tear-your-hair-out probability puzzle

castie · 30-07-2010 02:54PM

serendip wrote: »

Problem 1 (with answer):

I have two children. One of my children is a boy. What is the probability that my other child is a boy?

Answer: 1/3

Many people find this counter intuitive, but if you work through the cases, you'll see that it is indeed 1/3.

Outside factors affect whether your baby is a boy or a girl.

So realisticly the chance of having a boy increases if a boy has already been born.

i.e after a couple has a baby boy its no longer 50-50 for the next child

MathsManiac · 30-07-2010 04:16PM

I agree almost entirely with pH (although perhaps not with the very final assertion regarding most normal readings yielding 1/2. I would prefer to simply say that the problem has not been properly formulated and therefore cannot be properly answered, but that 1/2 is a reasonable answer.)

The reason this problem causes so many arguments can, as I see it, be summarised as follows:

1. Unsophisticated (and incorrect) analysis of question: answer = 1/2
2. More sophisticated and better informed analysis: answer = 1/3
3. Even more sophisticated analysis: problem is not well defined, but, under certain plausible assumptions about how you got the information, the answer is 1/2. Under a different set of assumptions, it's 1/3.

The difficulty is that people in category 2 are usually convinced that anyone who says 1/2 is in category 1. However, the manner in which you need to argue the case for 1/3 is quite different depending on which category of opponent you are arguing with.

On a previous thread on this topic a year or two ago, I was very much in camp 2, but ended up being convinced that the matter is a bit more complicated. In particular, having initially been certain that the manner in which the available information had been sought and given was irrelevant, I was subsequently convinced otherwise. The issue is as pH has said. Here's another way of looking at it:

Let T be the set of all women with two children. Leaving aside marginal issues about whether the two sexes are precisely equally likely and so on, let's all agree that, among all women in T, the four ordered combinations BB, BG, GB and GG are equally likely. That is, 25% of such women have two boys, 25% have two girls, and 50% have a boy and a girl.

Consider the subset S of T consisting of women who are inclined to walk up to complete strangers and make slightly bizarre statements about their children, and, in particular, statements of the form "I have two children, at least one of whom is a [boy/girl]". Let's suppose, again for the sake of argument, that these peculiar people are evenly distributed among the four categories in T. (i.e., the proportions of BB, BG, GB and GG in S are the same as in T.)

Let statement E be "I have two children, at least one of whom is a boy."
Let statement F be "I have two children, at least one of whom is a girl."

The BB women in S will walk up to strangers and make statement E.
The GG women in S will walk up to strangers and make statement F.
What of the remaining women in S? It is reasonable to assume (although we cannot say for sure) that these women are equally likely to make statement E as to make statement F.

Under the above conditions, the following apply:
P(BB,E) = 25%
P(BB,F) = 0%
P(BG,E) = 12.5%
P(BG,F) = 12.5%
P(GB,E) = 12.5%
P(GB,F) = 12.5%
P(GG,E) = 0%
P(GG,F) = 25%

So, if a woman you don't know anything about walks up to you in the street and makes statement E, (i.e., "I have two children, at least one of whom is a boy."), then Bayes' theorem applied to the above indicates that the probability that her other child is a boy is indeed 1/2.

hivizman · 30-07-2010 10:35PM

Nice analysis, MathsManiac.

Essentially, what the Case 2 people are doing is replacing your assumption that people with one girl and one boy are equally likely to make statement F as statement E with an assumption that such people will always make statement E.

This is basically the same distinction drawn by pH in post #12.

Perhaps the Case 2 people hear the problem as a sort of dialogue (pH basically makes this suggestion in post #31):

Mother: I have two children.

Statistician: Is at least one of them a boy?

Mother: Yes.

Statistician: We can rule out your having two girls. But it's twice as likely that you have one of each as that you have two boys. Then the probability that both your children are boys is 1/3.

Whereas the Case 3 people would represent the dialogue as:

Mother: I have two children.

Statistician: Tell me some more.

Mother: At least one of them is a boy.

Statistician: If both your children are boys, you have to say that, whereas if you have one of each, you could just as likely have answered that at least one of them is a girl. Therefore the probability that both your children are boys is 1/2.

c-note · 30-07-2010 11:35PM

wow, there are some subtleties in this question.

i have a question.
if it is worded:

"A woman has two children, at least one of whom is a boy, what is the probability that the other is a boy?"
is the answer 1/3

OR
what is the correct way to word the question such that the answer is 1/3

pH · 31-07-2010 09:25AM

c-note wrote: »

wow, there are some subtleties in this question.

i have a question.
if it is worded:

"A woman has two children, at least one of whom is a boy, what is the probability that the other is a boy?"
is the answer 1/3

OR
what is the correct way to word the question such that the answer is 1/3

Again, without knowing how you discovered "at least one of whom is a boy" then the question is at best ambiguous, you could argue for both the 1/3 and 1/2 answer.

If you put an explicit questioning step, then the answer is definitely 1/3.

Here's it all set out explicitly:

My neighbor has two children. Assuming that the gender of a child is like a coin flip, it is most likely, a priori, that my neighbor has one boy and one girl, with probability 1/2. The other possibilities---two boys or two girls---have probabilities 1/4 and 1/4.

Suppose I ask him whether he has any boys, and he says yes. What is the probability that one child is a girl? By the above reasoning, it is twice as likely for him to have one boy and one girl than two boys, so the odds are 2:1 which means the probability is 2/3. Bayes' rule will give the same result.

Suppose instead that I happen to see one of his children run by, and it is a boy. What is the probability that the other child is a girl? Observing the outcome of one coin has no affect on the other, so the answer should be 1/2. In fact that is what Bayes' rule says in this case. If you don't believe this, draw a tree describing the possible states of the world and the possible observations, along with the probabilities of each leaf. Condition on the event observed by setting all contradictory leaf probabilities to zero and renormalizing the nonzero leaves. The two cases have two different trees and thus two different answers.

This seems like a paradox because it seems that in both cases we could condition on the fact that "at least one child is a boy." But that is not correct; you must condition on the event actually observed, not its logical implications. In the first case, the event was "He said yes to my question." In the second case, the event was "One child appeared in front of me." The generating distribution is different for the two events. Probabilities reflect the number of possible ways an event can happen, like the number of roads to a town. Logical implications are further down the road and may be reached in more ways, through different towns. The different number of ways changes the probability.

http://research.microsoft.com/en-us/um/people/minka/papers/nuances.html

ZorbaTehZ · 01-08-2010 03:50PM

castie wrote: »

So realisticly the chance of having a boy increases if a boy has already been born.

i.e after a couple has a baby boy its no longer 50-50 for the next child

http://en.wikipedia.org/wiki/Gambler%27s_fallacy

Din Taylor · 04-08-2010 12:57AM

pH wrote: »

No. Even the "at least" wording is at best ambiguous, because as the question in the OP is worded it appears the information is being offered to you, you haven't been made aware of any explicit rule by which the "offerer" decided to give you the information.

For example, I toss two coins, keep one hidden, see the other one is a head and tell you in the form "I have tossed at least on head."

Everyone (I hope) can agree that the probability of the other coin being a head or tail is indeed 50/50.

This seems identical in wording to the boy girl question 1 - you've told me the sex of one of your children without specifying why you told me (and under what conditions you would have said nothing)

Note - I'm not disagreeing that you can get the question into the form where the answer is 1/3 - however to do so you need an explicit questioning step - for most normal readings of the question as asked by the OP (even including the words "at least") the correct answer is 1/2.

Not seeing the parallell with your coins example there. You're effectively tossing one coin there. It seems to me that it would only be parallel if you didn't tell the person deciding the probability which coin, that is the order, that is heads. From what you say it seems that the person knows exactly which coin is heads and the only possibilities (let the shown coin be the first coin, say) are HT and HH. And so the probabilities are 50/50.

Nowhere in the OP does it say that the first child or the second child is a boy. I think that this would need to be specified if this was the case and so the only sensible way that I can interpret it is if either the first child is a boy or the second child or both - or both at least (no pun intended:D) as long as he/she says "at least".

Maybe it's just because I've been hearing this one for years

but I still don't have too much of a problem with the wording. I'm probably just being lazy though and I'm known for being concise with my wording.

castie · 04-08-2010 10:17AM

ZorbaTehZ wrote: »

http://en.wikipedia.org/wiki/Gambler%27s_fallacy

I was more going on genetics but I understand what you thought I meant

pH · 04-08-2010 11:25AM

Din Taylor wrote: »

Not seeing the parallell with your coins example there. You're effectively tossing one coin there. It seems to me that it would only be parallel if you didn't tell the person deciding the probability which coin, that is the order, that is heads. From what you say it seems that the person knows exactly which coin is heads and the only possibilities (let the shown coin be the first coin, say) are HT and HH. And so the probabilities are 50/50.

Nowhere in the OP does it say that the first child or the second child is a boy. I think that this would need to be specified if this was the case and so the only sensible way that I can interpret it is if either the first child is a boy or the second child or both - or both at least (no pun intended:D) as long as he/she says "at least".

Maybe it's just because I've been hearing this one for years but I still don't have too much of a problem with the wording. I'm probably just being lazy though and I'm known for being concise with my wording.

Right, so let's be specific then, answer this one, take the text below as is, don't add any assumptions about any other questions or knowledge that I may have asked or have.

Mrs Smith is a neighbour of mine, I know she has two children, I've no idea of their ages or sex. One day I meet Mrs Smith and she has a kid with her, "This is my 8 year old son" she says.

Based on the facts stated above, the fact that you don't know if the son you're meeting is the oldest, youngest, fattest, cleverest or anything else, and how the information was obtained, all you know for certain is that she has 2 children , and at least one is a boy ... What is the probability of the other child being a boy/girl?

Din Taylor · 04-08-2010 03:38PM

pH wrote: »

Right, so let's be specific then, answer this one, take the text below as is, don't add any assumptions about any other questions or knowledge that I may have asked or have.

Mrs Smith is a neighbour of mine, I know she has two children, I've no idea of their ages or sex. One day I meet Mrs Smith and she has a kid with her, "This is my 8 year old son" she says.

Based on the facts stated above, the fact that you don't know if the son you're meeting is the oldest, youngest, fattest, cleverest or anything else, and how the information was obtained, all you know for certain is that she has 2 children , and at least one is a boy ... What is the probability of the other child being a boy/girl?

Right let's nip this one in the bud shall we.

The analysis that I think the OP is pertaining to:
A couple have two children and it is equally likely likely that each of these children is a boy or a girl. What is the probability one of the children is a girl on the condition that at least one of the children is a boy? This is the way that a pedantic and theoretical probability academic would state the problem. The convention in probablity is that the order is not important unless it is explicitly stated in the problem i.e. if the order was important "at least one of" would be replaced with "the first/second of which".

The first sentence says that there are four equally likely possibilities BB GG BG GB. The condition strikes out the GG so the only possibilities available are now BB BG GB. These are distinct and are all equally likely as stated at the start.

Now, can the OP be modelled by this analysis? The answer, to me anyway, is yes and no.

Yes - S/He is essentially describing an experiment. This an open internet forum so there is a certain amount of "artistic licence" with how the experiment is stated. I, for one, have no real problems with it.

No - You could first attack the "at least" one part of the model:
Does "one of them is a boy" mean either/or or both are boys or either/or but NOT both are boys. I remember arguing for over an hour with a professor about this - a first in a final mathematical economics module was at stake here:).

More easily though is to attack the "equally part" of the model:
This is a very important assumption to the problem.
In the experiment we could have:
-Hermaphrodites
-Transexuals
-Couples determined to have a boy/girl and will keep on having children until they get one. Are they having any more or are they finished.
Also is every newborn child equally likely to be a boy or a girl? I remain to be convinced of this een after a mathematical biologist spend a long time trying to convince me.
Each of these would refute the equally likely part of the model.
I'm sure smarter people than me could come up with better attacks of this part as well.

I don't think it is reasonable to attack the order part as I've already said that if the order was important it would be explicitly stated in the model AND the experiment.

HOWEVER no model describes an experiment perfectly. I think it was George Box that said
"All models are wrong, but some are useful"

I think this model comes under the useful category here. I'll need to have a better think to see if it is suitable for your experiment.

Finally, I think that there is a whole other thread on the suitability of models to describe a particular experiment. Another interesting problem/esperiment where this issue is debated vociferously is the Monty Hall problem http://en.wikipedia.org/wiki/Monty_Hall_problem.

pH · 04-08-2010 03:47PM

Din Taylor wrote: »

Right let's nip this one in the bud shall we.

The analysis that I think the OP is pertaining to:
A couple have two children and it is equally likely likely that each of these children is a boy or a girl. What is the probability one of the children is a girl on the condition that at least one of the children is a boy? This is the way that a pedantic and theoretical probability academic would state the problem. The convention in probablity is that the order is not important unless it is explicitly stated in the problem i.e. if the order was important "at least one of" would be replaced with "the first/second of which".

The first sentence says that there are four equally likely possibilities BB GG BG GB. The condition strikes out the GG so the only possibilities available are now BB BG GB. These are distinct and are all equally likely as stated at the start.

Now, can the OP be modelled by this analysis? The answer, to me anyway, is yes and no.

Yes - S/He is essentially describing an experiment. This an open internet forum so there is a certain amount of "artistic licence" with how the experiment is stated. I, for one, have no real problems with it.

No - You could first attack the "at least" one part of the model:
Does "one of them is a boy" mean either/or or both are boys or either/or but NOT both are boys. I remember arguing for over an hour with a professor about this - a first in a final mathematical economics module was at stake here:).

More easily though is to attack the "equally part" of the model:
This is a very important assumption to the problem.
In the experiment we could have:
-Hermaphrodites
-Transexuals
-Couples determined to have a boy/girl and will keep on having children until they get one. Are they having any more or are they finished.
Also is every newborn child equally likely to be a boy or a girl? I remain to be convinced of this een after a mathematical biologist spend a long time trying to convince me.
Each of these would refute the equally likely part of the model.
I'm sure smarter people than me could come up with better attacks of this part as well.

I don't think it is reasonable to attack the order part as I've already said that if the order was important it would be explicitly stated in the model AND the experiment.

HOWEVER no model describes an experiment perfectly. I think it was George Box that said
"All models are wrong, but some are useful"

I think this model comes under the useful category here. I'll need to have a better think to see if it is suitable for your experiment.

Finally, I think that there is a whole other thread on the suitability of models to describe a particular experiment. Another interesting problem/esperiment where this issue is debated vociferously is the Monty Hall problem http://en.wikipedia.org/wiki/Monty_Hall_problem.

I gave a simple question in my last post, did you answer it somewhere that I missed? If not why?

Din Taylor · 04-08-2010 03:59PM

pH wrote: »

I gave a simple question in my last post, did you answer it somewhere that I missed? If not why?

I said that I need a better think. If you just said you saw her with a boy and she said it was her son I would have no problem with modelling it as above as you still no nothing about the other child and have the BG GB BB possibilities.
This is ignoring the "non-equally likely" scenarios I mentioned.

The thing that are throwing me a little is the fact that you gave his age.

pH · 04-08-2010 07:05PM

Din Taylor wrote: »

I said that I need a better think. If you just said you saw her with a boy and she said it was her son I would have no problem with modelling it as above as you still no nothing about the other child and have the BG GB BB possibilities.
This is ignoring the "non-equally likely" scenarios I mentioned.

The thing that are throwing me a little is the fact that you gave his age.

OK fine, forget about the age, that was just a slight dig at the 2nd question giving the child's birth date, and was not intended to influence this question which was intended as a restating of the more simple original question.

I know Mrs Smith has 2 children, although I don't know the sex of either. One day by chance I meet Mrs Smith with a child, who she identifies as her son. What's the probability that her other child is a boy?

What I'm really interested in here is if anyone thinks the answer to the above is anything but 1/2 ?

Because you could start .. "before I knew anything she had either BB, GG, BG or GB, now I know it's not GG, therefore it's either BB, GB or BG each being equally likely therefore the probability of her having 2 boys is only 1/3"

c-note · 04-08-2010 07:36PM

pH wrote: »

Right, so let's be specific then, answer this one, take the text below as is, don't add any assumptions about any other questions or knowledge that I may have asked or have.

Mrs Smith is a neighbour of mine, I know she has two children, I've no idea of their ages or sex. One day I meet Mrs Smith and she has a kid with her, "This is my 8 year old son" she says.

Based on the facts stated above, the fact that you don't know if the son you're meeting is the oldest, youngest, fattest, cleverest or anything else, and how the information was obtained, all you know for certain is that she has 2 children , and at least one is a boy ... What is the probability of the other child being a boy/girl?

i believe (correct me if i'm wrong) the prob of the other child being a boy is 50/50, and heres why

you meet your neighbour on the street with a child.
it could be one of eight children from the 4 possible combinations
BB-BG-GB-GG
she then tells you its her 8yr old boy, it could now be one of four children
BB-BG-GB

in the BB situation it could be older boy
in the BB situation if could be the younger boy
in the BG situation it would be the younger boy
in the GB situation it would be the older boy

you are in one of the four situations outlined above,
in two of thoose situations the other child is a boy.

elementary · 05-08-2010 12:53AM

pH wrote: »

I know Mrs Smith has 2 children, although I don't know the sex of either. One day by chance I meet Mrs Smith with a child, who she identifies as her son. What's the probability that her other child is a boy?

What I'm really interested in here is if anyone thinks the answer to the above is anything but 1/2 ?

Yes, I make it 1/3, using the reasoning you presented.

Generalise the problem a bit: suppose you know she's the woman with seven kids, and you meet her one day with six boys. How many seven-kid families with 6 boys contain 7 boys - half of them?

To clarify it further, and remove the psychological effects from the problem, consider the problem as stated here:
You have a random binary string of length seven. You learn that six of the binary digits are 1s. What is the probability that the final binary digit is 1?

Clearly, as presented here, you have 128 such strings possible, all equally likely, before being given the information that 6 of the bits are 1s. The odds of 7 1s occuring is 1/128. With the extra information, there are now eight strings possible from the original 128, one of which is seven 1s, but are they all still equally likely?

It would be nice if the string with seven 1s were seven times more likely than the others. However, it's a trick of semantics and psychology rather than counting.

pH · 05-08-2010 01:26PM

elementary wrote: »

Yes, I make it 1/3, using the reasoning you presented.

Well the answer is 1/2.

Generalise the problem a bit: suppose you know she's the woman with seven kids, and you meet her one day with six boys. How many seven-kid families with 6 boys contain 7 boys - half of them?

Yes, if you 'meet' her with 6 random children and they all happen to be boys then the chance that her seventh child is a boy is still 1/2.

To clarify it further, and remove the psychological effects from the problem, consider the problem as stated here:
You have a random binary string of length seven. You learn that six of the binary digits are 1s. What is the probability that the final binary digit is 1?

Clearly, as presented here, you have 128 such strings possible, all equally likely, before being given the information that 6 of the bits are 1s. The odds of 7 1s occuring is 1/128. With the extra information, there are now eight strings possible from the original 128, one of which is seven 1s, but are they all still equally likely?

It would be nice if the string with seven 1s were seven times more likely than the others. However, it's a trick of semantics and psychology rather than counting.

And again here we have an ambiguous statement of the problem, making it impossible to determine the answer, because you have not specified how we learnt that 6 bits are 1s.

To make this absolutely clear there are 2 possible scenarios.

A) You get a computer to generate random 7 bit strings and check them for having "at least 6 1s" and throw the ones that pass into a set. And I agree for elements in this set the probability of the last bit being a 1 is not 50/50. This is equivalent to having a questioning step in the original 2 child problem - "Do you have at least 1 boy?" - This scenario has an explicit "Do you have at least 6 1s?" step.

You have a random 7 digit binary string, and you view 6 of the bits (you choose which ones to look at or pick them at random it doesn't matter). To your surprise you see that all the bits you've uncovered are 1s - NOW THE PROBABILITY THAT THE LAST BIT IS A 1 IS STILL 50/50 - this is equivalent to the way I stated the boy/girl problem above - you haven't asked if they have "at least one boy" you merely came across one of the children by accident.

What you've done in your post is to take my specific case where the information has been obtained randomly, fallen back to a new ambiguous version of the problem (where how you received the information isn't specified) and from there reasoned onward to a solution that only makes sense with an explicit questioning step.

Din Taylor · 05-08-2010 06:59PM

pH wrote: »

OK fine, forget about the age, that was just a slight dig at the 2nd question giving the child's birth date, and was not intended to influence this question which was intended as a restating of the more simple original question.

I know Mrs Smith has 2 children, although I don't know the sex of either. One day by chance I meet Mrs Smith with a child, who she identifies as her son. What's the probability that her other child is a boy?

What I'm really interested in here is if anyone thinks the answer to the above is anything but 1/2 ?

Because you could start .. "before I knew anything she had either BB, GG, BG or GB, now I know it's not GG, therefore it's either BB, GB or BG each being equally likely therefore the probability of her having 2 boys is only 1/3"

Righto I think we're getting somewhere like seeing eye-to-eye now.

Please first accept my apologies as I thought you were arguing about either the order of the children that is the boy or the fact that events were in some way independent. Must stop being lazy;)

I am assuming that you are arguing about how the boy is selected?

So some smart alec doesn't mention hermaphrodites or the like agan assume from now on that each child is equally likely to either a boy or a girl.

There are two possible models:

Model 1: A family with two children is selected (at random) they're asked (or state) that they have at least one son.
Here the original state space is derived from the family and is {BB GG BG GB}.
The fact that there is at least one boy decreases the state space to {BB GB BG}. Therefore the probability that the other child is a boy of 2/3.

Model 2: A boy from a set of two children families is selected (at random). The new state is now derived from the boy himself i.e. the possibilities are that the boy has:
An older brother
A younger brother
An older sister
A younger sister
Each of these is equally likely so the probability that the other child is a boy is 2/4 i.e. 50/50

It basically boils down to whether the family was selected first or the boy was selected on his own. The former being Model 1 and the latter being Model 2.

An easy way to decide which is which is by looking at the probability of each boy from a two child family being selected for the experiment. If each such boy is equally likely to be selected for the experiment then it is Model 2. In Model 1 a boy who has a sister is more likely to be selected than a bo with a brother (the former has a probability of 1 if his family is selected and the latter only has a proabability of 1/2 in the same circumstances).

Back to my original post in this thread. I still stand by it. The OP said "I have two children...". To me this suggests that the OP is selecting his/her family and is stating that s/he as at least one boy. Particularly as this is an open forum, we're not looking for mathemayical rigour here, and s/he gave the answer so it should be obvios that s/he is peraining to Model 1. Therefore it is not open to a Model 2 interpretation. I've an open mind on this and would be delighted if you could convince me otherwise.

When I first read the question I am replying to I immediately thought it was Model 1 as Mrs Smith's family was introduced first in the problem. However thinking about it more I think it is a bit ambiguous. It depends on how you interpret the meeting with Mrs Smith and her son. Are you meeting a boy who you know is in a two child family? Or are you learning that a two child family you know of has at least one boy. However I think you've stated in a later post that you meet the boy at random on the street so I'll say it is definitely Model 2 and so the probability that the other child is a boy is 1/2!!!

MathsManiac · 05-08-2010 07:37PM

Din Taylor wrote: »

...
Model 1: A family with two children is selected (at random) they're asked (or state) that they have at least one son.
Here the original state space is derived from the family and is {BB GG BG GB}.
The fact that there is at least one boy decreases the state space to {BB GB BG}. Therefore the probability that the other child is a boy of 2/3.
...

Still not quite seeiing eye-to-eye, I think, since you've rolled two very different scenarios into one: "They're asked" (and respond) is one thing. "They state" (unsolicited) is a very different thing, and is then subject to the objection pH raised and which I teased out in detail in post #33.

Din Taylor · 05-08-2010 08:36PM

Din Taylor wrote: »

When I first read the question I am replying to I immediately thought it was Model 1 as Mrs Smith's family was introduced first in the problem. However thinking about it more I think it is a bit ambiguous.

Right I'll scrap that part of the post. Been a while since I leanred or taught stuff so my details are a bit sketchy. The important point is that the boy is cosen randomly. Meeting the boy is akin to selecting him randomly. Even if I was stalking Mrs Smith to find out the gender of her children and subsequently met her son I could say that this is a new experiment and so it is now in the "prob = 1/2" scenario.

MathsManiac wrote: »

Still not quite seeiing eye-to-eye, I think, since you've rolled two very different scenarios into one: "They're asked" (and respond) is one thing. "They state" (unsolicited) is a very different thing, and is then subject to the objection pH raised and which I teased out in detail in post #33.

Not sure if I have. The important point is that it is the "prob = 1/2" scenario if the boy is chosen randomly. If they state that they have at least on boy a random selection of a boy?... Not sure, as I said I'm a little sketchy on the details. Will have a look at your post.

Din Taylor · 05-08-2010 08:44PM

Din Taylor wrote: »

Not sure if I have. The important point is that it is the "prob = 1/2" scenario if the boy is chosen randomly. If they state that they have at least on boy a random selection of a boy?... Not sure, as I said I'm a little sketchy on the details. Will have a look at your post.

Yeah I think you're correct. Essentially if you go to my point that if each boy, from a two child family, is equally likely to be selected then we're in the "prob = 1/2" scenario.

Now a family has selected themselves. We can say that they are the only two-children family without loss of generality. If they state they have at least one boy then (I think) this is the same as selecting one. Each boy in that family has an equal likelihood of being selected.

Think I'll have to relinquish my defence of the OP. Not a good day for me.:)

MathsManiac · 05-08-2010 08:55PM

Din Taylor wrote: »

Not a good day for me.:)

Au contraire! In my view, any day that you get a new perspective on an old problem is a good day!

pH · 05-08-2010 11:47PM

As MM outlines in post 33 the OPs phrasing is self identifying, and he examines the possibilities of who could make a statement "At least on of my children is a boy".

What I'm still not clear is how anyone could argue that 1/3 is a reasonable answer for this "self identifying" case as stated by the OP.

"I have 2 children, at least one is a boy, what's the probability that my other child is a boy".

The *only* way you can get 1/3 as an answer to this question is if you assume that all parents with both a boy and girl will make that statement - none will come up to you and say "I have 2 children, at least one is a girl ..."

BB - "I have at least one boy"
GB/BG - "I have at least one boy"
GG - "I have at least one girl"

I feel that this (even in the context of conventions of a probability puzzle) is an unreasonable assumption - as using the same assumption the answer to the next question is now plainly ludicrous.

"I have 2 children, at least one of them is a girl, what's the probability that the other child is a boy?"

Answer : ZERO !

So, I still feel that while the phrasing in the OP is ambiguous, it's needs a huge assumption to be made to get an answer of 1/3 - an assumption that no one would consider fair or reasonable when you look at the problem from the other side (my zero answer above).

I know that 1/3 is the clever and sophisticated answer, but for the phrasing used by the OP, I'm still entirely convinced it's wrong.

elementary · 08-08-2010 10:40PM

pH wrote: »

Well the answer is 1/2.

It's a bit more subtle than a simple appeal to authority - the reasoning behind the 1/2 is more interesting than the answer.

Yes, if you 'meet' her with 6 random children and they all happen to be boys then the chance that her seventh child is a boy is still 1/2.

Yes, but you are speculating on the process by which the six children are 'chosen' for this meeting. Taking the bare facts as presented, you have two pieces of information: She has seven children, and that of these seven children, six of them are boys.

And again here we have an ambiguous statement of the problem, making it impossible to determine the answer, because you have not specified how we learnt that 6 bits are 1s.

As stated above, we can determine an answer - under the reasonable working assumptions that the question is stated and interpreted unambiguously.

To make this absolutely clear there are 2 possible scenarios.

A) You get a computer to generate random 7 bit strings and check them for having "at least 6 1s" and throw the ones that pass into a set. And I agree for elements in this set the probability of the last bit being a 1 is not 50/50. This is equivalent to having a questioning step in the original 2 child problem - "Do you have at least 1 boy?" - This scenario has an explicit "Do you have at least 6 1s?" step.

You have a random 7 digit binary string, and you view 6 of the bits (you choose which ones to look at or pick them at random it doesn't matter). To your surprise you see that all the bits you've uncovered are 1s - NOW THE PROBABILITY THAT THE LAST BIT IS A 1 IS STILL 50/50 - this is equivalent to the way I stated the boy/girl problem above - you haven't asked if they have "at least one boy" you merely came across one of the children by accident.

There might even be other valid interpretations of the problem... but disregarding the uncontroversial [A] above, how do you justify the capitalised claim in ? It seems you've reinterpreted the problem to be equivalent to a sequence of fair coin flips, which would be independent events.

What you've done in your post is to take my specific case where the information has been obtained randomly, fallen back to a new ambiguous version of the problem (where how you received the information isn't specified) and from there reasoned onward to a solution that only makes sense with an explicit questioning step.

What do you mean by the information has been obtained randomly?

locum-motion · 12-08-2010 06:28PM

You can argue the mathematics of it till the cows come home, the question is a biological one, specifically genetic.
Barring some kind of fault such as hermaphroditism, the child about whom we have no information either came from an egg fertilized by a sperm carrying a Y - chromosome (boy), or came from an egg fertilized by a sperm carrying an X - chromosome (girl). The sex of the previous child is irrelevant. The Boy scenario and the Girl scenario are equally likely. The answer is 1/2.

pH · 12-08-2010 07:18PM

locum-motion wrote: »

The answer is 1/2.

Well that depends on the question!

If the mother of a 2 child family answers "yes" to the question "Is at least one of your children a boy?" then the probability that her other child is a boy is 1/3.

Rubecula · 13-08-2010 09:32AM

This problem highlights something I feel.

It shows that maths is not absolute.

Although I can follow both sides of the argument for 1/2 and 1/3 they are hardly the same thing. We have all had the same information, and yet nobody has come up with the definative answer.

Well maybe one of these is right and maybe not.....

But if a 'simple' problem like this can give two totally different and well discussed answers, what hope for complex maths?

Incidentally I still stick with my original answer of 1/2. The child not given is either a boy or a girl no matter that the first child is a boy.

Boy 50% chance Girl 50% chance ie 1/2
(Unless you are delving into the realm of genetics and biological events)

And no I am not arguing with those who say 1/3. I am saying what I believe is right, not what some mathematical fooling around states.

Incidentally for those who claim 1/3 might I ask one question.

One chance in three it is a boy, one chance in three it is a girl, what is the third option? Obviously it can not be either a boy or a girl as you have already covered those options.

pH · 13-08-2010 10:08AM

Rubecula wrote: »

This problem highlights something I feel.

It shows that maths is not absolute.

Although I can follow both sides of the argument for 1/2 and 1/3 they are hardly the same thing. We have all had the same information, and yet nobody has come up with the definative answer.

Well maybe one of these is right and maybe not.....

But if a 'simple' problem like this can give two totally different and well discussed answers, what hope for complex maths?

Incidentally I still stick with my original answer of 1/2. The child not given is either a boy or a girl no matter that the first child is a boy.

Boy 50% chance Girl 50% chance ie 1/2
(Unless you are delving into the realm of genetics and biological events)

And no I am not arguing with those who say 1/3. I am saying what I believe is right, not what some mathematical fooling around states.

Incidentally for those who claim 1/3 might I ask one question.

One chance in three it is a boy, one chance in three it is a girl, what is the third option? Obviously it can not be either a boy or a girl as you have already covered those options.

It really just shows that spoken (or written) English is not a great way of tightly and unambiguously stating a maths problem.

Anyway back to the 1/3 answer, bear with me here, follow the steps and let me know where you think it goes wrong.

1/ Imagine a room full of 100 mothers (each of 2 children)

2/ If they were picked randomly we'd expect them to be distributed as follows:
25 would have 2 boys (BB)
25 would have 2 girls (GG)
25 would have an older boy, younger girl (BG)
25 would have an older girl, younger boy (GB)

3/ or put another way:
25 would have 2 boys, 25 would have 2 girls and 50 would have a boy & a girl.

4/ We ask "Please stand if you have at least one boy" - We'd expect 75 mothers to stand - only the 25 with 2 girls will be seated.

5/ Now of the 75 standing, we can see that they're made up of 25 women with 2 boys, and 50 with a boy and a girl.

6/ Therefore if you picked one them at random to find out what her "other" child was was, the chances that it's a boy is 25/75 (1/3) while the chances it's a girl are 50/75 (2/3).

Rubecula · 15-08-2010 10:36AM

Believe me I do follow your logic pH, but look at it this way too.

No mention of age is in the question, therefore the options are:

1) the mother has a second child that is a son.
2) the mother has a second child that is a girl.

In this instance it is 1/2

As I said, although not expertly, maths can be confused if the EXACT data is not available.

Maths can be used to prove almost any scenario.

As an example of what I mean.

It is possible, though not likely that the child could be neither girl nor boy, ie a Hermaphrodite. Not common I know but bear with the idea.

We could have, BB, BG, BH, GG, GB, GH, HH, HB, HG.

If one of the children is a boy for certain, then we are left with:

BB,BG,GB,BH,HB. Which would give a possiblity of 1/5 by your method.

However the chances of a hermaphrodite are so small as to be negligable.

There is no allowance in the question for negligable occurances so it has to be taken into account by those rules.

Yet we do not include it.

The question has no referral to age of the children, so BG and GB are the same.

So we have BB, GG and BG(GB) which gives 1/3

Yet we all know a child can only be a boy or a girl (if we ignore the negligable).

So we can ignore the order the children are born in which gives us either B, or G. Which is 1/2.

Coupled with this is the biological percentages (which I don't actually know off hand) Perhaps it is more likely a girl will be born than a boy, perhaps it is the other way around. I don't know. So we give each an even chance. And again, this in biological terms rather than mathematical terms, gives us a 1/2 probability.

Biology says 1/2 and one way of doing maths gives 1/2. So being as logical as possible I also go with 1/2.

That is not saying I am right of course.

Your example was much better written than mine so in argument alone I think you may have won this one.

However forgive me if I remain unconvinced.;)

Colours · 15-08-2010 04:26PM

The way I see it, the answer 1/3 only features when you're speculating on all the different scenarios there could be and that is still arrived at based on the fact that the birth of a Girl or a Boy are equally likely and are independent of eachother.

However in the example posed in the OP and then later on by pH, the question is simply given that one of a pair of children is a boy what is the likelihood of another boy and that is still an independent likelihood of 1/2 and is regardless of if the parent already had umpteen other boys to begin with.

I think using the analogy of a coin may be easer to appreciate how one result in a flip has no bearing on the result in the next consecutive flip.

elementary · 17-08-2010 07:35PM

In response to this formulation of the problem:

I know Mrs Smith has 2 children, although I don't know the sex of either. One day by chance I meet Mrs Smith with a child, who she identifies as her son. What's the probability that her other child is a boy?

What I'm really interested in here is if anyone thinks the answer to the above is anything but 1/2 ?

One way to explain an erroneous 50-50 probability for this is that of the set of parents with two children who have at least one son, one is twice as likely to encounter a parent with two sons. However, the fallacy lies here.

Given that we have met a parent of two children, at least one of whom is a boy, it is now twice as likely that this boy has a sister rather than a brother.

tear-your-hair-out probability puzzle

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