I was walking along the street and saw a woman playing with two children of different ages. One of the children was a girl. What are the chances that both children were girls?

If you have not tried to solve it, have a go now. For everyone else the answer is after the break.

So, there are four possible combinations in which two children could be born:

BG

GB

BB

GG

We know that BB is impossible, so we are left with:

BG

GB

GG

If the girl was born first then the possible combinations are

GB

GG

If the girl was born second then the possible combinations are

BG

GG

Either way, the answer is 50:50

Did you solve it?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called **PUZZLED** and is available for the **Kindle** (UK here and USA here) and on the **iBookstore** (UK here in the USA here). You can try 101 of the puzzles for free here.

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Shouldn’t it be simpler than that? If one child was a girl, then the possibilities are girl-girl or girl-boy, regardless of who was born first. 50% chance it was girl-girl then

Exactly. It doesn’t matter which child was born first.

In the population of two children families, 50% will be boy-girl (both orderings) and 25% each will be boy-boy & girl-girl. So if you were to pick a family randomly from those with at least one girl, you would expect 1/3 to be girl-girl.

Yup. Like many questions of this type, the “correct” answer largely depends on how you understand the question.

In this case, I think that the “official” answer is correct. The impression I got was that the first child that you notice is a girl. Hence, the probability of the other one also being a girl is 1/2.

But if you closed your eyes, walked up to the woman, asked if at least one of them was a girl and (assuming she didn’t run away or call the police) she answered “yes”, then the probability of both being girls is 1/3.

I also assumed, incidentally, that the “different ages” qualification was merely to eliminate the possibility of twins or a child playing with a friend. Either of these possibilities would skew the statistics. Since twins and childhood friends are typically of similar ages, we can safely rule them out.

I don’t think the question says anything about families.

But why not just stop at the stage where you have three possibilities (BG, GB, GG), and so the answer is 1/3?

By splitting these into two cases you are effectively looking at four possibilities (GB, GG, BG, GG), counting the GG case twice and getting the wrong answer.

Though this is an incredibly controversial puzzle – see http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

As decourse says, it is a well known controversial question. One way to look at it is to consider how the initial observation is made. If it was the first child you saw that matters and you were going to select the gender of interest based on the gender of the first child (i.e., if he were a boy you would ask what is the probability of both being boys), then it would be 1/2. If you went around rejecting families until the first is a girl then it would be 1/3. Also assuming all families have 2 children.

Sorry – as Ed says, it is a well known controversial question.

Indeed. The problem is in the reporting. What was the observer reporting? The colloquial use would be “I’ll tell you the gender of the first child I saw”, or “I’ll tell you the gender of one of the children I saw at random” or something.

But another possibility was “If either of the two children had been a girl, I would have said I saw a girl. Otherwise I would have said something else.” then the odds of the other child being a boy become 1/3. (GG BG GB are the 3 possibilities, equally likely in this case).

This is why it’s so important to state your experimental hypotheses carefully.

You count GG twice because the girl referred to by the questioner could be either Girl 1 or Girl 2.

It is a badly worded answer from Ricardo howver.

Of the course the other answer is zero. Since the observer actually saw the two children and then stated that one was a girl it must mean the other was a boy.

I was with the 1/3 crowd, because I interpreted “at least one girl …” I the dry mathematical sense, similar to Gardner’s second question, and because it is not stated that the observer reported the oldest or youngest child to be a girl.

The enumeration above is specious – just because the probabilities for both birth orders are 50/50 does

notmean that for an unknown birth order it is 50/50 also. In the latter case the options are BG GB GG.The Wikipedia article that Ed cited above is most helpful.

Maybe you missed one important point of this article :

“Gardner initially gave the answers 1/2 and 1/3, respectively; but later acknowledged that the second question was ambiguous.[1] Its answer could be 1/2, depending on how you found out that one child was a boy. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Bar-Hillel and Falk,[3] and Nickerson.[4]”

Yat,

Thank you – yes, I did read the entire article.

I raise you Vos Savant’s example number two, a random woman’s “at least one” child vs a random man’s “older child”. Richard used the former formulation, neutering his prefatory statement about different ages.

The “the older child” case is perfectly unambiguous and leads to 50/50, that does not change anything about the ambiguity of stating that at least one child is a girl. The point is that we could imagine two perfectly reasonable protocols leading to the description of the meeting, one of them gives 1/2, the other gives 1/3, and nothing in the way Richard stated the problem allows to decide which one is right.

It may well remain ambiguous and we understand why. I am stressing 1/3 as more credible because the formulation “at least one” was used.

The other point I made is a separate one, in that the reason given for 1/2 is based on a specious enumeration.

I’m afraid Richard’s answer is incorrect, as GG falls into both “If the girl was born first”, and “if the girl was born second”. This means that there is an overlap in the probability tree. For the cases where at least one of the children is a girl GB, BG, GG, only one of them is GG, so the answer is 1/3.

Yeah, the explication is not very satisfying.

This solution is correct if we consider for example that Richard just saw one of the kids first, and talked accordingly. In that case, if he had seen a boy first he would have stated that one of the children was a boy. So, with a little Bayesian magic, you land on 50/50.

Richard’s answer is, indeed problematic as he has counted the GG combination twice. So the answer is 2/3rds. However, this needs to be qualified in noting that his stating of the problem is ambiguous. If he has stated that the first child he’d seen was a girl, then the answer would, indeed, be 1/2. However, that’s not how the question was put – a perfectly reasonable interpretation of the wording was that at least one of the two children was a girl.

So it’s impossible to come to a firm conclusion based on this wording. There is lots of literature on this problem, and it again goes to show that humans aren’t good at putting problems unambiguously, and we often have incomplete information on which to base probabilities.

Think some of you need to get out more.

What a bizarre puzzle. How could it have been anything other than 50:50?

It could have been, for example, if it was a dialogue :

Richard: I was walking along the street and saw a woman playing with two children of different ages

Me: Was at least one of the children a girl ?

Richard: Yes

In that case, there is only a 1/3 chance that both children are girls.

In that time is infinitely divisible I took the view that everyone has a different age (no one is born at precisely the same time) so I ignored the age factor – it seemed to add nothing to the problem.

Neither does the ‘women’ mentioned, although if you substitute ‘man’ in the sentence you get a totally different problem (!)

I thought this was a very poorly worded version of an interesting puzzle, it was just a puzzle with no interest at all ! Seeing this answer really made me smile !

Indeed. It can also be read as “precisely one is a girl”, meaning theat the probability of a boy is 100%. In fact I rather assumed he meant that, as a kind ofjoke, as we have argued about the “at least one is a girl” and the “at least one is a girl who was born on Tuesday” versions before.

Yup, unfortunately it continues the recent trend of poorly worded versions of interesting puzzles….

What a surprise to see you here Yat. Hope you had a good weekend. I am also surprised by your comment that “it was a puzzle with no interest” given the amount of time you have spent discussing it.

I think we finally agree on something though. Richard’s explanation could have been better. Then again, I think his questions and answers are deliberately worded to provoke a reaction and discussion. He is a psychologist after all, and is interested in human behaviour.

I did 50:50…the probability it’s alwais the same..

I found it quite easy. I dont know much about probability except it only applies to the unknown quantity. I the question was about the probability both children were girls and both sexes were unknown the probability would be smaller.

soz-speling error!

I found it quite easy. I don’t know much about probability except it only applies to the unknown quantity. If the question was about the probability both children were girls and both sexes were unknown the probability would be smaller.

Martin Gardner’s answer to this problem: 1/3

Just for the record.

So he got it wrong too.

Martin got it right.

my ans is this and it is not the same as mr wisemans

prob one child is girl: 100% – this is the given

prob other child is girl: 50% – basic birth stats given mr wise has ruled out identical twins.

therefore prob that both are girls: is average of both cases – (100 + 50) / 2%

this ans then is 75%. simple when you think laterly.

You really should pick up a basic primer on probability and study up.

The order of birth has nothing to do with it, since the question didn’t provide any information about the relationships of the children, to each other or to the woman. If we want to introduce the assumption that they were siblings, why not assume they are female siblings and make the result 100%. The ages of the children do not matter, unless the question also specifically introduces further pertinent information, which it does not. Even the fact that there was some woman involved is irrelevant, because the actual question is only about the children.

We might feel we are invited to draw some inference about sibling relationships, but none is really provided. So, there may be ambiguity in the situation being described that draws us in, but there was no ambiguity in the actual question. The question very clearly does not mention relationships. If the information isn’t there, don’t make it up.

The only relevant information is that there were two children, and one was a girl. In this situation, asking what is the probability of both being girls is merely a way of asking the probability that the other child was a girl. With no further information we can only assume independent probabilities. And, without any information on any other aspect of the situation we can only assume 50% B/G for the second child.

But, the question didn’t give a 50% B/G probability, or state that the probabilities were independent so how can we introduce these factors? The only external information we can bring to the question is a crude population probability of 50%, along with independent probabilities for lack of other (e.g. relationship) information. This is reasonable enough. If we want to be more precise: http://en.wikipedia.org/wiki/List_of_countries_by_sex_ratio.

Without these basic population assumptions the question loses all meaning and is unanswerable.

the controversy over whether the answer to certain permutations of this problem relates both to how the question is phrased and to the independence of the variables involved.

the answer here is far simpler than richard stated it. the gender of one of the two children is known, so her gender is totally irrelevant* to the problem, which is simply, “what is the probability that someone is female?”**

*in real life, couples are genetically predisposed to having children of a particular gender. thus, a family who has one girl is more likely (by how much, i don’t know) to have another girl than another boy.

**and in real life, the answer is about 51%.

Real life????!!!!!

When has anything in these threads had anything to do with “real life”?

Get a (real) life dude

When he says “one of the children was a girl”, it seems to mean that he was referring to a *particular* individual, as if to point to someone and say “this is a girl”. Then the sex of the other individual is independent, and therefore 1/2 chance of being a girl. Thus, in this interpretation of the question, Richard’s answer and explanation are both correct. If, however, it is interpreted to mean that the family had *a* girl, then it would be 1/3.

“Martin Gardner’s answer to this problem” depends on which problem it is. Which depends on the randomizing procedure. Their ages and the woman are irrelevant. If “One of the children was a girl” means “I looked at one of the children and noticed she was a girl”, you get 1/2. “I looked at both children and noticed at least one was a girl” gives you 1/3. “I looked at both children and noticed exactly one was a girl” gives you 0.

I assume this was one of those puzzles that provokes an intuitively incorrect impulse in some. I’m not immune to the odd irrational leap myself, but don’t see where the puzzle was on this one?

The correct answer is 1/3. Assuming that probability for B/G is 0.5 and first and second child gender are independent, the 3 possibilities BG, GB, and GG are equaly likely. If you think otherwise, just run a simulation on a computer to convince yourself.

Richard, you are wrong here. You are correct up to the point were you identify GB, BG, and GG as the three possibilities. However, when you break that up into first child and second child, you are counting GG twice! The correct answer is 1/3.

Anonymous.

Correct me if I’m wrong, but I’m not sure you’ve read the question correctly:

“What are the chances that both children were girls?”

By your calculations, surely the combinations are either (GB or BG) or GG. Surely that’s a 1/2.

Ta.

For crying out loud folks, the question states that he saw two children and ‘one of the children was a girl’ not ‘the first child’ or its like … so the answer to the question is zero. R

Ha ha! Just shows how ambiguous language can be!

Regardless of the gender of the one child, the chance the other child being another girl is always 50:50. There is no need for listing all the scenarios.

…I guess Richard W is just warming us up to feed us “Monty Hall” and “Sleeping Beauty” the coming weeks.

48.31%, There are 107 boys for every 100 girls.

Tony, you are correct in stating there are more boys born than girls. For those who say real life doesn’t count ,noone said the puzzle took place in fantasyland.

He’s still wrong, of course.

Why?

Nowhere in the question does it state that the children are siblings, or taht the woman is their mother. It does say “one” is a girl; tehrefore the other MUST be a boy. Plus the place is not named. Where male children are highly prized in Asian countries, it would have skewed the results. Also a country where the males are off to war or work at an early age would also skew the outcomes. This is just a poorly worded puzzel whose sole purpose would seem to be to provoke discussion in a classroom type setting.

1/3. BB is the only eliminated pairing, and of the remaining three possible pairings, only one yields a second girl. Birth order or any other such arbitrary distinction NOT MENTIONED IN THE ORIGINAL QUESTION is entirely superfluous.

OK. So whenever there is controversy like this, I like to plug it into excel… Here’s what I did, and perhaps my interpretation of the text meant that my scenario is incorrect.

Step 1: created a bunch (16,384) random pairings of boys and girls.

Step 2: counted how many of those had at least 1 girl (approx 12,000). This was the set I was working from (this is as much information as we are given)

Step 3: Counted how many of those were both girls (approx 4,000)

So we know that out of 12000 possibilities where one person is a girl, the chance of both being a girl is 4000. Around 1/3.

Is this right?

Ah, I see. Mr. Wiseman’s statement “One of the children was a girl” indicated the gender of a specific child, rather than a statement about the situation as a whole. This, to me, is not the most natural interpretation of that statement.

It’s right if the question was “at least one of the children was a girl”. However, if the question was “the first child was a girl” then the answer is 1/2. Your Excel sheet models the first scenario, not the second. Really it’s impossible to arrive at a single answer for this question as it is ambiguously worded. It’s also a classic problem over which statisticians have argued for decades.

Harry Phillips

Thanks for your monkey related post

Spot on

The answer is 0%, there are 2 children, ONE is a girl.

Ok my answer was only correct if the question is interpreted as there is ONLY one girl.

To the people that think that the answer is 1/3. Instead of thinking of it as counting GG twice think of it as

GKGX

GXGK

Where K is the “known” girl and the X is the variable, therefore 1/3 is wrong, sorry but just accept it.

The other way to think about it is:

“Two creatures, one was a monkey the other was a child, what are the chances that the child is a girl?”

Still 1/3 now? I don’t think so.

Harry,

Your analogy only works if you tell us that the monkey is female, and you ask us ” What is the probability that they are both female?”

Oh wow, not again. I wish Richard would steer clear of these cans of worms! Anything to do with Bayesian/frequentist arguments or clarity of question statements regarding probability is a minefield.

It’s one third. I happen to have a copy of Ian Stewart’s The Magical Maze right beside me, which has exactly the same puzzle (copy cat, stole a rat, etc). His answer is 1/3 – as there are three combinations with at least one girl, and only one where both are girls. Come on, Richard.

You have a copy of Ian Stewart’s The Magical Maze right next to you?

Awesome.

That’s like when Mr Wallace sends in the Wolf in Pulp Fiction.

Game over dude.

We are not worthy etc.

Harry,

Your analogy only works if you tell us that the monkey is female, and you ask us ” What is the probability that they are both female?”

Bahhhhh ha ha ha… why you think the sex, the species or ANY attribute of the first object baffles my mind.

“There are 2 objects one is a rock (or a balloon, or a digital watch, or a book, or a bird, or a toe nail clipping) the other is a mammal, what are the chances the mammal is a female?”

You are reading WAY WAY too much into the question.

See the link below, this should convince all the sceptics. The key here is the wording of the question.

http://mathforum.org/dr.math/faq/faq.boy.girl.html

This is all very well Jerry, but your link takes us to a different question from the one which Richard asked.

That’s a good link. It shows reasoning for both scenarios quiet well. I believe the wording of the question here suggests the first scenario more than the second, though, and I’m sticking with 1/3.

Richard and the mathforum ask similar question. It would be the same as if you said “A man tosses a coin twice. One (out of the two) is a head, but I’m not tell you what the other toss yielded. What is the probality that the other is also a head? 1/3! The key here is that you have no information on the other (1st or 2nd) toss, or child.

Another attempt:

Any two children of different ages are born in one of these orders:

i) bg

ii) bb

iii) gb

iv) gg

1) What’s the prob of gg, with no other conditions? It’s the result of two independent trials, each of 1/2: 1/4

2) Given that one is a girl (condition A), what’s the probability that the other is a girl, irrespective of age? 1/3, from bg, gb, gg, because it can’t be bb by condition A.

3) Given that one is a girl (condition A) and she is the older of the two children (condition B), what’s the prob of gg? Only (iii) and (iv) satisfy both conditions, so it’s 1/2.

Although Richard mentions the age difference it’s irrelevant, because he doesn’t say whether the stated girl was the older child or not – so (3) does not apply. So, on the surface, it looks like (2) is the right framing and answer: 1/3.

But, the probabilities considered so far only apply as a means of estimating the probability prior to *any* test, but assuming that one outcome of the test is that one of them is a girl. But that’s not how the problem is stated. This isn’t like Monty Hall, where a subsequent choice takes into account the prior probabilities. Richard states up front that there is a woman and a girl, and some other child. These are givens in this problem. They are not uncertain. Their probability is 1.

Harry is right in the sense that he is using the monkey to illustrate the independence of the two variables. The first ‘object’ spotted with the woman happens to be a human female child, but it could have been anything.

Look at it this way. Richard could also have asked, “There’s a woman and a female child, and another child. What’s the probability that there is one woman and two female children?” Let’s use the probability of the woman being present (as opposed to a man) in the same way we are using the probability of the stated girl being present.

i) mbg (m = man)

ii) mbb

iii) mgb

iv) mgg

v) wbg (w = woman)

vi) wbb

vii) wgb

viii) wgg

Using the reasoning of (2), the probability of wgg, given that there is a woman and one girl, only (viii), wgg, satisfies this requirement and the answer is 1/6.

What other information is in the question that we could introduce probabilities for? He was on a street, but he might not have been. Let’s make up an argitrary probability that he was (s) or was not (x) on the street when he saw the event, of 1/2. Now we have:

xmbg

xmbb

…

swgg

Now the probability that both children were girls, given that he was on a street and that there was a woman, … is 1/12. This becomes rediculous. Everything in the question is independent of the other child’s gender, and the probability of that is 1/2. Other probabilities only apply to questions where these other factors are also unknowns (as in Monty Hall). The street, the woman, the first girl, are all independent of the gender of the other child, and they are all known.

Here’s another example. (A) I throw a pair of dice. What’s the probability of throwing a 6+6: 1/36. (B) But, if I roll one dice and get a 6, I then need a 6. What’s the chance of rolling a 6 with the second dice? 1/6 (C) Or, suppose instead of rolling the first dice I’m always allowed to set it down to a 6, and then only have to roll the second 6? 1/6. This latter description (C) best matches Richard’s question. The stated girl is no more a factor in the gender of the second child than a fixed 6 is in throwing a second 6.

Here’s how Richard’s question could have been phrased for the answer to be 1/3, “I’m going onto the street shortly, where I’m informed there is a woman with two children. What’s the probability that both children will be girls given that one will be a girl?”

Or, “I’m going out onto the street, where someone has tossed two coins. What’s the probability both will be heads, if at least one of them is heads?”

In these formulations we are calculating probabilties on two independent variables, prior to any actual trial, but given that we presume the outcome of one of them.

But, if Richard walks onto a street and says, “Ah! Here’s pound coin [woman], oh, and a penny [child], heads up [girl]. Ah, my lucky day, I can see another penny [child] down the street, but it’s dirty so I can’t see which way up it is [girl/boy]. What’s the odds it will be heads up [girl] like the other penny?” This is how Richard’s puzzle question is formulated.

The common notion that there is an even distribution in any population of boys and girls is both the minimum and the maximum we can assume. It’s the minimum because without it we can’t give any answer at all, and it is approximate to actual distributions, and so is reasonable to assume. It’s maximum because we have no other useful information in the question. For example, we don’t know that the children are related so we can’t use any pupulation statistics about the probability of a second child being the same gender as the first. Although we are told they are of different ages we don’t know which is the older, so that doesn’t help. We are stuck with 50% and independence.

As I said earlier, the situation is vague, in that we have only the information given. The only unknown is the gender of the other child, and the knowns are independent of that. Richard’s question is quite clear about what the unknown is. Our knowledge of the situation, as stated in the question, adds nothing more. So the question is clear in the nature of the problem (1 independent unknown of 50%), though vague about the description of the situation. We are being persuaded to consider situational information simply because it is offered.

Your example introducing the gender of the adult is useful, however I think your conclusion based on the reasoning of B) is wrong. There are 8 possibilities, as you say.

i) mbg (m = man)

ii) mbb

iii) mgb

iv) mgg

v) wbg (w = woman)

vi) wbb

vii) wgb

viii) wgg

We know the genders of the adult and one of the children. So that eliminates all but wbg, wgb and wgg from the set. Using the reasoning of B) (which I believe more closely resembles the wording of the puzzle) the chance of wgg is still 1/3.

I agree, though, that this puzzle is about interpreting that wording more than it is about probability. I just disagree that the coin or monkey examples reflect that wording.

Reasoning of 2) sorry, not B)

Wow… and I thought that I had spent far too much time on this. You guys take the biscuit.

I favour 1/3, purely on the basis that we can’t assume to know how Richard chose to state “One of the children was a girl”. We can only work with what we do know – there are two kids, and they are not both boys.

BB/BG/GB/GG.

We know that BB is impossible, leaving 3 possible outcomes, one of which is successful (GG).

Quite right Rob on my duff analysis. Thank you. It does remain 1/3.

But I’m can’t see how the penny example fails. In any of these examples of 1/3 v 1/2 the main difference is about what unknowns there are and what predictions we are making, or what expectations we can have.

The 1/3 is the combined probability of GG, when the genders of both children are unknown, but presupposing one will turn out to be a girl.

The 1/2 is the independent probability of the single unknown – the remaining child’s gender. We already know the gender of the other child, and that there’s a woman, and that they are on the street.

The introduction of ages into the question is what fools us into wanting to look at the full set of ordered pairs. This is why the coin example helps. We ignore the age and look at whether we need to consider all ordered pairs or not.

The tossing of two coins can result in the ordered pairs TT, TH, HT, HH. The probability of HH is 1/4. The probability of HH, given that one result is H (we don’t know which of the ordered pair it will be) is 1/3.

But, if we know the first coin is H, what is the probability of HH, then the domain for this trial is HT, HH, or by simplification, T, H. Both TT and TH are excluded. This reduces to a single toss of T or H. We have already removed the need to consider the full set of ordered pairs because we already know the outcome of the first of the actual pair – we know part of the outcome.

Similarly we can exclude the street, the woman, the first child, from the analysis (the point I was trying to make but fluffed). The probability that the first child noted is a girl is 1. The probability that the first child noted is a boy is 0, because we know it is a girl. It is not an unknown.

So:

A) Richard sees a woman and a girl and another child, what’s the probability of BB? Pb x Pg = 0 x 1/2 = 0

B) Richard sees a woman and a girl and another child, what’s the probability of GG? Pg x Pg = 1 x 1/2 = 1/2

But:

C) Richard sees a woman and a girl and another child, what’s the probability of the second child being a boy? Pb = 1/2

D) Richard sees a woman and a girl and another child, what’s the probability of the second child being a girl? Pg = 1/2

(B) reduces to (D). The phrasing of the question as (B) is masking the real problem, which is (D). (A) and (C) help to illustrate related questions that follow the same lines. I think everyone would agree that the answer to (A) is zero: there is zero probability of it being BB if one is already known to be a girl. This is because we have a known. The 1/3 applies only when both are unkown – and then only when we presuppose one of the outcomes is G, otherwise the GG probablity is 1/4.

You almost convinced me. But then this exchange jumped onto my head:

“I saw 2 children today. What is the chance they were both girls?”

“There are 4 possibilities, so 1/4”

“Oh I forgot to mention, one of them was a girl”

“That eliminates one of the possibilities, so 1/3”

If it became 1/2, which 2 possibilities does the extra info eliminate? BB and…?

Rob,

“I saw 2 children today. What is the chance they were both girls?”

You’re mixing up what happened, and hence what you now know, with what your *estmation* was prior to seeing them.

Suppose you’ve seen them, as you say, and you ask me, “Ron, what is the chance I saw two girls?”

I’d say, before you saw them your *estimation* should be 1/4. Remember, your estimation at that time is your assesment of a probability model based on the fact that you know there are two children, you don’t know their genders, but you know that B/G is 50% independent.

But, having seen them, let’s say you actually saw two girls, then for you the estimation of the probability of two girls is 1, because you know; but without your confirmation my estimation of the probability is still 1/4.

If you now tell me that, “Well Ron, one of them is actually a girl, what do you suppose the probability is of the other being a girl too; that is two girls?” I’d then say that since we now both know that one of them is a girl my estimation of the probability of the other being a girl is 1/2. So the chance of them both being girls, knowing that one is a girl, is 1/2. And, if as I said, you knew they were actually both girls then your estimation of the probability is 1. There is no uncertainty upon which *you* need to construct a more tentative probabilistic model, though for *me* there is.

Let’s change the facts. If, instead, you saw that one was a boy, but let’s say the second child was obscured by the woman and you never got to determine the gender of the second child, then for you the probability of the second child being a girl is 1/2, because you are uncertain and are relying on a probabilistic model of even gender distribution. But, what’s your estimation of the probability that they are both girls? Zero, because you know they cannot *both* be girls, even though you don’t know the gender of the second. My esitmation of two girls is still 1/4; unless you let me in on the fact that the first was a boy, in which case my estimation of two girls matches yours and is zero.

What we know now, i.e. what facts we are given, is different from what we estimate before we know. This is why time travel would be distasterous for the betting industry: future uncertainties would become known future facts, which is of no use to bookies.

“That eliminates one of the possibilities, so 1/3? If it became 1/2, which 2 possibilities does the extra info eliminate?”

It illiminates BB and BG, leaving GB and GG, because by declaring one of them is a girl you are creating your own ordered pairs, where the ordering is: 1st known child and 2nd unkown child. Just as tossing two coins together or tossing them in sequence does not matter, you have to assess the ordered pairs of coin 1 and coin 2. You are declaring your first child is G, so the only possibilities with the first child being G is GB or GG, so the chance of GG is 1/2. Or, looking at it the other way, you can ignore the ordered pairs because the first child is already known, you only need to know the chance of the second child being a girl: 1/2.

And note, the phrase ‘the second child’ here has nothing to do with their ages. I feel sure that it’s the spurious inclusion of ‘different ages’ and Richard’s own answer that refers to age that is causing additional confusion for some. If you toss two pennies does it matter that one was a 2005 penny and the other a 2010? We have to forget penny age (forget birth order).

Concentrate only on what is known when the question is asked, and who has that information: when Richard states it *we* have the information about one child, but not about the other, so what is *our* estimation of the second being a girl too; i.e. both being girls?

I think we need to remember here that probability is just a modelling tool for when we don’t have concrete information, for when there is some undertainty. When we have the data we don’t need probability, because we know. What’s the probability that 1 + 1 = 2? 1! What’s the probability of HH from two coins if I lay them on the table heads up? 1! What’s the probability of two coins being HH if I lay one on the table heads up and toss the second? 1/2!

Also, probability is entirely useless when we have no data, when we have no probability statistics (or are unable to derive them analytically the way we can for coins and dice). In that situation we are using ‘guesses’ of ‘probabilities’ to ‘estimate’ an outcome – a common methodology for estimating project time scales in the software industry in which I work. Richard was walking down the street; what’s the probability he saw two girls with a woman? Well, did he see a woman? Did he see a woman with children? How many children? 100 children? Then he probably saw two at least two girls, yes? Well not if they were all Boys Scouts….. No data, no probabilities, just hopeless guessing.

What actual concrete data do we have – probability 1! To what extent can we only estimate or analyse probabilities – probability X! What if we have no data – probability unknown!

Yes Photon Boy, I should be getting on with my day job. 🙂

It’s 1/3. There are four possibilities, equally probable. The information that we have eliminates one of those possibilities.

That’s right! I think Richard is more interested in selling his books than correcting his obvious misktake.

Please explain why ANY attribute of the first object is relevant to the question, he is asking the sex of an unknown person, the sex of the first has zero relevance to the second.

What if the first object seen was a rock, now what are the chances the second object is female when the second object is a child?

Still 1/3?

He is not asking for the sex of an unknown person. He is asking about the distribution of sexes in a *pair* of persons, and something about that *pair* is known.

The problem is identical if I have a pair of things, each of which has a 50/50 chance of being a rock or a child, and I tell you that at least one of them is a rock.

I’ve come around to the 1/2 camp. The trick is eliminating the knowns and working out the probability of the unknowns. (Ron hinted at this earlier.)

The only unknown is the gender of the remaining child and what we are really being asked is the probability of that child being a girl . To take into account the gender of the known child, we would essentially be calling the law of diminishing returns.

If I toss 100 coins and get 100 heads, what is the probability that the next coin will be a head? The answer is 1/2 because each coin toss is independent. That 101st coin doesn’t know what the other 100 did, so we eliminate the coins we have already tossed. If I had asked “What is the probability that, having already tossed 100 heads in a row, I could toss 101 heads in a row?” The answer would be 1/2, although the wording of the question might lead me to think otherwise.

So what is the chance the unknown child is a girl (thus making them both girls)? 1/2

Rob, Rob Rob. You are missing the point. In your example of 101 coins you are given that the first 100 are heads. As probability has no conscience the probability of the 101th coin being head is 1/2. But, if you are told that out of 101 coin tosses, 100 were heads but you are given no information on the 101th coin, what is the probability that it was also a head?

Consider your example with only three tosses. Two are heads, and you know nothing about the 3rd. What is the probabilty it is also a head? 1/2 you think. Nein, nada, нет. It is 1/4.

There are eight possibilities: HHH, HHT, HTT, HTH, THH, THT, TTT, TTH. Only four of these has the indicated 2 given heads: HHH, HHT, HTH, and THH. The probaility that all four are heads is 1/4. QED!

Try reading Richards new book, “Did You Spot the Mistake”.

Jerry, in your 3 coin case that matches Richard’s question there are not eight ‘possibilities’, there are two certainties and one uncertainty: HH?. We already know that two coins are heads, so where is the uncertainty? It is in the last coin only.

Unless you change the problem, as you do:

“There are eight possibilities: HHH, HHT, HTT, HTH, THH, THT, TTT, TTH. Only four of these has the indicated 2 given heads: HHH, HHT, HTH, and THH.”

This is a different question from Richard’s. Here you are saying that there two heads, but they may appear in any position, but we don’t know where *prior to doing the test*. So by simply stating that the sample space must include those options with at least two heads you are reducing the sample space to those last four, *and then doing the test*, and in this case then yes the chance of HHH is 1/4.

But Richard’s question is already stating that he as seen a girl. He already knows that a specific child is a girl and he is only asking us what the gender of the second might be, that is, what’s the chance that the second child is also a girl, making two girls. This is the same as you saying that the first two coins are heads, *having already seen that two are heads*, so what is the chance that the final coin will be heads, making HHH? 1/2.

For Richard’s question to require a 1/3 probability of two girls it would have to be worded something like this: “I’m told there are two children on the street, but they are definitely not two boys. What’s the chance they will be two girls?” Now the sample space is GB, BG, GG, and GG becomes 1/3.

Rob & Ron:

In my example of three coins tosses I am stating that 2 of the 3 are heads, but NOT the order. I am not saying that toss # 1 and toss #2 are heads. I am saying we don’t know which two of the three tosses yield heads. This is the same as Richard’s question. We do not know which of the two children is a girl; we only know that one of them is.

THIS is the vital issue here. 1/3 for the second girl and 1/4 for my third coin.

Jerry,

As soon as you see the girl, or two heads, they are already ordered, and there is only one remaining unknown.

Only prior to any toss or any sighting is the full sample space of all ordered pairs considered because they are all unknown outcomes, so 1/8 for 3 heads from 3 coins, or 1/4 for 2 heads from 2 coins, or 1/4 2 girls from 2 children. And only then are we eliminating certain parts of the sample space. Calculating the estimate ahead of any trials, but eliminating part of the sample space changes to odds to 1/4 for 3 heads from three coins given 2 heads, or 1/3 for 2 girls from two children given at least one girl.

Think of this latter case as the selection of one ball from three (1/3). The sample space is three balls to choose from. This is like encoding balls A, B, C in binary as 01, 10, 11. Though it’s binary we are eliminating 00 by the definition of the problem. The two child case is eliminating BB (00), and allowing only the sample space BG(01), GB(10), GG(11).

The former case, as in Richard’s question, is a quite different sample space: GB or GG. Even if we want to think of them as ordered pairs the first girl has already been seen and is not in doubt. Even if you want to call the first girl the second girl, this is only a change of label assignment. You are not introducing any more uncertainty, as you are when estimating prior to tossing or prior to seeing, where the probabilities are from a sample space of two independent variables, but then, by definition, restricting the sample space further, *but still prior to the trial*.

Ron: It seems you do agree with me. 1/3 for two girls. I hope Richard will be convinced.

Jerry,

I agree with you that from a sample space of BG, GB, GG that the probability of GG is 1/3.

But that is not the sample space of Richard’s question, which is GB, GG. Or if we include the woman: WGB, WGG, or if we include the street: SWGB, SWGG. And for all of these the probability of GG (or WGG or SWGG) is 1/2. Do we agree on that?

I love how we keep re-writing the scenario to fit our arguments. Here are a few more.

I saw a red haired child and a blonde child. The red haired child was a girl. What are the chances that both are girls?

Or

I saw two children today, one from the front and one from the back. The one whose face I saw was a girl, but I couldn’t tell with the other one. What are the chances they were both girls?

I think, because I am identifying the children in this way, I am creating an order and as such the chance, in both cases, is 1/2. I’m pretty sure, Jerry, that you would agree there.

Now try this one:

I saw two children today: one whose gender I could identify and one whose gender I couldn’t. The one whose gender I could identify was a girl. What are the chances they were both girls?

By identifying one of the children as a girl you are creating that order. So the answer, again, is 1/2.

I might go read that link you posted again…

R&R:

You are both hell-bent on identifying the children as #1 and #2.

Richard’s original question was, “I was walking along the street and saw a woman playing with two children of different ages. One of the children was a girl. What are the chances that both children were girls?”

It does NOT say, “Child Number One” was a girl. It says, “One of the childen was a girl”. That means the set of equally probable possibilities is GB, BG, or GG. [Pr] of GG is 1/3!

The statement “…of different ages” implies they are not twins, which would complicate the issue.

Now, R&R, we all need to get some R&R (Rest & Relaxation)!

Jerry,

“of different ages” has no bearing in it. Even twins are of differrent ages, unless they are co-joined. The phrase is of no more use than tossing coins and making a comment that one coin is older than the other. And, there is no mention of the relationship between the children, so whether they are twins or not can’t be decided, and any age difference is uninformative. In fact, other than being co-joined twins the only way they could be of the same age is if they were born at the same time to different mothers. The “of different ages” does not rule out un-joined twins. Real twins often make a note of who was born first and by how long. In this problem it is completely useless information.

On top to that I am specifically saying that the order is what should be ignored. This is why I say the sample set appears to be GB,GG but reduces to B,G. You are making an issue of the difference between BG and GB, when there is none to be had. You are considering permutations when we only need to consider combinations. The BG, GB distinction, the ordering of pairs, the permutation, is only pertinent under some different questions, not Richard’s question. Under Richard’s question once he identifies one child as being a girl the only remaining uncertainty is the gender of the other child. The order of the children should be ignored just as the permutations of other items (woman/man, on-street/off-street) are ignored because they are knowns.

I have two coins. One I’ve laid down as heads. The other one I’m going to toss. What’s the chance the second coin will be heads, matching the first; that is, given that you already know that one is heads, what’s the chance of two heads at the end of the one and only flip? Can you see that there is no probability concern, no uncertainty, about the state of one of the coins because I’m not even flipping it. I am not concerned with the order because what order I might create out of two coins has already been accounted for: the first is heads, the second I will flip. Richard already tells us that he has seen a girl. There is no doubt. He is not offering any uncertainty about it. And which child it is does not matter.

Richard’s question only requires a combination, and the combination can be GB,GG if we simply label the seen girl as the first child (which is the option I’ve been using); or BG,GG if you choose to arbitrarily label the seen girl as the second. We don’t care which it is really since Richard asks only about one unknown. But the question is disguising the problem by asking the probability of there being two girls, and by mentioning an age difference without asking more about it, and mentioning a woman that is disguising the any relationship or none that might exist between the children, leading some people to consider them siblings.

From combined independent probabilities it’s:

P1 = probability of seen child being a girl: 1 (It must be. We are told it is a girl)

P2 = probability of second child being a girl: 1/2

Combined probability = P1 x P2 = 1/2

Anonymous said it was 1/3 and added “If you think otherwise, just run a simulation on a computer to convince yourself.” Well, the problem there is that if you model it wrong you’ll get the wrong answer. Try this model, on a computer.

The model performs mulitple trials using a random number generator to feed a line of children through a door out onto the street, one a day at 12 noon. It just so happens that at 12 noon there is always an additional child, a girl, waiting to be picked up by her mum at 12:30. When the random child comes out the door he/she plays with the girl until the girls mother comes along and starts to play with the two children. And it just so happens that the random child always appears to be a different age to the lone girl. At 12:32 each day Richard walks along the road and sees a woman with two children. From this point on, each day, Richard blogs: “I was walking along the street and saw a woman playing with two children of different ages. One of the children was a girl. What are the chances that both children were girls?”

Readers of his blog start to wonder about the ages of the children, the order of their births, think about the implication of them being twins, think that teh woman must be the mother of both children, etc. They quite rightly eliminate the possibility of two boys. But they have no furher information about this scenario from which to draw other conclusions. They are told quite clearly that there is always one girl, but somehow assume this means that because they have eliminated BB that they must consider BG and GB as permutations as well as GG. But clearly the model above matches Richard’s blog post perfectly, and in this model the odds of GG will be 1/2 because there is only one random variable, and the computer simulation will show that. Any doubt?

The posted answer, 1/2, was correct; but the solution itself was wrong. It is wrong because it counts one of the three “possible” cases, GG, twice. Think about it – should you really be able to count up to four from among the three cases (GB, BG,GG)?

But the answer is right because you don’t want to merely count the cases, you want to use the probabilities that this person who was walking down the street would tell you about a girl ineach case. That probability is 100% for the GG case, 0% for the BB case, but 50% for the BG and GB cases. So the answer is (100%)/(100%+50%+50%+0%)=1/2.

It turns out that the 1/2 camp and the 1/3 camp “can” both be correct. This paradox has been around since the Big Bang. For those with continued interest see:

http://en.wikipedia.org/wiki/Boy_or_girl_paradox

As someone who, over the course of this discussion, started in the 1/3 camp but then switched I can say I am most definitely right, then.

🙂

Consider all women who have 2 children with them with at least one of them being a girl. If one of these women is chosen at random, the second child has [Pr] = 1/3. My scenario.

Consider all women who have 2 children with them. If one of these children is chosen at random and identified as a girl, the second child has [Pr] = 1/2. Your scenario.

The argument becomes heuristic. Read the Wikipedia article.

“I was walking along the street and saw a woman playing with two children of different ages. What are the chances both children were of the same gender if…

1) … I tell you that one of the children was a girl?

2) … I tell you that one of the children was a boy?

3) … I tell you that I know the gender of one of the children?

4) … I don’t tell anything?

Jerry, all of these questions have to have the same answer. #2 is obviously the same as #1. #3 conveys the totality of knowledge in #1 and #2; and since they must both be same, it must match that answer. But #3 conveys no actual information to you, so it is the same as #4.

This might be easier to see with more possibilities. Say I take one card at random from two different decks. The chances both are the same card are 1 in 52, right? Now, suppose I tell you that one is the Seven of Clubs. Is it more reasonable to believe the chances both are the same is still 1 in 52 (the answer that compares to 1/2 in this puzzle), or that it changes to 1 in 103 (compares to 1/3 = 1/(2N-1))?

In probability, there is a difference between OBSERVING a fact after selection, and REQUIRING that fact before selection. 1/3 (or 1 in 103) is correct if and only if you asked me “Was one a girl?” (or “is one the Seven of Clubs?”) and I said “yes.” 1/2 (or 1 in 52) is correct anytime it was possible for me to have said something else when there was a girl (or a Seven of Clubs).

So ask yourself: is there something in the statement of this problem to prevent, if I had seen a boy and a girl playing together, from saying “one was a boy” ? If you can’t find an explicit reason to prohibit it, the answer is 1/2. I can’t see one.

BTW, the original set of four questions I asked is a variation of Bertrand’s Box Paradox – just use three possible combinations instead of four. It’s called a paradox because the answer to #4 is clearly 1/2, but it seems, to those who use your line of reasoning, that it must be 1/3. Joseph Bertrand used this example, in 1889, to warn people about the error implicit in ignoring the difference between OBSERVING and REQUIRING.

Jerry,

Your 1/3 solution isn’t the solution to Richard’s question, which is he main point here. Yes, 1/3 can be the answer to a different problem. But try getting away with that on a math test: What’s the probability of throwing a six in one throw with a single fair dice? 1/6. But I wouldn’t offer as a solution to *that* problem the solution to the probability of throwing heads from a single toss of a fair coin, 1/2, and then explain to the examiner, “Well, yes, but my answer is right for *my* scenario! We’re both right!”

“Consider all women who have 2 children with them with at least one of them being a girl. If one of these women is chosen at random, the second child has [Pr] = 1/3. My scenario.”

Yes, your scenario, not Richard’s question. Your scenario is about the probability of two unknowns, with an extra condition that at least one will be a girl. This is the answer that Richard would have to come to if he asked himself your question before going onto the street – his estimate of the probability, given that he did not at that time know the actual outcome. But his actual question is posed such that he does know the outcome of one of the children, and he is telling us that.

The only remaining uncertainty *for us* (the one’s asked the question) is the gender of the other child.

And, we don’t know if Richard knows the gender of the other child because he does not tell us, so Richard may know with certainty whether there are two girls (zero probability if it’s actually GB, or 1 if it’s GG), or he too may be uncertain about the other child, perhaps because his view was obscured, or for some other reason, in which case his estimate, like ours (should be) is 1/2.

Jerry,

Your link to the boy girl paradox is about your scenario, not Richard’s question.

From your wiki link First Question:

“…but later acknowledged that the second question was ambiguous. Its answer could be 1/2, depending on how you found out that one child was a boy. ”

But Richard’s question is not ambiguous. He says very clearly he saw a girl. There is only one remaining ucnertainty.

Note this one from the wiki article:

“Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?”

This has the same answer as Richard’s question, but for quite different reasons. How many sides has a coin? Two. How many ears have I got? Two. You can have the same numerical answer to two quite different problems. This ‘Jones’ question is still unlike Richard’s in that it is still asking the probability *prior to seeing the children*, but with the condition in this case that it specifically says the older child is a girl. In this case the older child is acting as a label, a marker, to identify a subset of the original sample space, so eliminating not only BB, but also BG, prior to the actual test. Richard’s wording not only eliminates BB (which everyone gets) but also removes uncertainty from one child, which is definitely 100% a girl.

Second Question:

“Gardner argued that a “failure to specify the randomizing procedure” could lead readers to interpret the question in two distinct ways”

In Richard’s question the randomisation of one child is set: 100% girl. There is only one remaining random variable, the second child, which is merely phrased in the context of the first known child, so leading you to misread the question to think the first child is significant in the calculation. Ether the first child is not significant and it’s the probability of the gender of the other child, or if you want to take account of the identified girl and make it a product of two independent probabilities it’s 1 x 1/2 = 1/2.

The discussion that follows is about the ambiguity in *those questions* in that section. There is no such ambiguity in Richard’s question.

Analysis of Ambiguity:

“Thus, if it is assumed that both children were considered while looking for a boy, the answer to question 2 is 1/3”

Both are not being considered in Richard’s question. One is certain.

Next: “However, if the family was first selected and then *a random, true statement* was made about the gender of one child …” This, as is explained below the figure gives the answer 1/2. But this is still not Richard’s question because it is still a *random true statement* – in other words it is asserting a required outcome, it is conditioning the sample space, prior to actually looking. Richard has already seen and knows for certain the gender of one child.

“Note that this is not necessarily the same as reporting the gender of a specific child, although doing so will produce the same result by a different calculation.” Precisely. Richard is reporting the gender of a specific child and we are getting the same answer by a different calculation (or more than one calculation: P(100%) x P(50%), or simply P(50%) )

The Bayesian analysis gives similar sets of results and the 1/2 part is most similar to Richard’s question but still not quite the same because they are still discussing prior probabilities but with sampling that introduces conditional probabilities. There is no conditional probability in Richard’s question. Though it asks us the probability of two girls that probability is only dependent on the outcome of the remaining unknown gender and is not conditional on the first, unless, as above, we trivially introduce the first child with a gender probability of being a girl of 1.

Variants of the question:

“Bar-Hillel & Falk use this variant to highlight the importance of considering the underlying assumptions. The intuitive answer is 1/2 and, when making the most natural assumptions, this is correct. However, someone may argue that “…before Mr. Smith identifies the boy as his son, we know only that he is either the father of two boys, BB, or of two girls, GG, or of one of each in either birth order, i.e., BG or GB. ”

The “before Mr. Smith identifies the boy as his son” clearly states that the 1/3 case is about prior estimates when we don’t have specific information.

“The intuitive answer is 1/2 and, when making the most natural assumptions, this is correct.” Closer. I am assuming that when Richard says he has seen one girl that he is telling us the truth. There is no uncertainty. This is what I’ve been trying to point out in several comments, the difference between what is known, who knows it, what the remaining uncertainties are. It’s the difference between making estimates of probabilities with some conditions applied and making estimates about probabilities of specific knowns and unknowns.

And so on… in all the examples considered in the wiki page they are talking mostly about prior probabilities under different scenarios.

But read Richard’s actual question. “I …saw… One of the children was a girl. What are the chances that both children were girls?”

I don’t know how it can be any clearer that when you remove the fluff about the street, the woman, the age difference, the only thing this question telss us is that there were two children, one was a girl and the gender of the other is unknown. One unknown.

As an additional note, since Richard gives a quite different explanation that includes a reference to the birth order it may be that Richard himslef intended to ask one question but actually asked another. I am referring only to the question that Richard actually asked, irrespetive of what he thought he was asking.

Richard arrives at the same answer of 1/2 because his use of birth order also applies in getting to the same result. As in one of the wiki scenarios he is using birth order as a marker. But I think that his result is conincidentally the same result using the same sample space breakdown. I don’t think it’s actually valid because the age relationship is incidental. The solution I’m offering would still apply to the same question even if the age relationship is ignored, and even if the question left out the age phrase.

All we need to know is that there is one unknown and it has a probability of of being a girl of 1/2. Stating it along side the certainty of the other child’s gender adds no more uncertainty that does the gender, age, hair colour of the adult.

One good reason that Richard told us that the children were of different ages would have been to rule out the possibility of identical (maternal) twins, which must be both boys or both girls.

I was thrown off by Richard’s incorrect explanation, which I knew was wrong because he counted GG twice. “jeffjo” brought that up, and he is right.

If ya’ll would really like to frustrate yourselves, check out “The Sunrise Problem” at:

http://en.wikipedia.org/wiki/Sunrise_problem

Meanwhile, I quickly solved last Friday’s puzzle, which is nowhere as contentious.

Given our days are measured by the rising of the sun, surely the probability is 1/1. If the sun doesn’t rise then there is no tomorrow. So the fact the question implies a tomorrow means that there will be a sunrise!

There are lots of ways people rationalize the answer they want to be correct, by imagining why reasons are “good.” You do it is well, in your “Consider all women who…” scenarios. Because the proper one – the only one that both:

1) Matches the problem statement (“a child chosen at random is identified as a girl” does not because no specific child is identified), and …

2) Does not require a filter placed on the selections (“with at least one of them being a girl”) …

2A) Except “who have two children” because that does not affect the answer, …

Is this one:

Consider all women who have 2 children with them. For each, identify one gender for a. Choose one at random from those where you identified the gender “girl” (NOT a specific girl). What is the probability this woman has two girls.

This is 100% consistent with the problem, with no filtering assumptions. The answer is 1/2. You get 1/3 applying a filter that is not implied by the problem. This answer simply cannot be correct. Ron Murphy’s solution is wrong (even though it gets the right answer) because there is no “other girl” for half of the women who are identified with a girl.

The so-called “Sunrise problem” is an example of mental self-gratification. It cannot be expressed as a probability problem because there is no frame of reference where it is subject to chance. If one want one that seems to be problematic (but really isn’t), try the http://en.wikipedia.org/wiki/Sleeping_Beauty_problem

Regarding “The Sunrise Problem”, there is a clear inevitable scenario when the sun will not rise. Our sun will eventually become a red giant and engulf the Earth. We know this today, but did not two hundred years ago. Astrophysics teachs us about the “lifetime” of stars that govern the probability of sunrises on the planets that orbit them.

There is also the philosophical aspect of “The Sunrise Problem” Technically, the sun does not “rise”. The Earth rotates, giving the appearance of a sunrise.

Jeffjo: You are doing what you are claiming I am doing, creating your own scenario. Both “The Second Child Problem” and “The Sunrise Problem” have implications that transcend the obvious.

Jerry, where does it say, or imply,that the person walking down the street could not, under any circumstances, tell you he saw a boy playing? Because such a requirment is needed to get the answer 1/3. I do not assume that,or any othe scenario. You do. I only assume the lack of one.

Hi jeffjo.

Could you clarify what I’m wrong about. I’ve not provided a solution to that specific scenario “All women who…” or mentioned “other girl”. I have provided a solution to Richard’s question regarding one girl and another child which I think is correct.

Ron said “All we need to know is that there is one unknown and it has a probability of of (sic) being a girl of 1/2. Stating it along side the certainty of the *other* child’s gender …”

Here, your “one unknown” is the gender of the ubiquitous “other” child, which you indeed did mention in relation to the “identified child.” You just switched which was called “other.” If there is also a boy, you can deduce which child was “identified,” and which is the “other.” But if there are two girls you cannot distingish the two, no matter what labelsyou put on them.

This problem is just like the Monty Hall Problem, where the trick is to acknowlegde that:

(1) You could have learned the opposite fact from what is exemplified in the problem. Here, you could have learned “boy” instead of “girl.” In the Monty Hall Problem, the host had two doors he could have opened.

(2) In some of the cases where you would have learned the opposite fact, the fact you did learn was also true. Here, that means the possible families with one boy and one girl. In Monty Hall, it means games where you picked the right door to begin with.

(3) So, not only must you remove from consideration the instances where the fact you learned does not apply, you must remove the ones where it does apply but you would have learned something else.

You did not do this.

So, say 100 people go out walking, and each tells you about the first woman they see playing with two children. 25 of the women will have two girls, 25 will have two boys, and the remaining 50 will have one of each. But 50 of the people will tell you about a boy, and 50 will tell you about a girl. The correct solution has to make the (25:50:25) split of women agree with the (50:50) split of people, and the only way is to have only 25 of the 50 people who saw a boy an a girl actually tell you about the girl.

Jeffjo,

The use of ‘other’ in that statement was only in relation to the one I was focusing on in that statement. Read my other comments. One known: girl. One unknown: the ‘other’ child. But in the context of a sentence if I’m talking specifically about the unknown child it’s legit to talk about the known girl as the ‘other’. Not the clearest use of language on my part but no change to the logic.

And this is nothing to do with Monty Hall. In Monty Hall the contestent starts with all unknowns and hence 1/3 for selected door and 2/3 for pair of unselected doors. Even after MH excludes a door the contestent still has two unknowns, but still 1/3 and 2/3 hence he should switch. But, still two unknowns. And in the MH problem selection (i.e permutation) is significant.

Richard’s question is quite different. We are given one known, a girl, and one unknown gender child. Two independent variables. First P(girl) = 1. Second P(girl) = 1/2. Combined = 1 x 1/2 = 1/2. And we only state it like that because the question is phrased as asking the prob of two girls. The same answer would apply had the question asked for prob of one woman and two girls: 1 x 1 x 1/2 = 1/2 because gender of adult and one child are given.

“The use of ‘other’ in that statement was only in relation to the one I was focusing on in that statement.”

Right. The point is that you cannot always establish a relationship like the one you describe here. There isn’t always a “one” you can “focus on.” If there are two girls, neither is the “known” girl. We are not “given one known” girl, we are given that there is a girl amongst the two. The second (not “other”) gender is not indepenent of the first. No “first” or “second,” only one between the two.

Until you can try to understand this difference, there is no point in discussing it. I’m not blowing you off, I’ve just had too much experience with people who refuse to address the difference.

I think, though, the fact that Richard in this scenario was a witness, then there is a “known” isn’t there?

Richard is saying “I saw two children and one of them was a girl”. So he’s an eyewitness to an event, and he’s describing what he saw. That gives the children an order, doesn’t it? The “first” child is “the one Richard identified as a girl”, the “second” child is the “one Richard didn’t identify”.

So the question, to me, implies a single unknown child, hence 1/2.

I can see the difference between there being a “one” and not, but I think the way this question is worded implies a “one”.

As you say, there isn’t always a “one” like this, and in those cases it is 1/3. So “what are the chances of someone with two children having two girls?” The answer is 1/4. “What are the chances if one of those is a girl?” That eliminated BB so the answer becomes one in 3. But we are still talking in hypothetical “if” statements.

The way this puzzle is worded (and yes, I realise it is still a puzzle, and still didn’t actually happen) is talking about a witnessed event, and as part of the eye witness report, a “one” is identified.

Rob:

1) There is an order, whether or not Richard could determine it or reported it. But “order” isn’t what is important – the two children are individuals, and identifying one as “older” (or “taller,” “faster,” “smarter”, “named first alpabetically,” etc.) is just a way to identify how each individual fits in the math. Think of it this way: Three people try to determine experimentally the statistics for rolling two dice: one sees a red and a green die, and orders them. One sees two idenical dice, an can’t order them. One has a red and a green die, but is color blind and can’t see the difference and so can’t order them. Does anbody really think the results should be different for any of them?

If Richard dosn’t tell us an order, we are comparable to the color blind person. But…

2) Richard isn’t identifying a child. He is saying that there is a girl amongst the two; and if both are, neither he nor we can say “it was Alice who was identified.” Of the four possible ordered (whether or not we can place a child in the oder) pairs, three remain possible. But …

3) Remaining possible isn’t enough. In two of those three pairs, Richard could have told us about a boy. So we don’t count three full cases, we count one full case and two half cases. The answer is 1/(1+1/2+1/2)=1/2. And…

4) Yes, Ron, this is just like Monty Hall. Only here we start with three cases (the places the prize could be) instead of four, and two remain possible. The one where the prize is behind the remaining door counts as a full case, because Monty Hall had to open the door he did in that case. The one where the car is behind the contestant’s door counts half, because Monty Hall had two choices (just like Richard had two choices in a boy-girl family). The answer is 1/(1+1/2)=2/3.

jeffjo,

I walked into a room and saw a woman with two coins of different ages. One coin was on the table heads up. The woman said she was about to flip the second coin. What is the probability both coins would be heads up after the flip?

Note what you don’t know…

You don’t know if the woman flipped the first coin or intentionally laid it down heads up, so you can’t assume it was flipped and so can’t include its tails value in any sample space. You do not need to consider the probability of this coin being heads up, other than saying it has a probability of 1.

You don’t know whether the coins are members of the same family. If you think they are not then you think this because ‘family’ makes no sense with regard to coins. But note there is no contradicting claim in the problem statement that they are related. Why read family relationship into Richard’s question when there is none?

Note what you do know that is incidental and irrelevant to the problem…

You know there is a woman with the coins not a man so you need not introduce any probability regarding her, other introduce a probability of 1 if you want to include that in some equation.

You know that the coins are of different ages. But how does this help? Are you letting this imply family relationship in Richard’s question when there is none stated?

You know that there cannot be two tails. But this is knowledge based on logical elimination, because if one coin is already heads both cannot be tails. This is not elimination based on prior probabilities; but you could use the trivial logical probability of zero for tails, or the certainty probability of 1 that the first coin is heads.

Note what you do know that is relevant…

One coin showing heads, with prob 1 if you insist on using that fact. One coin about to be flipped – i.e. the one uncertain outcome with prob 1/2.

This coin example is logically the same as Richard’s question. There is one known P(H/G) = 1, and one unknown P(H/G) = 1/2.

You seem to be mistaking one case, (a) a prior unknown about two variables with the prior elimination of one permutation, with the entirely different case, (b) of one known of prob 1 and one unknown of prob 1/2. Your insistance on thinking family or age has anything to do with it is puzzling and unexplained. You do not explain why you think this is related to MH at all. It isn’t.

“I walked into a room and saw a woman with two coins of different ages. One coin was on the table heads up. The woman said she was about to flip the second coin. What is the probability both coins would be heads up after the flip?”

Contrary to your claims, this is a different problem. And that difference is my whole point. It is different because you have identified one coin (the one that has been flipped, and is showing heads), so there is an “other” coin as well. We don’t need to know which is older for you to have done this, we just need to know that you have associated “heads” with a specific coin.

You can make the problems the same by having Richard say “The child wearing a red shirt was a girl.” This associates “girl” with a specific child, but ignores age. You can make the problems the same by saying the woman told you she was hiding two coins, which she had just flipped, under her hand which you see flat on the table. She simply tells you that one coin landed on heads, so you can’t identify a specific coin this way.

What is the probability that both (not the “other,” because there is no “other”) coins are heads in that scond comparison? The answer is 1/(1+2Q), where Q is the probability she would tell you “heads” when there was a heads and a tails. If you had asked her “Is there a heads,” then Q=1 and the answer is 1/3. If she volunteered the information, you can’t assume Q is anything except 1/2, and the answer is 1/2. This “Q” is an important concept, because it points out exactly how the two debated answers can be correct.

I have never mentioned “family,” or associated the children as a unit in any such way. I have never insisted on “family” or “age.” I have said that age is a convenient way to specify a child, but I offered others. Because you are right, both “family” and “age” are totally irrelevant except age as a specifier.

I am making no mistakes, but you are. First, by imagining reasons why I am making mistakes (“insistance on thinking family or age has anything to do with it”). Second, you are assigning the value “girl” to one of your two unknowns, when all you know is that one or both have that value. You can’t make an assignment to either variable. That’s my point: you have to solve it without assigning a value to a variable.

Nope. It’s the same problem. Which coin is heads? That one, the one identified. Which child is the girl? That one, the one identified. That sort of definite identification is just good as naming for the purpose of this problem.

The diff between 1/2 and 1/3 is whether the question is asked after knowing and being certain about ‘that’ child (1/2) or prior to knowing and being uncertain (1/3). The known gender excludes BB logically, leaving only GB or GG. The prior unknown case is a permutation, with BB excluded as a condition.

“Which child is the girl? That one, the one identified.”

Except there was not one identified.

I can see you aren’t interested in finding the right answer, only yours. Have fun.

jeffjo,

Sorry I attributed ‘family’ stuff to you. I’m working from a phone and clearly mixed comments I was responding to.

Try this.

I was walking down the street and saw a woman with a girl. What’s the probability the child was a girl?

This is not asking what probability you *would* have asigned before I told you the gender. Even if I had known it was definitely a girl, had I not told you that fact it would have been reasobable of you to use a probabilistic model. But because I have told you it’s a girl you do not need to apply a probabilistic model to it; or if you insist on doing so the probability is 1.

Probabilistic models need only be applied when we don’t know. But even then there use may appear to be justified but later turn out to be mistaken. So, I can use a probabilistic model to predict the odds of H/T when a friend tosses a coin, but if I then find out it was a double headed coin I will realise my use of a probabilistic model was mistaken.

In Richard’s question there is only one unknown. It does not matter which child is the girl he identified. It is the very fact he is not interested in which specific child is the unknown that we do not need to consider both BG and GB permutations, along with the fact that there is a girl plus a child of unknown gender.

Can you explain why you think a probabilistic model should be applied to anything other than the child with the unspecified gender? We are not being asked to identify the child, so which one it was does not matter.

“In Richard’s question there is only one unknown.”

No. There are two unknowns. We know a value that applies to one of them, BUT WE DON’T KNOW WHICH ONE, so we still have to treat both probabilistically.

“It does not matter which child is the girl he identified.”

No child was identified. As a result, the answer isn’t always 1/2, which it would be by your methods. If we had asked Richard “is there a girl,” or there was something else that prompted him to tell us about a girl and not a boy, then he does not have to pick a child when there are two girls. And the correct answer is 1/3. Your method, of identifying a child when Richard didn’t, can’t get this answer.

If Richard had volunteered the information, he only has to pick a GENDER, not a CHILD, to tell us about. And while we can deduce a child indirectly “identified” this way when there is both a boy and a girl, we cannot when there are two girls. We actually need to consider all four combinations GG, BG, GB, and BB; but we assign a probability to the event R, that Richard tells us about a girl, in each case (One of those will be zero). Such a probability is called a conditional probability; for example, P(R|GB) is the conditional probability that Richard would tell us about a girl if he saw a GB set of children. The correct Mathematical solution comes from a proven relationship called Bayes’ Theorem:

P(GG|R)=P(R|GG)*P(GG)/[P(R|GG)*P(GG)+P(R|GB)*P(GB)+P(R|BG)*P(BG)+P(R|BB)*P(BB)]

Since all the P(XY)’s are 1/4, we can divide them out:

P(GG|R)=P(R|GG)/[P(R|GG)+P(R|GB)+P(R|BG)+P(R|BB)]

= 1/(1+1/2+1/2+0) = 1/2.

Why assign a probability that Richard tells us about a girl? He did tell us about a girl. Probability 1. You would only introduce probability about this if you were uncertain – perhaps Richard could be mistaken or perhaps on some days of the week he tells lies, etc. Then it would make sense to for probabilities such as P(R|GG) – the probability that Richard will tell us about a girl if there are two girls. Without this added uncertainty P(R|GG) = 1.

However, P(R|BG) = P(R|GB) = 1 and P(R|BB) = zero.

Note also here that you are introducing a distinction between GB and BG when you need not…

… you are providing ordering when you need not: BG, GB, but choosing to ignore the ordering gG, Gg, where G is the girl that Richard saw.

Label the children x and y. Then Richard saw xG and yb or xG and yg. Or alternatively and arbitrarily he saw xb yG, or xg yG. Or, using order alone to represent x and y child: Gb or Gg; or bG or gG.

You can if you wish consider the total sample space of xGyb, xGyg, xbyG, xgyG. Again, using order to represent x and y: Gb, Gg, bG, gG.

We can either consider the entire sample space, or arbitrarily choose one subset, ignoring the specific identity. In either case the probability of both being girls is 1/2.

And this is because we are given definite information about there being one girl. We are asked to estimate probability after being given this information.

We only need to consider {BB, BG, GB, GG} if we are told both child genders are unknown, but that something is preventing BB, which does then leave {BG, GB, GG}. The interpretation of this could be as follows.

The ordering here could represent birth order. It could be a family that do not want two boys and choose to abort any second boy. But we don’t know the birth order. In this case the first outcome could be the result of the following pregnancies: BbG, or BbbbbbbG, or any other outcome that results in BG, where Bb results in b being aborted.

Another possibility is that the children are not related but where the woman can choose from a class of girls and boys but refuses to take two boys because two boys always fight.

Note here that what actually happens to determine the actual outcome does not matter. This is only about our estimate of probabilities of two unknowns based on limited knowledge.

In the birth case the women in the family may be predisposed to give birth to only girls, and so if we could take this into account we would alter our probabilities. In the second case the woman might choose randomly from a class of one boy and fifty girls, in which case it is more likely that she will have two girls with her. In this case she is choosing without replacement so if she picks the boy first she has to pick a girl second, but if she picks a girls first she has a slightly improved chance of picking the boy second. So now the probabilities we should estimate are not independent of order.

The point is that this is an entirely different problem. Both genders are not unknown to us, but something of the sample space is (not BB). Richard’s problem is simple by comparison. We know there is a girl, so what is the probability that the *other* child is a girl, making both children girls?

And the use of *other* here is entirely legitimate means of arbitrarily identifying the *other* (i.e. not the one noted girl). Your objection to it is unreasonable.

“Why assign a probability that Richard tells us about a girl?”

Because the fact that he told us about a girl does not rule out the possibility. You don’t “eliminate” the possibility of a BB pair, you assign it a probabiltiy of Zero because it is impossible Richard woud tell us about a girl in that case. You don’t “keep” the possibility of a GG pair, you assign it a probabiltiy of One because it is impossible Richard woud tell us anything else in that case. By treating BG and GB like GG, you are implicitly assigning it a probability of One also, which is the wrong probability.

I’m just recognizing that what you yourself are doing is assignihng a probability, and making it the correct probability.

I am not “providing” order, I am recognizing that it exists and using it to distinguish the two unknowns you admit exist. I’m just refusing to apply a value a specific one, which you do, and is in fact the only “use” of order here.

Try this again, another way, since you skipped it before. Say Richard goes out walking for 100 stratght days, in different areas just to keep things random. Each day, he takes notes on the qualifying pair of children he meets, and tells you one gender.

1) How many pairs do you expect to have the matching genders? I hope you say 50, or else there is no point in discussing it with you.

2) How many times do you expect him to tell you about a girl? A boy? I hope you sy 50 each, or else explain why you think Richard is biased (and why you would get a different answer if the original question had been about boys).

3) Of the 50 times he said “girl,” in how many could he have said “boy?” Of the 50 times he said “boy,” in how many could he have said “girl?”

4) So, what is the conditional probability P(R|BG)?

In short, what we know about the sample space is not “not BB.” It is “Richard told us there is a girl.” He didn’t identify one, he didn’t use order, but in two of the three possible cases he had a choice. We have to address that choice.

Your 100 days example is an instance of sampling from random data. You are estimating prior probabilities on two random independent variables. This is logically different to Richard’s problem.

The woman and two children Richard met might well have been a random or chance meeting with just one of many women with two children walking the streets that day, and if Richard simply asked what the chances are of meeting such a woman with two girls our best approach would have been to estimate 1/4 based on our knowledge of the population distribution of boys and girls. We could be totally wrong, since this particular woman might be going to or from a girls dance club in the street so skewing the data – so that the prob of seeing two girls is highet than usual. The point being that in this case the probabilities come only from what we estimate it to be and not from what actuslly occurs.

But, it does not matter about the chance nature of the meeting with regard to one of the children of the pair he actually met because he tells us that one is in fact a girl. This meeting has ceased being a random sample from all women with children he might meet and has bcome an actual sample with actual children. There is definitely a girl and a boy, or there is definitely a girl and a girl. Richard is simply not telling us which of these it is so our estimated probability for it being two girls is 1/2.

It does not matter whether Richard does or does not know the gender of the other child. He may know with certainty that it is two girls or a girl and a boy. The important point is that is we the hearers of Richard’s statement that are left with the uncertain gender of one child. We don’t know which child it is we are uncertain about and we don’t care because we are not asked that. So the important random variable is the unknown gender of one child which we should, lacking any other data, estimate as 1/2.

If we want to include the redundant information about the known gender child and the known gender adult we can include them as P(W) = P(G) = 1. Though both P(W) and P(G) are redundant, as are any other known facts Richard might choose to include, we might be inclined to include P(G) and not P simply because of Richard’s phrasing of the question that asks the chance there being two girls: P(G) x P(g) = 1 x 1/2.

“Your 100 days example is an instance of sampling from random data. You are estimating prior probabilities on two random independent variables. This is logically different to Richard’s problem. ”

I am not sampling from random data (in which case we probably would not get exactly 50), I am estimating from ideal distributions and using the ideal count to illustrate what that accomplishes. My 100-days problem is an example of applying conditional probabilites to such distributions, and it is identical to Richard’s problem. Unless you can see a part where only women accompanied by at least one girl were allowed to be seen by him. And in that case, the answer is 1/3.

Look, we get the same answers. The only thing I am arguing about (I have no idea what you think you are,only that you object to my saying you were wrong one a point you haven’t replied to), is how you are identifying “the” child so that the “other” child’s gender is an independent random variable, and that is wrong. We need a joint distributiuon for two children, and then we condition it on “Richard observed to us that one was a girl.” If there was a reson why he would only mention girls this way, and never boys, the answer is 1/2. If he coud mention boys,and the problem provides no reason he couldn’t, the answer is 1/2 for different reasons that you support.

Your method can’t get both of these answers, so it must be wrong.

Try looking up “Bertrand’s Box Paradox,” and change the number of boxes from three to four. It’s the same problem.