Sometimes places become famous not because of what’s there, but because of the good times people have there.

There’s a somewhat historic bar in Singapore called the Colbar. Apparently that’s short for “Colonial Bar”. It’s nothing to look at: pretty primitive, basically a large shed with no air conditioning and a roofed-over patio made of concrete. Its main charm is that it’s “locked in a time warp”. It used to be set in the British army barracks, but it was moved in 2003. According to a food blog:

Thanks to the petitions of Colbar regulars and the subsequent intervention of the Jurong Town Council (JTC), who wanted to preserve its colourful history, Colbar was replicated and relocated just a stone’s throw away from the old site. Built brick by brick and copied to close exact, Colbar reopened its doors last year looking no different from what it used to be.

It’s now in one of the few remaining forested patches of Singapore. The Chinese couple who run it are apparently pretty well-off; they’ve been at it since the place opened in 1953, even before Singapore became a country.

Every Friday, a bunch of philosophers go there to drink beer, play chess, strum guitars and talk. Since my wife teaches in the philosophy department at NUS, we became part of this tradition, and it’s a lot of fun.

Anyway, the last time we went there, one of the philosophers posed this puzzle:

You know a woman who has two children. One day you see her walking by with one. You notice it’s a boy. What’s the probability that both her children are boys?

Of course I instantly thought of the probability puzzles we’ve discussed here. It’s not exactly any of the versions we have already talked about. So I thought you folks might enjoy it.

What’s the answer?

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As ever the answer is “it depends”. What it depends on is why you saw the boy.

Are you in a country where a woman with two boys is required by law to keep one of them with her at all times? Are you in a country where girls are not allowed out of the house? Did you see her and her son at the Slugs and Snails Naughty Boys’ Toy Shop, where no girl would agree to be taken?

Different people can give different answers to this problem, all of which are self-consistent.

Once you remove the social context and ask a mathematical question, the answer is easy to come by, but then the apparent paradox does not appear and the problem is not nearly as interesting!

Are you in a country where a woman with two boys is required by law to keep one of them with her at all times? Are you in a country where girls are not allowed out of the house? Did you see her and her son at the Slugs and Snails Naughty Boys’ Toy Shop, where no girl would agree to be taken?

Suppose I don’t answer any of these questions. Suppose all you know is what I said in my statement of the problem:

You know a woman who has two children. One day you see her walking by with one. You notice it’s a boy.

What odds would you consider fair for a bet that both children are boys? That’s what I’m asking.

Of course you’re allowed to refuse, saying there’s not enough information to answer this question.

And of course there are still different answers, at least if you’re a subjective Bayesian, since you might have different priors. So the fun part of your answer is not the number but your explanation of that number.

Thanks for pointing us to your article about this problem! If you haven’t seen at it yet, you might enjoy a look at our discussion of related puzzles here.

Here’s a different way of contextualising: suppose you own a factory which produces widgets of price q. You know woman who has 2 children and has a son who is a habitual thief who’ll pocket one of anything he walks past. You see them walking out of your factory (where there’s some valid reason for them to have been) into a subway station that has a 1 euro minimum ticket price. You could follow them into the subway (paying the fee) and then, if it turns out this is the “son who’s a thief”, reclaim your property. Or you could leave it and maybe have lost q euros.

At what widget price level q would you decide to go into the subway?

Mathematically this is the same question, except it highlights the way you can sometimes be in a position where even “inaction” may have an effect.

D’oh. I mean that there’s an equation relating q to the probability p asked for. So ignoring the choice of output variable, their mathematically the same problem.

I really shouldn’t post late at night, as I use so many incorrect wordings. What I should have said in place of “You see them walking out of your factory” was “You see the woman and a boy (who’s clearly a child of hers) walking out of your factory”. Also, I need to really think a bit to see if the basic event probability (from which you can compute q) is actually the same as John’s original question.

The bigger picture was that, when doing inference, you’re sometimes put in the position where you every action (including inaction) you take has consequences, so you may as well come up with some assumptions, even if they’re the numerical weight of those assumptions vastly dominates the “known” probabilities in the problem. (Of course, there’s the result that, barring starting with pathological priors, all Bayesians will converge on the same probability distribution in the limit of increasing numbers of observations. So the issues subside as the problems get progressively more complex.)

In this case, assuming we don’t know anything else about the family than has been disclosed, there are three stochastically independent variables: the sex of her first child, the sex of her second child, and whether we saw the first child or the second child. Potentially that gives eight combinations, but only the following four are compatible with what we saw:

BB1
BB2
BG1
GB2

Of these four, half involve both children both being boys. So P=1/2.

The precise information given is perfectly equivalent to the problem where a parent says “I have (at least) one boy” All we know in both cases is that the parent has two children and at least one is a boy. It is a paradox that the answer to the previous case was 1/3 and in this case it was 1/2.

The resolution to the paradox is that in both cases we are not given sufficient information about a prior conditional pobabilites to solve the problem, e.g. we dont know the probabilites of someone saying “I have a daughter” if they have one, or of someone walking in public with their children if they are boys or girls.

However we assume natural values for these probablilities based on what we think the background tells us. In the original problem we assume that the parent would never have told us that he has a daughter. In this case we assume that the parent would be walking with either of his children equally often whether they are boys or girls.

The result of the differing assumptions is different answers. In truth any assumptions for the assumed probablitlites are possible in both cases so different answers are also valid.

A nice question! I think we can take it as read that the paradox is that Greg’s argument above looks correct, but so (at least initially) does the argument that the information we are being given is that the woman has at least one boy, so the probability that the other child is a boy is 1/3 (since the three possibilities BB, BG, GB are equally likely).

To resolve the paradox, we need to understand what extra information could be being conveyed by seeing the woman walking. One way of thinking about that is to ask the following variant. Suppose you already know that the woman has at least one boy and then you see her out walking with one boy. Does anything change?

If you assume that she is equally likely to have gone for a walk with either of her two children, then the fact that you see her with a boy does indeed alter the conditional probability that she has two boys rather than a boy and a girl (since that is certain to have happened if she has two boys and happens with probability only 50% if she has a boy and a girl). More to the point, it is intuitively clear that it does, so the air of paradox is somewhat dispelled.

Refining the probability based on additional info fits naturally in the Bayesian framework. Suppose someone flips two fair coins (to isolate from the social brain wiring used to consider children), and reports that at least one is a head. Then the probability of two heads is
(0) q = p(2H | at least 1H) = 1/3.

Now the person shows you one of the two coins, and it is a head. Using Bayes’ thm, this can be used to refine the estimate of the probability of two heads:
(1a) p(2H | see H) = p (see H | 2H) q / p(see H).
= q / p(see H)
The denominator can be expanded as usual
(1b) p(see H) = p(see H | only 1H) p(only 1H) + p(see H | 2H) p(2H)
= p(see H | only 1H) (2/3) + 1 (1/3)
The value of p(see H | only 1H) depends on the protocol employed by the person who showed you the H.

A. If the protocol is the person looks at the two coins and always shows one with an H (if there is one, which is always the case here), then p(see H | only 1H) = 1, and
(1b’) p(see H) = 2/3 + 1/3 = 1,
and you’ve learned nothing (you already knew there was at least 1 H):
(1a’) p(2H | see H) = q / p(see H) = q = 1/3

B. But if the coin shown is instead sampled, say the person shows coin 1 with probability x and coin 2 with probability 1-x for x in [0,1] (this includes as special cases all of random sampling, x=1/2, or always showing first or second coin, x=0,1), then p(see H | only 1H) = 1/2 , and
(1b”) p(see H) = (1/2) (2/3) + 1/3 = 2/3
so that
(1a”) p(2H | see H) = q / p(see H) = q / 2/3 = 1/2

This framework makes explicit how the probability estimate is refined based on what one learns from random sampling, beyond already knowing that there was at least one H.

For a hybrid sampler, say using protocol A with probability y and protocol B with probability 1-y, then p(2H | see H) will interpolate between 1/3 and 1/2 as y decreases from 1 to 0.

OK, it’s simple. We know the sex of the child she has with her; it’s a boy. So P(C1B) = 1, P(C1G) = 0.

We’re being asked to speculate on the sex of the child she doesn’t have with her; for both children to be boys, the child she doesn’t have with her has to be a boy.

All things being equal, societal predilections in favor of boys not applying, there’s an equal chance the child she doesn’t have with her is a boy or girl, i.e. P(C2B) = P(C2G) = 1/2.

One way to account for the apparent paradox when comparing this scenario to the one where the women simply answers “Yes” to the question “Do you have at least one boy?” (which clearly has P(two boys)=1/3) is to realise that although this current scenario “projects down” to the same subsets as that one — when we ignore the variable describing the random choice of which child we saw her with — that’s no reason to think of the two scenarios as the same.

The four cases {BB1, BB2, BG1, GB2} of which {BB1,BB2} involve two boys have projections {BB,BG,GB} of which only {BB} involves two boys. So neglecting the third variable can suggest a resemblance between the “Seen with one boy” scenario, with P(two boys)=1/2, and the “Do you have at least one boy?”/”Yes” scenario, with P(two boys)=1/3. But of course a projection won’t generally preserve probabilities.

Those are clearly different questions. The probability she has two boys given that at least one is a boy, is clearly 1/3; but it’s entirely different than asking what’s the probability that her second child is also a boy. I’m amused by the elaborations, when the answer is pretty straightfoward, regardless of the order. If we ask what’s the probability her second child is a boy, the answer is 1/2, regardless of the sex of the first child. If we ask what’s the probability her first child is a boy, the answer is also 1/2, regardless of what comes next. This assuming a flat distribution. The question is just: what’s the likelihood the boy sitting at home is a boy, and the answer cannot be other than 1/2!

Great! Thanks, Greg, Tim, Philip, and Henry — I think you clobbered it!

A few basic remarks for everyone else:

When I heard this puzzle, I was confused — and it wasn’t just the beer. It was because I knew the answers to many similar puzzles!

In what follows, just to keep the math simple, let’s pretend twins don’t exist, pretend boys and girls are equally likely and uncorrelated, and ignore leap years.

The two classics:

A woman has two children, and at least one is a boy. What is the probability that both are boys?

Answer: 1/3.

A woman has two children, and the eldest is a boy. What is the probability that both are boys?

Answer: 1/2.

And some intermediate variants:

A woman has two children, and at least one is a boy born on a Tuesday. What is the probability that both are boys?

Answer: 13/27 ∼ 0.48

A woman has two children, and at least one is a boy born on June 12th. What is the probability that both are boys?

Answer: 729 / 1459 ∼ 0.49966

A woman has two children, and at least one is a boy born on June 12th, 1961. What is the probability that both are boys?

Answer: 1/2.

Somehow all this made me panic like a deer in the headlights of an oncoming car when I heard:

A woman has two children. You see her with one and it’s a boy. What is the probability that both are boys?

But I think our esteemed panel of experts has nailed it. This is also the answer given by Mark D’Cruz, who posed the puzzle to me in the Colbar.

I hate the feeling that the more I know, the less able I am to figure out these puzzles. “Wait, is this one of the variations where the obvious answer is correct, or is it the twist where the answer is actually 1/π? And if the other child is behind one of the three doors, and Monty opens door 3, do I stay with my current choice or switch . . .”

I got positive feedback explaining the Monty Hall problem with a higher number of doors, like 100, which is a trick that obviously other people came up with, too, see wikipedia.

Different explanations work for different people; the problem “clicked” intuitively for me when I was coding a Monte Carlo simulation to demonstrate the payoffs of different strategies. I didn’t even have to run the program. While I was in the middle of writing the darn thing, it all Just Made Sense — the effort of setting everything out explicitly made it work inside my brain.

Richard continued, “What I mean is that if you really want to understand something, the best way is to try and explain it to someone else. That forces you to sort it out in your own mind. And the more slow and dim-witted your pupil, the more you have to break things down into more and more simple ideas. And that’s really the essence of programming. By the time you’ve sorted out a complicated idea into little steps that even a stupid machine can deal with, you’ve certainly learned something about it yourself. The teacher usually learns more than the pupil. Isn’t that true?”

— Douglas Adams, Dirk Gently’s Holistic Detective Agency

A while back, Oxford UP sent me an advance reader copy of Jason Rosenhouse’s The Monty Hall Problem (2009), which I enjoyed — it gathers together lots of explanations and variations, together with the history of the puzzle as well as some psychological research on how and why people get so flummoxed by it.

Hello John, it’s Mark from the Colbar. I found Azimuth! I notice you were drawing the thread to a close but I enjoyed reading the mathematical analyses above (as well as those in the first thread) and hope you don’t mind another comment.

Another way to approach the problem is to notice that “the child I saw” identifies a specific child as much as “the eldest child” does. So the current variant of the puzzle should be equivalent to the eldest-child variant and the answer should indeed be 1/2.

I remember you said to be careful about “specific identification” talk, but what follows is some confirmation (I hope) that it is okay.

We can make out a neat relation between the current variant of the puzzle and the Tuesday-boy variant, since in the Tuesday-boy variant, “born on a Tuesday” identifies a specific child IFF exactly one of the two children was born on a Tuesday.

So if we knew further that exactly one was born on a Tuesday, we would essentially have the eldest-child variant and an answer of 1/2. But if we knew instead that both were born on a Tuesday, the info about Tuesday would be “empty”, and we would have the classic answer of 1/3.

So the answer to the Tuesday-boy variant should be the weighted average:

p/3 + (1-p)/2

where p = P(Both are born on a Tuesday | One is a boy born on a Tuesday).

As for finding p, this is just a puzzle of the same sort, solvable in the same way. Consider:

A woman has two children, and at least one is born on a Tuesday. What is the probability that both are born on a Tuesday? (Answer: 1/13.)

A woman has two children, and the eldest is born on a Tuesday. What is the probability that both are born on a Tuesday? (Answer: 1/7.)

By parallel reasoning, p should now be the weighted average:

q/13 + (1-q)/7

where q = P(Both are boys | One is a boy born on a Tuesday).

So the two puzzles are “duals” of each other and can be solved simultaneously!

q = p/3 + (1-p)/2
p = q/13 + (1-q)/7

yielding:

p = 1/9 and q = 13/27. This is the calculation I was trying to effect at the Colbar under the guise of a second, blinded ruminant mammal.

Thanks for showing me the Tuesday-boy puzzle and Greg Egan’s diagram, from which all the answers can be obtained of course. I also enjoyed his deeper puzzle about the Bayesian vs. the frequentist, but this seems also to have been cleared up in the comments.

Hi, Mark! Glad you made it to the discussion here. We enjoyed your puzzle (except for Blake Stacey). I’ll have to ponder your comments sometime soon… I’m a bit burnt out right now.

I put this question to my son, who found it obvious that the answer was 1/2 (more obvious than I did — I had something like John’s experience). His reasoning boiled down to something similar to yours, and one could put it like this. When we see the woman with her son, there are two equally likely cases: that she is with the elder of her two children or that she is with the younger of her two children. So we can restrict to one of those two cases without any loss of generality, and then it really is obvious that the answer is 1/2. And after one has been through that thought process one realizes that the fact that there is some order to the ages of the two children is completely irrelevant. So it really must be correct to say “I now know that THAT child is a boy. Clearly the probability that the child who is not THAT child is a boy is 1/2.”

Sounds right. All these puzzles about “one is a boy born on a Tuesday” or “one is a boy born on June 12th” show that the closer we come to specifying a particular child and then saying that one is a boy, the closer the probability of them both being boys comes to 1/2. Saying “the one I saw the other day is a boy” is, apparently, a complete specification. Somehow we are less comfortable with that when “the eldest is a boy”.

The question of a complete specification can be quantified as follows: suppose there are B species of boy and G species of girl (“species” includes day of week born, in which case B=G=7). First suppose you know there’s a boy but you don’t know species it is, then there are B*B ways of having two boys, and 2BG ways of having a boy and a girl, so
p(2 boys | at least one boy) = B*B / (2BG + G*G).
This is equal to 1/3 when B=G.

Now suppose instead you learn the species of the boy: now there are 2(B-1) ways of having two boys of different species, plus 1 one way of having two of the same species (a total of 2B-1), and there are 2G ways of having a boy of fixed species and a girl. So
p(2 boys | at least one boy of known species) = (2B-1) / (2B-1 + 2G).
when N=B=G, this reduces to
p(2 boys | at least one boy of known species) = (1/2)(1 – 1/(4N-1)),
and hence monotonically increases from 1/3 for N=1, to an asymptote of 1/2 as N goes to \infty. (For the intermediate case N=7, it gives 13/27.)

There’s the converse way of looking at it. For the “seeing a boy case”, it’s a probability conditional on zeroth-order statement (albeit one you could view as having particular variable of which child you see is chosen by a random process). On the other hand, saying/knowing someone has boy is implicitly involving an existential quantifier (“there exists one child who’s a boy”) so it’s conditional on a first order statement. It doesn’t seem unnatural to me that a first order statement should yield different results to a zeroth order one.

It seems like the unease comes from the way natural language doesn’t make it that obvious there’s a first order quantifier in one case and not the other (along with possibly the order in which we’ve considered the problems).

The bit I don’t have an intuitive grasp on is how adding more “problem orthogonal” constraints to an existentially quantified statement makes it “tend to” a zeroth order statement with only the problem relevant constraints.

I’m glad to find myself in complete agreement with both of you. I don’t know the source of the puzzle but I do remember that I saw it here, dated Sep. 2007:

I’m glad! It shook me up too. But I’ve been thinking about this for a week or so, on and off, and it now makes sense. That’s the good thing about ‘paradoxes’ — they pop you into a new reality.

The above “philosopher-posed” problem here can also be found in this popular book, Innumeracy: Mathematical Illiteracy and its consequences, by J.A.Paulos (1989, 2001) p. 86:

Consider a randomly selected family of four that is known to have at least one daughter. One possible way you may come to learn this: You’re in a town where every family includes a mother, father, and two children, and picking a house at random you are greeted by a girl. You’re told that in this town a daughter, if there is one, always answers the door.
In any case, given that a family has at least one daughter, what is the conditional probability that it also has a son? The perhaps surprising answer is 2/3, since there are three equally likely possibilities — older boy, younger girl; older girl, younger boy; older girl, younger girl — and in two of them the family has a son. The fourth possibility — older boy, younger boy — is ruled out by the fact that a girl answered the door.

By contrast if you were simply to run into a girl on the street, the probability that her sibling is a boy would be 1/2.

The first part is well-specified. The second assumes no further conditions (e.g., as “Tom” said in response #1 answer could be “it depends”, and would change to probability 1 if in this town families were overcautious specifically if they had two girls and never let one venture out on the street alone…).

The book covers many examples of elementary mathematical misconceptions for the general public.

What I do not understand is why in this analysis, order is taken into account. Why should we care about first child/second child?

We know that one of the child is a boy. If we know the lady has 1 boy, we can rule out the GG case. So we are left with two possibilities: the other child is either a boy or a girl. The cases are not BG, GB or BB but rather B or G. the probability that the lady has 2 boyes is 1/2.

Not saying that the analysis is wrong, just that I dont understand it (I seem to always go wrong somewhere when trying to solve probability problems.) I just can not see where in this case, so if someone could point it out for me?

Relative to the Monty Hall Problem.
One way to see that A should at least reselect (in some way) is to consider a Player B.
Suppose B comes on stage after player A has selected, Monty has opened 1 door AND A’s door is reclosed.
Now what is the probability of success for B? Oviously it is 1/2. B can pick randomly from the 2 remaining doors. But wait how can B get 1/2 from the current situation while A is at 1/3.
Why wouldn’t A just try again randomly on the 2 remaining doors.
But as we know A can do better than 1/2.
Since A knows that the prob(D1) + prob(D2) equals 1. And since A already knows his door has probability 1/3 (its prob couldn’t have changed because of Monty’s actions) the other door must have prob = 2/3!

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As ever the answer is “it depends”. What it depends on is why you saw the boy.

Are you in a country where a woman with two boys is required by law to keep one of them with her at all times? Are you in a country where girls are not allowed out of the house? Did you see her and her son at the Slugs and Snails Naughty Boys’ Toy Shop, where no girl would agree to be taken?

Different people can give different answers to this problem, all of which are self-consistent.

Once you remove the social context and ask a mathematical question, the answer is easy to come by, but then the apparent paradox does not appear and the problem is not nearly as interesting!

I wrote about exactly this issue here:

http://blogs.ft.com/undercover/2010/07/the-boy-problem/

Tom wrote:

Suppose I don’t answer any of these questions. Suppose

all you know is what I saidin my statement of the problem:What odds would you consider fair for a bet that both children are boys? That’s what I’m asking.

Of course you’re allowed to refuse, saying there’s not enough information to answer this question.

And of course there are still different answers, at least if you’re a subjective Bayesian, since you might have different priors. So the fun part of your answer is not the number but your explanation of that number.

Thanks for pointing us to your article about this problem! If you haven’t seen at it yet, you might enjoy a look at our discussion of related puzzles here.

Here’s a different way of contextualising: suppose you own a factory which produces widgets of price q. You know woman who has 2 children and has a son who is a habitual thief who’ll pocket one of anything he walks past. You see them walking out of your factory (where there’s some valid reason for them to have been) into a subway station that has a 1 euro minimum ticket price. You could follow them into the subway (paying the fee) and then, if it turns out this is the “son who’s a thief”, reclaim your property. Or you could leave it and maybe have lost q euros.

At what widget price level q would you decide to go into the subway?

Mathematically this is the same question, except it highlights the way you can sometimes be in a position where even “inaction” may have an effect.

D’oh. I mean that there’s an equation relating q to the probability p asked for. So ignoring the choice of output variable, their mathematically the same problem.

I really shouldn’t post late at night, as I use so many incorrect wordings. What I should have said in place of “You see them walking out of your factory” was “You see the woman and a boy (who’s clearly a child of hers) walking out of your factory”. Also, I need to really think a bit to see if the basic event probability (from which you can compute q) is actually the same as John’s original question.

The bigger picture was that, when doing inference, you’re sometimes put in the position where you every action (including inaction) you take has consequences, so you may as well come up with some assumptions, even if they’re the numerical weight of those assumptions vastly dominates the “known” probabilities in the problem. (Of course, there’s the result that, barring starting with pathological priors, all Bayesians will converge on the same probability distribution in the limit of increasing numbers of observations. So the issues subside as the problems get progressively more complex.)

In this case, assuming we don’t know anything else about the family than has been disclosed, there are three stochastically independent variables: the sex of her first child, the sex of her second child, and whether we saw the first child or the second child. Potentially that gives eight combinations, but only the following four are compatible with what we saw:

BB1

BB2

BG1

GB2

Of these four, half involve both children both being boys. So P=1/2.

The precise information given is perfectly equivalent to the problem where a parent says “I have (at least) one boy” All we know in both cases is that the parent has two children and at least one is a boy. It is a paradox that the answer to the previous case was 1/3 and in this case it was 1/2.

The resolution to the paradox is that in both cases we are not given sufficient information about a prior conditional pobabilites to solve the problem, e.g. we dont know the probabilites of someone saying “I have a daughter” if they have one, or of someone walking in public with their children if they are boys or girls.

However we assume natural values for these probablilities based on what we think the background tells us. In the original problem we assume that the parent would never have told us that he has a daughter. In this case we assume that the parent would be walking with either of his children equally often whether they are boys or girls.

The result of the differing assumptions is different answers. In truth any assumptions for the assumed probablitlites are possible in both cases so different answers are also valid.

A nice question! I think we can take it as read that the paradox is that Greg’s argument above looks correct, but so (at least initially) does the argument that the information we are being given is that the woman has at least one boy, so the probability that the other child is a boy is 1/3 (since the three possibilities BB, BG, GB are equally likely).

To resolve the paradox, we need to understand what extra information could be being conveyed by seeing the woman walking. One way of thinking about that is to ask the following variant. Suppose you already know that the woman has at least one boy and then you see her out walking with one boy. Does anything change?

If you assume that she is equally likely to have gone for a walk with either of her two children, then the fact that you see her with a boy does indeed alter the conditional probability that she has two boys rather than a boy and a girl (since that is certain to have happened if she has two boys and happens with probability only 50% if she has a boy and a girl). More to the point, it is intuitively clear that it does, so the air of paradox is somewhat dispelled.

Refining the probability based on additional info fits naturally in the Bayesian framework. Suppose someone flips two fair coins (to isolate from the social brain wiring used to consider children), and reports that at least one is a head. Then the probability of two heads is

(0) q = p(2H | at least 1H) = 1/3.

Now the person shows you one of the two coins, and it is a head. Using Bayes’ thm, this can be used to refine the estimate of the probability of two heads:

(1a) p(2H | see H) = p (see H | 2H) q / p(see H).

= q / p(see H)

The denominator can be expanded as usual

(1b) p(see H) = p(see H | only 1H) p(only 1H) + p(see H | 2H) p(2H)

= p(see H | only 1H) (2/3) + 1 (1/3)

The value of p(see H | only 1H) depends on the protocol employed by the person who showed you the H.

A. If the protocol is the person looks at the two coins and always shows one with an H (if there is one, which is always the case here), then p(see H | only 1H) = 1, and

(1b’) p(see H) = 2/3 + 1/3 = 1,

and you’ve learned nothing (you already knew there was at least 1 H):

(1a’) p(2H | see H) = q / p(see H) = q = 1/3

B. But if the coin shown is instead sampled, say the person shows coin 1 with probability x and coin 2 with probability 1-x for x in [0,1] (this includes as special cases all of random sampling, x=1/2, or always showing first or second coin, x=0,1), then p(see H | only 1H) = 1/2 , and

(1b”) p(see H) = (1/2) (2/3) + 1/3 = 2/3

so that

(1a”) p(2H | see H) = q / p(see H) = q / 2/3 = 1/2

This framework makes explicit how the probability estimate is refined based on what one learns from random sampling, beyond already knowing that there was at least one H.

For a hybrid sampler, say using protocol A with probability y and protocol B with probability 1-y, then p(2H | see H) will interpolate between 1/3 and 1/2 as y decreases from 1 to 0.

Very nice! Nice application of Bayes’ theorem!

Belatedly read the earlier thread:

same simple Bayes theorem treatment of earlier (day-of-week) problem is now added here.

OK, it’s simple. We know the sex of the child she has with her; it’s a boy. So P(C1B) = 1, P(C1G) = 0.

We’re being asked to speculate on the sex of the child she doesn’t have with her; for both children to be boys, the child she doesn’t have with her has to be a boy.

All things being equal, societal predilections in favor of boys not applying, there’s an equal chance the child she doesn’t have with her is a boy or girl, i.e. P(C2B) = P(C2G) = 1/2.

If she’s Chinese, these probabilities won’t work out quite like this (see http://www.indyposted.com/17479/china-faces-looming-girl-shortage-of-30-million/). There are a lot of Chinese people in Singapore.

But, if all things are taken to be equal, the probability that both children are boys is:

P(C1B)P(C2B) = 1.0×1/2 = 1/2

One way to account for the apparent paradox when comparing this scenario to the one where the women simply answers “Yes” to the question “Do you have at least one boy?” (which clearly has P(two boys)=1/3) is to realise that although this current scenario “projects down” to the same subsets as that one — when we ignore the variable describing the random choice of which child we saw her with — that’s no reason to think of the two scenarios as the same.

The four cases {BB1, BB2, BG1, GB2} of which {BB1,BB2} involve two boys have projections {BB,BG,GB} of which only {BB} involves two boys. So neglecting the third variable can suggest a resemblance between the “Seen with one boy” scenario, with P(two boys)=1/2, and the “Do you have at least one boy?”/”Yes” scenario, with P(two boys)=1/3. But of course a projection won’t generally preserve probabilities.

Those are clearly different questions. The probability she has two boys given that at least one is a boy, is clearly 1/3; but it’s entirely different than asking what’s the probability that her second child is also a boy. I’m amused by the elaborations, when the answer is pretty straightfoward, regardless of the order. If we ask what’s the probability her second child is a boy, the answer is 1/2, regardless of the sex of the first child. If we ask what’s the probability her first child is a boy, the answer is also 1/2, regardless of what comes next. This assuming a flat distribution. The question is just: what’s the likelihood the boy sitting at home is a boy, and the answer cannot be other than 1/2!

Great! Thanks, Greg, Tim, Philip, and Henry — I think you clobbered it!

A few basic remarks for everyone else:

When I heard this puzzle, I was confused — and it wasn’t just the beer. It was because I knew the answers to many similar puzzles!

In what follows, just to keep the math simple, let’s pretend twins don’t exist, pretend boys and girls are equally likely and uncorrelated, and ignore leap years.

The two classics:

Answer: 1/3.

Answer: 1/2.

And some intermediate variants:

Answer: 13/27 ∼ 0.48

Answer: 729 / 1459 ∼ 0.49966

Answer: 1/2.

Somehow all this made me panic like a deer in the headlights of an oncoming car when I heard:

But I think our esteemed panel of experts has nailed it. This is also the answer given by Mark D’Cruz, who posed the puzzle to me in the Colbar.

This question is very similar to the following one

Question

http://richardwiseman.wordpress.com/2010/07/09/its-the-friday-puzzle-67/

Answer

http://richardwiseman.wordpress.com/2010/07/12/answer-to-the-friday-puzzle-60/

I hate the feeling that the more I know, the less able I am to figure out these puzzles. “Wait, is this one of the variations where the obvious answer is correct, or is it the twist where the answer is actually 1/π? And if the other child is behind one of the three doors, and Monty opens door 3, do I stay with my current choice or switch . . .”

Chuckle :-)

I got positive feedback explaining the Monty Hall problem with a higher number of doors, like 100, which is a trick that obviously other people came up with, too, see wikipedia.

Different explanations work for different people; the problem “clicked” intuitively for me when I was coding a Monte Carlo simulation to demonstrate the payoffs of different strategies. I didn’t even have to run the program. While I was in the middle of writing the darn thing, it all Just Made Sense — the effort of setting everything out explicitly made it work inside my brain.

A while back, Oxford UP sent me an advance reader copy of Jason Rosenhouse’s

The Monty Hall Problem(2009), which I enjoyed — it gathers together lots of explanations and variations, together with the history of the puzzle as well as some psychological research on how and why people get so flummoxed by it.Hello John, it’s Mark from the Colbar. I found Azimuth! I notice you were drawing the thread to a close but I enjoyed reading the mathematical analyses above (as well as those in the first thread) and hope you don’t mind another comment.

Another way to approach the problem is to notice that “the child I saw” identifies a specific child as much as “the eldest child” does. So the current variant of the puzzle should be equivalent to the eldest-child variant and the answer should indeed be 1/2.

I remember you said to be careful about “specific identification” talk, but what follows is some confirmation (I hope) that it is okay.

We can make out a neat relation between the current variant of the puzzle and the Tuesday-boy variant, since in the Tuesday-boy variant, “born on a Tuesday” identifies a specific child IFF exactly one of the two children was born on a Tuesday.

So if we knew further that exactly one was born on a Tuesday, we would essentially have the eldest-child variant and an answer of 1/2. But if we knew instead that both were born on a Tuesday, the info about Tuesday would be “empty”, and we would have the classic answer of 1/3.

So the answer to the Tuesday-boy variant should be the weighted average:

p/3 + (1-p)/2

where p = P(Both are born on a Tuesday | One is a boy born on a Tuesday).

As for finding p, this is just a puzzle of the same sort, solvable in the same way. Consider:

A woman has two children, and at least one is born on a Tuesday. What is the probability that both are born on a Tuesday?(Answer: 1/13.)A woman has two children, and the eldest is born on a Tuesday. What is the probability that both are born on a Tuesday?(Answer: 1/7.)By parallel reasoning, p should now be the weighted average:

q/13 + (1-q)/7

where q = P(Both are boys | One is a boy born on a Tuesday).

So the two puzzles are “duals” of each other and can be solved simultaneously!

q = p/3 + (1-p)/2

p = q/13 + (1-q)/7

yielding:

p = 1/9 and q = 13/27. This is the calculation I was trying to effect at the Colbar under the guise of a second, blinded ruminant mammal.

Thanks for showing me the Tuesday-boy puzzle and Greg Egan’s diagram, from which all the answers can be obtained of course. I also enjoyed his deeper puzzle about the Bayesian vs. the frequentist, but this seems also to have been cleared up in the comments.

Hi, Mark! Glad you made it to the discussion here. We enjoyed your puzzle (except for Blake Stacey). I’ll have to ponder your comments sometime soon… I’m a bit burnt out right now.

I put this question to my son, who found it obvious that the answer was 1/2 (more obvious than I did — I had something like John’s experience). His reasoning boiled down to something similar to yours, and one could put it like this. When we see the woman with her son, there are two equally likely cases: that she is with the elder of her two children or that she is with the younger of her two children. So we can restrict to one of those two cases without any loss of generality, and then it really

isobvious that the answer is 1/2. And after one has been through that thought process one realizes that the fact that there is some order to the ages of the two children is completely irrelevant. So it really must be correct to say “I now know that THAT child is a boy. Clearly the probability that the child who is not THAT child is a boy is 1/2.”Sounds right. All these puzzles about “one is a boy born on a Tuesday” or “one is a boy born on June 12th” show that the closer we come to specifying a

particularchild and then sayingthatone is a boy, the closer the probability of them both being boys comes to 1/2. Saying “the one I saw the other day is a boy” is, apparently, a complete specification. Somehow we are less comfortable with that when “the eldest is a boy”.The question of a complete specification can be quantified as follows: suppose there are B species of boy and G species of girl (“species” includes day of week born, in which case B=G=7). First suppose you know there’s a boy but you don’t know species it is, then there are B*B ways of having two boys, and 2BG ways of having a boy and a girl, so

p(2 boys | at least one boy) = B*B / (2BG + G*G).

This is equal to 1/3 when B=G.

Now suppose instead you learn the species of the boy: now there are 2(B-1) ways of having two boys of different species, plus 1 one way of having two of the same species (a total of 2B-1), and there are 2G ways of having a boy of fixed species and a girl. So

p(2 boys | at least one boy of known species) = (2B-1) / (2B-1 + 2G).

when N=B=G, this reduces to

p(2 boys | at least one boy of known species) = (1/2)(1 – 1/(4N-1)),

and hence monotonically increases from 1/3 for N=1, to an asymptote of 1/2 as N goes to \infty. (For the intermediate case N=7, it gives 13/27.)

There’s the converse way of looking at it. For the “seeing a boy case”, it’s a probability conditional on zeroth-order statement (albeit one you could view as having particular variable of which child you see is chosen by a random process). On the other hand, saying/knowing someone has boy is implicitly involving an existential quantifier (“there exists one child who’s a boy”) so it’s conditional on a first order statement. It doesn’t seem unnatural to me that a first order statement should yield different results to a zeroth order one.

It seems like the unease comes from the way natural language doesn’t make it that obvious there’s a first order quantifier in one case and not the other (along with possibly the order in which we’ve considered the problems).

The bit I don’t have an intuitive grasp on is how adding more “problem orthogonal” constraints to an existentially quantified statement makes it “tend to” a zeroth order statement with only the problem relevant constraints.

This has really shaken me up.

I’m glad to find myself in complete agreement with both of you. I don’t know the source of the puzzle but I do remember that I saw it here, dated Sep. 2007:

forum.richarddawkins.net …

Bruce wrote:

I’m glad! It shook me up too. But I’ve been thinking about this for a week or so, on and off, and it now makes sense. That’s the good thing about ‘paradoxes’ — they pop you into a new reality.

The above “philosopher-posed” problem here can also be found in this popular book,

Innumeracy: Mathematical Illiteracy and its consequences, by J.A.Paulos (1989, 2001) p. 86:The first part is well-specified. The second assumes no further conditions (e.g., as “Tom” said in response #1 answer could be “it depends”, and would change to probability 1 if in this town families were overcautious specifically if they had two girls and never let one venture out on the street alone…).

The book covers many examples of elementary mathematical misconceptions for the general public.

Thank you for sharing this reference.

What I do not understand is why in this analysis, order is taken into account. Why should we care about first child/second child?

We know that one of the child is a boy. If we know the lady has 1 boy, we can rule out the GG case. So we are left with two possibilities: the other child is either a boy or a girl. The cases are not BG, GB or BB but rather B or G. the probability that the lady has 2 boyes is 1/2.

Not saying that the analysis is wrong, just that I dont understand it (I seem to always go wrong somewhere when trying to solve probability problems.) I just can not see where in this case, so if someone could point it out for me?

Relative to the Monty Hall Problem.

One way to see that A should at least reselect (in some way) is to consider a Player B.

Suppose B comes on stage after player A has selected, Monty has opened 1 door AND A’s door is reclosed.

Now what is the probability of success for B? Oviously it is 1/2. B can pick randomly from the 2 remaining doors. But wait how can B get 1/2 from the current situation while A is at 1/3.

Why wouldn’t A just try again randomly on the 2 remaining doors.

But as we know A can do better than 1/2.

Since A knows that the prob(D1) + prob(D2) equals 1. And since A already knows his door has probability 1/3 (its prob couldn’t have changed because of Monty’s actions) the other door must have prob = 2/3!