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#1




Probability question re: boy/girl riddle
If we encounter someone with exactly two children, given that at least one of them is a boy, what is the probability that both of her children are boys?
Now, I understand why the answer is 1/3. What I can't wrap my head around is this: If I am going to visit a couple that has two children, and a son answers the door, the probability that his sibling is a boy is 1/2, right? Known child: boy Unknown child has an equal chance of being: boygirl Why can't I use the same logic in the riddle? Known child: boy Unknown child has an equal chance of being: boygirl 
#2




I think I got it:
In the riddle, we're assuming we were randomly given the information about one of the children, in the case where a son answers the door, we know the sex of a specific child? 
#3




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You didn't visit the family yourself. A friend of yours did, and reported: "The couple has two children. One answered the door. The other was out in the backyard playing at the time. At least one of them is a boy." So, you have the following possibilities:  Boy at the door, girl in the backyard  Girl at the door, boy in the backyard  Boy at the door, boy in the backyard Whereas if you knew that the boy answered the door, you have only possibilities 1 and 3 still in line with the evidence. That's the general rule. If you know that one specific child is a boy, it is different from knowing that 'the kids are not both girls', (which is the same thing as saying that at least one of the children is a boy but not specifying which in any way.) Like a lot of cases with probabilities, it's easy to trip up. If your friend, in the previous example, WANTED to be able to report to you that a boy had been in one place, (at the front door or in the backyard,) and would have chosen the place accordingly if there were one child of each gender, then you're not at 1/2 probability anymore. Hope that this helps. 
#4




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The child we know the sex of: girl The child we don't know the sex of: girl/boy 


#5




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For full independence, we should ideally find a way of distinguishing the kids and then arbitrarily choose which one we want to test before any information is gathered. Last edited by chrisk; 07242008 at 06:44 PM. Reason: corrected bad phrasing into the italicized words 
#6




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If at least one of them is a boy, then you could have GB BB What other option is there? 
#7




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The 1st is picking the physical possibilites from the expanded set of all possible permutations, while the 2nd is zeroing in on a specific individual. To put it another way, the 1st is looking at % of total population (which you can easily imagine like that of a census), while the 2nd, down to the granularity of an individual person, will always be a binary state  excepting cases like Michael Jackson & Amy Winehouse, of course. Basically, taking % of a sample population <> one binary state value for an individual. 
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#9




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In the son answering the door situation with the answer "1/2", the probability distribution "widens" a bit, and you also have further information to conditionalize on: now the probability distribution isn't just over the genders of the kids in twochildren families, but also over the genders of the kids who answer the door; furthermore, the information you have isn't just that the family has at least one son, but also that a son answered the door. That is, you start with eight equiprobable possibilities SS[older S answers], SS[younger S answers], SD[S], SD[D], DS[D], DS[S], DD[oldest D], and DD[youngest D], and the information you have filters out all the ones where a daughter answers the door, leaving four equiprobable situations, two of which are all boys. The key is that you're looking at more in the second case, and also, in doing so, take advantage of the opportunity to have more information; specifically, the information of who answers the door. And further information can always cause probabilities to shift wildly. Last edited by Indistinguishable; 07242008 at 06:56 PM. 


#10




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The usual implicit model of birth genders, however, is that every birth has independent equiprobable likelihood of being a boy or a girl. Using that fact, we can categorize twochildren families into four situations, based on the gender of the oldest child and the gender of the youngest child, and be justified in concluding them all equiprobable (being able to make this conclusion is key for "probability as counting"). Then, we conditionalize on the information we know, which is that at least one child is male. That tosses out one of the four equiprobable situations, leaving three equiprobable situations, in only one of which both children are male. Last edited by Indistinguishable; 07242008 at 07:03 PM. 
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#12




Is this essentially the "Let's Make A Deal" thing? Because I never, ever, no matter how it was explained to me, understood that one.

#13




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#15




If a boy answers the door, the probability that the other child is boy or girl is the same in both cases, 1/2. Because in the first case, "at least one of them is a boy", you would have eliminated GB, leaving just BG and BB.
You are comparing the situation before somebody opens the door to the situation after somebody opens the door. They are different, because you know more after you learn the gender of the first child. 
#16




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Last edited by Indistinguishable; 07242008 at 07:20 PM. 
#17




In a way, the usual boygirl problem is the opposite of the Monty Hall problem. The typical naive answer to the boygirl problem is "Two boys and mixed gender are equiprobable" instead of the correct "Mixed gender is twice as likely" because people are mistakenly thinking of the situation where they have slightly more information; that where they not only know that the family has at least one boy, but also that a boy was chosen in some random selection from the two children. (The OP's second case is the situation where they actually do have such extra information)
The typical naive answer to the Monty Hall problem is "Switching and staying are equiprobable to win" instead of the correct "Switching is twice as likely to win" because people are mistakenly thinking of the situation where they have slightly less information; that where they know that no prize is behind Door #X, but they don't also know that Monty Hall (when required to choose a losing door other than the contestant's) chose Door #X. (If, after picking a door on the Monty Hall show, you use magic Xray glasses to reveal that the second door has no prize, then staying or switching to the third door are equally good; this is the situation where you actually do lack such extra information) Last edited by Indistinguishable; 07242008 at 07:43 PM. 
#18




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#19




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Roll a die twice. What's the probability that it will add up to 5? Well, there's 36 possibilities: 11, 12, 13, ..., 64, 65, 66. Four of them add up to 5. That gives a ratio 4/36. Though, order doesn't matter... so that gives 21 possibilities. Two of them add up to 5. That gives a ratio 2/21. Though, actually, nothing matters but the sum. So that gives 11 possibilities (from 2 through 12). One of them is 5. That gives a ratio 1/11. Though, actually, nothing matters but whether the sum equals 5. So that gives 2 possibilities (it does or it doesn't). One of them is "it does". That gives a ratio 1/2. Well, all that is true. But not all those ways of classifying outcomes use equiprobable possibilities. So not all of those "probability as counting" arguments give the actual probability. In this case, only in the first categorization are all the possibilities equiprobable (a conclusion supported by the implicit assumption that die rolls are independent, with each face equiprobable). Similarly, when dealing with children's genders, it's not the case that "two boys" is equiprobable with "a boy and a girl"; however, it is the case that "two boys" is equiprobable with "an older boy and a younger girl" and with "an older girl and a younger boy" (a conclusion supported by the implicit assumption that birth genders are independent, with each gender equiprobable). Last edited by Indistinguishable; 07252008 at 01:09 AM. 


#20




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#21




Sure, in some sense, although that's not really it. In another sense, there's very specifically only one child whose sex we don't know: the child who didn't answer the door. Though, we don't know whether that child is the older or the younger one...
Symmetrically, though, we might consider the case where we know of a family that the older child is a son, but we don't know what the gender of the younger child is, nor whether the older child is the one who just left to answer the door or if that was the younger child. There's nothing distinguished about categorizing children according to "older"/"younger" vs. "dooranswerer"/"nondooranswerer". The key difference between the OP's case 2 and case 1 is that there is more information available in case 2: the information about the gender of who answered the door, on top of the information that the family has at least one son. Concerns about rigid designators and so on are an orthogonal issue. Last edited by Indistinguishable; 07252008 at 02:40 AM. 
#22




To perhaps illustrate it: suppose you already know of Family X that they have two children, and at least one son (but know nothing more). Before ringing their doorbell, then, conditionalizing on your available information (against the conventional distribution of families), the probability that they have two sons is 1/3.
After ringing their doorbell, suppose a son shows up to answer the door. Have you learnt nothing new, since you already knew they had at least one son? No, you've learnt plenty new; now, the probability that they have two sons is 1/2^{*}. The fact that it was a son who answered the door rather than a daughter is significant new evidence to conditionalize upon, even though you already knew there was a son in the house. And nothing in this reasoning involves any invocation of relative ages. *: Specifically, we can write it out using Bayes' Theorem as P(two sons  son answers door) = P(son answers door  two sons) * P(two sons)/P(son answers door) = 1 * (1/3)/P(son answers door) = (1/3)/[P(son answers door  two sons) * P(two sons) + P(son answers door  only one son) * P(only one son)] = (1/3)/[1 * 1/3 + 1/2 * 2/3] = 1/2. [This deduction is based, of course, on the model where one of the two children is selected at random to answer the door] Last edited by Indistinguishable; 07252008 at 02:58 AM. 
#23




I think you miss my point. In the post to which I was responding, Tazman was talking about the first situation and suggesting there were two options. I was pointing out that one of those options was really two ie three total. Sorry if I didn't explain myself in detail.

#24




My apologies; I think I did misinterpret your intent.
I'd say, yes, there's three options, but the OP's not wrong that there are two options too. And his confusion is the question as to why he should split one of his options up into two, when, in other cases, he'd be perfectly fine to leave it as it is. I think the issue of "How many possibilities?" has been suitably addressed already, but I'd point out, also, that in the OP's first case ("the riddle"), we are not primarily told of any designated child that they are male. That is, we are told "Family X has at least one son", rather than "The child of Family X who [is oldest/answered the door/whatever...] is male". I guess this may have been your point; that there is just information about the family, but not about any child in particular. (This sort of thing is why I take issue with a traditional, but cheatingly broken, phrasing of the riddle: "Family X has at least one son. What's the probability that their other child is a boy?". The incoherence of this type of phrasing is illustrated by thinking about the analogous "Adam and Eve have at least one son. Is their other child a murderer?" or "The famed Nintendo plumber siblings include at least one male. Does the other one wear red?" or so on.) Last edited by Indistinguishable; 07252008 at 03:58 AM. 


#25




It's the equivalent of something like this.
Your best friend invites you to a Christmas party at a workmates place. You figure if you go you will take a token gift for any kids so you ask if the workmate has any kids. "Yeah two." "Boys, girls or one of each?" "I don't know but one is named Dave so he's a boy. The other is Tony or Toni, don't know which though. But at least one boy." When you arrive there a child answers the door and is obviously a boy. So it's not Toni. If he is Dave then there is still only a 50:50 chance that the kid you haven't met is Toni. It's just as likely that the unmet kid is Tony. But if it is Tony standing in front of you then the other kid is Dave. And there is no Toni. Before you got there the door kid/unmet kid could have been: Dave/Tony.........................Dave/Toni......................Tony/Dave...............Toni/Dave seeing the kid at the door eliminated the last combination. 
#26




References:
Last edited by zut; 07252008 at 06:57 AM. 
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#28




Of course, if the possibility of ambiguous genitalia is included, then the riddle becomes tricky.

#29




Ah! I've got it. In this thread, Zut says
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#30




OK, think about it this way. If you have no information at all about the gender of the kids, there are four equally likely possibilities:
1) BB 2) BG 3) GB 4) GG Since we know one of the kids is a boy, it's not #4. The other three, however, are still all equally likely. In only one of these cases are both kids boys. 
#31




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What does clear it up for me is this part of Zut's post Quote:

#32




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BB  25% BG  25% GB  25% GG  25% Now we can combine BG and GB because for our purposes we don't care about order: BB  25% GG  25% One of each  50% If you learn that at least one of the kids is a boy that eliminates GG. But note that "one of each" is twice as likely as "boyboy". That ratio is unaffected by eliminating "girlgirl". So if we renormalize: BB  33% One of each  66% Last edited by The Hamster King; 07252008 at 11:20 AM. 
#33




Generally I agree with the consensus here. However, I can see one hypothetical case where "At least one is a boy" does not lead to 2/3 odds. It's a bit like the Monty Hall case, where Monty Hall knows which door hides the prize.
If the person giving the information is known to be equally likely to say, with a given family, that at least one is a boy OR that at least one is a girl, then: (1) With BB, they will say "At least one is a boy." (2) With GG, they will say "At least one is a girl." (3) In the other case, they are equally likely to say either. Then, if they say "At least one is a boy", then there is a 50/50 chance of the other being a girl. 
#34




Right. Because in that case, you have more information than just that "At least one is a boy"; you also have the additional information that the semirandom selection process transpired so as that the person giving the information came to tell you that at least one is a boy. Knowing this second fact goes above and beyond simply knowing the first fact.
But, generally, with these problems, you shouldn't assume some extra process (and corresponding extra information) like that unless told about one. Last edited by Indistinguishable; 07252008 at 01:13 PM. 


#35




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What matters is that the two children are separate entities. That is what makes BG (child X is a boy, child Y is a girl) a different situation from GB, and that's why the two cases should be considered separately. 
#36




Eh. You don't ultimately have to consider the two cases separately, and I am wary of fuzzy concern over invocations of separate entities or such (going with the dooranswering version, the two children are dooranswerer and nondooranswerer, and the only two respective gender possibilities are BB and BG, since the dooranswerer is known to be a boy. No GB here... And yet, if we stipulate the way dooranswering works is that a girl cannot answer the door if there is a boy around, and the eldest eligible child always answers the door, then the probability of two boys is still 1/3, even though there are only those two possibilities).
I think Pochacco put it well in post #32; you can categorize possibilities however you like, and ignore whatever differences you don't care about. However, the question remains of how to determine the probability of each possibility (as they may not all be equally likely). One way to do so is to start off with a situation where all possibilities are known to be equiprobable, and then start coalescing them into the coarser categories one cares about. Last edited by Indistinguishable; 07252008 at 01:42 PM. 
#37




I'm not sure what you're getting at. I confess that I found your first paragraph rather difficult to understand. But I was just trying to illustrate why BG is different from GB.

#38




What I mean is that BG is different from GB the way BGwithboyborninJune is different from BGwithboyborninDecember. It's not wrong to say that the difference isn't relevant, and to collapse the two into a single case; after all, if all you knew was "The family has one boy and one girl", but not which was child X and which was child Y or whatever, you'd still know enough to answer the question "Does the family have two boys?".
One might make the mistake of unwarrantedly assuming that that particular single case ("The family has one boy and one girl") is as likely as other particular single cases, whatever one takes them to be (e.g., "The family has two boys", "The family has two girls"). But it's not, in itself, a mistake to treat it as a single case and not care about the distinction between BG and GB. Last edited by Indistinguishable; 07252008 at 02:24 PM. 
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