Probability riddle

sd123

Here's a riddle that involved probability and I heard the answer but I'm not really sure about it....

Random woman meets you and tells you she has two children, at least one of which is female, What is the liklihood of the other child being male?

Some help and/or explanation of the answer would be great. thanks:)

Thomas_S_Hunterson

Maybe I'm missing something but it should be 0.5

The first one is definitely a female.

The second could be either a male or a female, each assumed to be equally likely to be conceived, also assuming both to be independant (i.e. neglecting such occurances as twins etc, which may reduce the probability slightly.)

/edit sorry, thinking about it again, i think it's 2/3

Ignoring what the woman says, the probabilities of the different combintions are as follows:
boy+boy = (0.5)^2 = 0.25
girl+girl = (0.5)^2 = 0.25
boy+girl = 2 * (0.5)^2 = 0.5

We can discount the first case as the woman tells us that at least one of teh children is a girl, this leaves us with the remaining probabilities:
girl+girl = .333
boy+girl = .667


I'm open to correction though

sd123

Sean_K wrote: »

Maybe I'm missing something but it should be 0.5

The first one is definitely a female.

The second could be either a male or a female, each assumed to be equally likely to be conceived, also assuming both to be independant (i.e. neglecting such occurances as twins etc, which may reduce the probability slightly.)

Of course that's what I thought, but apparantly you have to look at the original probabilities. I'm not saying it's right but it was explained to me like so....

knowing that the first girl is female can lead us to different possibilties for the second child, in order of birth.

girl, older sister
girl, younger sister
girl, older/younger brother ( which is unrelated to the first girl so is just counted as one boy)

So there is twice as much chance as it being a girl than a boy so it's 67-33% prob, I can definitely see the flaws in it though....

Pherekydes

The sample space for two children is:

{BB,GG,GB,BG}

The sample space for this particular woman is:

{GG,GB,BG}

Knowing one child is female means that the probability of the other child being male is indeed 2/3.

Probability can be very counter-intuitive.

Fremen

This confused the hell out of me when I first heard about it. The probability of (two girls) given (at least one girl) is one third, as slow coach illustrated.

However, the probability of (two girls) given (the eldest is a girl) is one half. Why should the age of the girl affect the probability? Similarly, the odds change if we know that the girl is the tallest, smartest, or if we have any other kind of informatiohn which relates her to her sibling.

Really, I think the only way to deal with this problem is to look at the space of possibe events and decide which ones are still allowed given the information we have.

Can anyone guess what the probability of having at least one boy is, given that if there is a girl, she has the blondest hair?

2Scoops

It makes perfect sense to me: if you already know that one sibling, in a starting group of two, is female, it is reasonable that the other is more likely to be male, assuming a probability of .05 for a given sex (surprisingly, it isn't a perfect .5 in many populations!).

2Scoops

Fremen wrote: »

Can anyone guess what the probability of having at least one boy is, given that if there is a girl, she has the blondest hair?

100%

thetaxman

Has to be 0.5, assuming that the woman is concieving throught the use of normal methods. The first child has no effect on the second, although some biology student may know otherwise.

2Scoops

thetaxman wrote: »

Has to be 0.5, assuming that the woman is concieving throught the use of normal methods. The first child has no effect on the second, although some biology student may know otherwise.

But the starting set is 2 children, then you are given information about one child. The probability of any one child being a boy is indeed (approximately) .5, but in a starting population of 2 where one is a girl, the probability of the other child being a boy is .67. Make sense?

hivizman

I'd go back to the original probability space:

{BB, BG, GB, GG}

The probability of at least one boy (assuming that the probability of any child being a boy is 0.5, and the gender of any child is independent of the gender of any other child) is 0.75. But suppose that the order of the children in the probability space is significant - let's assume that the child with blonder hair is the first child in each pair and the child with less blond hair is the second child in each pair (and also that hair colour is independent of gender, so girls don't have a higher probability of being blonde than boys).

If we specify that the girl referred to by the mother has blonder hair than the other child (whatever its gender) then the cases BB and BG are not possible, the probability space becomes {GB, GG}, so the probability of a boy is 0.5.

But in this case, we are told that IF there is a girl, she has the blondest hair.

Now, does the condition "if there is a girl, she has the blondest hair" lead to the elimination of any of the original four possibilities? It may depend on how you interpret this condition - I would interpret it as "if one of the children is a girl, then that child has the blondest hair". On this interpretation, the case of two boys (BB) is still possible, because, if neither child is a girl, the condition is simply irrelevant. The case of GB is still possible, because there is a girl, and she is the blondest. The case of two girls (GG) is still possible (unless you read the word 'she' in the condition as implying that there can be only one girl). However, the case of BG is not possible, because there is a girl, but she is not the blondest (this is the boy). So the condition changes the probability space to {BB, GB, GG}, and the probability of at least one boy is 2/3 (0.67).

2Scoops

2Scoops wrote: »

100%

Oops - WRONG! 75%

Fremen

Yeah, I originally intended it as hiviz reasoned it out (i.e. 2/3), but I guess I should have been a bit clearer. There is some ambiguity in the phrasing of the question.

thetaxman

The answer is not 75%. It is 50%. Independent events. The first event has no effect on the second.{B, G} 1ST CHILD{B,G} 2ND CHILDIt is as simple as that.

2Scoops

thetaxman wrote: »

The answer is not 75%. It is 50%. Independent events. The first event has no effect on the second.{B, G} 1ST CHILD{B,G} 2ND CHILDIt is as simple as that.

So, by that rationale, in all families with 2 children, having 2 girls is equally as likely as having 1 boy and 1 girl - do you agree with that?

The census certainly doesn't.

PS: the 75% one is the probability that there will be at least one boy in a family with 2 children. It's 67% when it's the likelihood of the 2nd child being a boy in a family of 2 children when you know one child is a girl.

thetaxman

As i said in my original post, there are other factors affecting child sex. These factors include genetic makeup of the parents, hereditary traits and so on. Therefore this is not a purely mathematical problem. I know many families where there are three girls and no boys etc. I would not call this a mathematical coincidence, rather a genetic coincidence.You need to consider the inputs in this problem before the solution is considered. If the probability of a childs sex is 50/50 then the result of this problem is 0.5, the sex of the first child has no affect on the second.

2Scoops

thetaxman wrote: »

As i said in my original post, there are other factors affecting child sex. These factors include genetic makeup of the parents, hereditary traits and so on. Therefore this is not a purely mathematical problem. I know many families where there are three girls and no boys etc. I would not all this a mathematical coincidence, rather a genetic coincidence.You need to consider the inputs in this problem before the solution is considered. If the probability of a childs sex is 50/50 then the result of this problem is 0.5, the sex of the first child has no affect on the second.

Well, the genetic make up or the parents has a rather small effect on the sex of the child - other than the whole XY chromosome thing.

Assume, for a moment, that this is indeed a purely mathematical problem. A family can still have 3 children of the same sex. Probability does not say it is impossible - only that it is less likely than having some combination of boys and girls. The evidence of this is all around you.

You're looking at this the wrong way, I think. Each child has a roughly 50% chance of developing into either sex, this we can agree on. The sex of the first child does not alter the sex of the next child born. BUT the likelihood of someone having lots of children of the same sex is LESS LIKELY than having a mixture BECAUSE each child has a roughly 50% chance of developing into either sex

You are correct in what you're saying - but you're answering a different question to everyone else. The sex of the child has no bearing on the sex of the next child at birth. But we are dealing with a set of children, already born, and are trying to work out the probability of sex distribution based on available information.

Mark200

2Scoops wrote: »

But the starting set is 2 children, then you are given information about one child. The probability of any one child being a boy is indeed (approximately) .5, but in a starting population of 2 where one is a girl, the probability of the other child being a boy is .67. Make sense?

No because you know for a fact that she had one daughter. So if you want to do it in a string set thingy, then you put the probablility of the first child being a girl as 1, because you know that it definately is. Which will have no effect on the outcome because you will multiply it by the odds of the other being male.

And also, no point getting into biology....it's a maths problem, not a biology one.

thetaxman

2Scoops, I my opinion people are making this much more complicated than need be. If as Mark200 suggests this is to be treated as a maths problem, then the answer is 0.5 than the child will be male. I cannot see any other answer.

Thomas_S_Hunterson

thetaxman wrote: »

2Scoops, I my opinion people are making this much more complicated than need be. If as Mark200 suggests this is to be treated as a maths problem, then the answer is 0.5 than the child will be male. I cannot see any other answer.

The probability of a boy being born in a given birth is 0.5

The probability of a girl being born in a given birth is 0.5

The answer to the question posed is most certainly 2/3

This question caught me out at first in the 50/50 trap, and, as was mentioned above, statistics can be counter intuitive sometimes, but for a start make sure you're dealing with the same question as everyone else;)

Pherekydes

thetaxman wrote: »

2Scoops, I my opinion people are making this much more complicated than need be. If as Mark200 suggests this is to be treated as a maths problem, then the answer is 0.5 than the child will be male. I cannot see any other answer.

The sample space for two children is:

{BB,GG,GB,BG}

Can you agree with that?

The sample space for this particular woman is:

{GG,GB,BG}

Can you agree with that?

Once she tells you one child is a girl you can cross off one G and you are left with

{G,B,B}

Can you agree with that?

It's quite simple, not complicated as you are making out.

Put in plain English:
A random woman you know nothing about tells you she has 2 children. You would expect, by the law of averages, that she would have one boy and one girl. Yes? Now she tells you one is a girl. Would you not be inclined to expect the other child to be a boy?

P.S. You should base your expectation on the law of averages, not on one family that has 3 girls or 5 girls.

P.P.S. 2/3 is the right answer; it isn't a matter of opinion.

thetaxman

Fundamentally flawed argument. These events are independent. How does the fact that she has had a girl increase the chance that she will have a boy for second child. Answer this, are the births independent? If you answer yes, the solution to this problem is 0.5. If you answer no, I have no idea where you are coming from.</p>

Pherekydes

thetaxman wrote: »

How does the fact that she has had a girl increase the chance that she will have a boy for second child.

The Law of averages. You do realise that roughly half the population is female and roughly half is male, right?

So you would expect two children to be one boy and one girl, right?

If not, what would you expect?

Do you understand what is meant by the term expectation?

thetaxman

Yes, I understand the term, you may not.You are defeating your own argument through this last post. Half the population is male, I agree. Therefore the chance of her second child being male is 0.5. If i toss a coin and get heads, will the chance of getting tails with a second toss reach 0.67? No. Answer my question, are the events independent?

Fremen

Take a look at this:

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

The events ARE independent, but that doesn't mean the information you are given doesn't change the probabilities.

2Scoops

thetaxman wrote: »

Yes, I understand the term, you may not.You are defeating your own argument through this last post. Half the population is male, I agree. Therefore the chance of her second child being male is 0.5. If i toss a coin and get heads, will the chance of getting tails with a second toss reach 0.67? No. Answer my question, are the events independent?

You're equating this situation with the gambler's fallacy but it's not what the question was! Yes, the births are independent events. They do not affect each other. But what you're arguing is that groups unbalanced for sex are equally likely to happen as those that are not - which is patently untrue!

And on that bombshell, population statistics the world over and I respectfully bow out of this discussion.

thetaxman

Again wrong. You are taking a sample size of 6.7 billion people, thats when things balance out. Two wrongs can make a right, in this case a two girl family in china may balance out a two boy familt in stoneybatter.

thetaxman

Fremen wrote: »

Take a look at this:

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

The events ARE independent, but that doesn't mean the information you are given doesn't change the probabilities.

That was no Leonardo Da Vinci who wrote that piece. He/she is totally incorrect.

2Scoops

thetaxman wrote: »

Again wrong. You are taking a sample size of 6.7 billion people, thats when things balance out. Two wrongs can make a right, in this case a two girl family in china may balance out a two boy familt in stoneybatter.

But, on average, how many 2 child families will have a mixture of boys and girls compared with exclusively boys, for example?

You're going to kick yourself when you realize that you're completely wrong.

thetaxman

To be honest I don't know, and once again thing balance on the macro level, but not necessarily on the micro. You are all incorrectly cancelling out 1/4 of the probability tree, if you look on that link, they are mistaken in moving onto conditional probability, the link of second child from first is non-existant in real life.

Fremen

Yep, the maths PhD is wrong and you're right :rolleyes:
This is beginning to look like a troll to me

thetaxman

Say for example, I have a family with two female children. This is totally out of balance with nature. You on the other hand have a family with two sons, again totally out of balance with nautre. Yet put the two togeather and total harmony returns. This untimately does boil down to a macro/micro issue in my opinion.
Troll? I bet the mathematics forum is a pure hotbed of trolls. Debate what I have said, and incidentially, you should consider my background when comparing me unfavourably to a Dr Mathematics website.