't well-defined.
I'm not saying that the problem isn't a good illustration of an unintuitive idea - it's just that it's usually not worded very well. The Three Prisoners problem illustrates the same idea with no ambiguity.
John,
"The Man: did you actually bother to read my post?"
Yes, I did read your post, and I understood your point. I just disagree with it, and I thought you might be interested in a different point of view. No disrespect intended.
"My entire point is that we are given no reason to assume that the host always opens a door."
We do have a reason to assume that the host always opens a door: he opens the door in the question. The default assumption should be that the host will continue to open a door, not that the contestant will open a door or that the host will cancel the show. That's why I disagree with your point.
"Hosts on gameshows aren't always so obliging. Monty Hall certainly wasn't."
The actual behavior of hosts on real game shows is irrelevant. Why would anybody assume that the question is about a real game show? It's a logic puzzle. Most people immediately understand that.
"[W]hether or not the host always offers you a switch changes the optimal course of action completely."
Yes, but this question is not part of the problem. The problem doesn't give us any reason to believe that the host will vary his behavior from the given setup. One of the conditions of the problem is that the host offers a switch. You can't change that after the fact. Yes, if you do, you will get different answers. That's another reason why doing so makes no sense. The problem is better under the default assumption of constancy, because it yields only one answer, whereas your assumption undermines the question and is unhelpful.
"I'm not imagining other scenarios..."
Then why are the following statements in your post?
• "Imagine it's Monty's last day..."
• "Or imagine Monty is in a really bad mood..."
• "Or imagine Monty just likes..."
Anyway, I do understand your point. But I think that "if the hidden assumptions in the question are more clearly stated," there will be only one answer: switch. Why? Because the hidden assumptions are not that Monty is moody or that it's his last day or whatever.
The tacit assumptions that people rightly make are that Monty is honest, will always offer a switch, etc. There's no reason to spell these assumptions out. They're already understood (by most people, anyway).
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