[Closed] The boy-girl puzzle

Go and get two coins.

What are the chances of the first one being a head? 50:50
What are the chances of the second one being a head? 50:50

If you dont know any of the results then all 4 permutations have a 25% chance of happening.

After the first result is known they have a 50% chance.

Look Smee, you have a friend... 😀

By the way, you never did answer me about whether an older girl/younger boy is really the same as an older boy/younger girl...

Yes yes Smee those are independent events. Everyone agrees that, as explained 16 times already.

the first result doesn't influence the second. No one disputes that either.

But you don't know the "first result", you know [u]one[/u] of the results.
That is very different.

Try my 100 coin toss. Explain the results on my spreadsheet.
Or come up with an experiment that matches the question.

How about this then: you agreed above that before you know any results then all four coin combinations have a 25% chance.

So how about I throw the coins and I then tell you "I didn't throw two tails"
What are the odds on the other combinations now?

[i]What are the chances of the first one being a head? 50:50
What are the chances of the second one being a head? 50:50

If you dont know any of the results then all 4 permutations have a 25% chance of happening.

After the first result is known they have a 50% chance. [/i]

I can't believe you don't grasp the coin one.

Come on lads, this isn't over, we can stretch it for another day!

Smee, at some point we need to meet up and do this with coins. I'll bring 50 £1 coins and you bring 50 £1 coins.

If there's two tails, you get to keep the coins. If there's one of each, I get to keep them. If you're right, we'll both come out even.

If you're right, we'll both leave with ~£50. If the rest of the planet is right, I'll leave with ~£67 and you'll leave with ~£33.

right, final answer here
http://mathforum.org/dr.math/faq/faq.boy.girl.html

Surely smee would not dare doubt the wisdom of Dr Math!?

Come on lads, this isn't over, we can stretch it for another day!

250 replies is respectable, but I'm still hoping for my first Triple Century posting. 😀

What happens if we use 20p peices instead of £1 coins?

Would this reduce the odds to 20% or 66%?

From the Dr Math site: "We can also visualize this problem using a probability tree"

Gosh, they look familiar. Mine were prettier though.

[i]right, final answer here
http://mathforum.org/dr.math/faq/faq.boy.girl.html [/i]

But it's wrong the maths wrong, Smee said so.

I know why he's confused I was at first but I never claimed to be right just couldn't see why it wasn't 50/50 1:1 50%. Then it suddenly sunk in that's it's the lack of specify which child that increases the odds, the same way the order the lottery numbers comes out doesn't matter therefore increasing the odds.

Hummm...

I think I'm beginning to understand how the global economy became f*****.

Population too high? Not enough parents with two girls? I think I'm with you here.

I think I'm beginning to understand how the global economy became f*****.

I don't think you can blame Smee for that one.

I think I'm beginning to understand how the global economy became f*****.

I don't think you can blame Smee for that one.

I think you can if he's been teaching the banks how to do sums.

when you know that one is a h you have:
one of ht or th - not both.

if you have 1 x 10p and 1 x 2p, then quite clearly 10p heads, 2p tails is not the same as 10p tails, 2p heads. hence two different ways of getting one heads and one tails. therefore twice as likely as getting 10p heads and 2p heads. is this getting through to you?

Lets get it over 300.

It can be 50%, here's how.

What you want to know is whether "she has a girl and a boy?". You can find this out by finding out the gender of both children. As you already know the gender of 1 child you only need to know the gender of the other child. As the only unknown is the gender of 1 child where the two possible outcomes each have a probability of 50% you can determine whether "she has a girl and a boy?" with a probability of 50%.

Andy.

What you want to know is whether "she has a girl and a boy?".

Nope - that's not what the question asked.

The question is about whether her responding that she has at least one girl changes the probability that she is in the group that have one child of each sex. And it does.

If the question said [i]"Here is a woman with a daughter, what is the gender of her other child?"[/i] [u]then[/u] it would have been a straight 50:50 - but that is a very different question.

Also.. Rona is a girl's name. 😛

hi just for the next post

GrahamS is quite correct.

The way the question is asked/worded provides 2 very different answers.

Oooh reading back a few pages, quite a few people are correct apologies.

Yeah, it's pretty much just been Smee that hasn't seen the light so far.
Everyone else has got there - as I'm sure Rona will 🙂

Funny how this is Rona's first ever post.

Drac, been here a very long time.

Graham,

"What you want to know is whether "she has a girl and a boy?".

Nope - that's not what the question asked."

No, indeed but that's why probability is only probability, if you ask a different but relevant question to get the information you need you may get a different probability, i.e. as pointed out before the odds that the remaining child is a boy are 50:50 by simply asking that question the probability of her having a girl and a boy are 50:50.

Andy.

not too many to go now. . .

[i]Drac, been here a very long time.[/i]

Took you a month to post though.

It's been pretty dull.

Sorry I'm being sceptical spent too long on many forums and always cautious when a dormant member suddenly starts posting on dying threads.

Well yes, probabilities are based on the information you have. if you ask different questions then you gain different information and you get different probabilities.

You could have simply asked her "Do you have a son and a daughter?" and, assuming that she is telling the truth then the probability is 100% that she does. That doesn't mean that if you'd asked her the original question instead that the probability would be anything other than 66%.

[img] [/img]

Yes, but why would you ask her the original question? If you were speaking to her you'd ask a much more sensible question which would confirm the probability is 50% :-).

Andy.

I've got three kids and they are all boys, does that help at all?

[i]Yes, but why would you ask her the original question? [/i]

For the interests of science.

Yes, but why would you ask her the original question?

So you can find out who the less intelligent people are.

42.

Smee! You're back. There is hope for this thread yet.

Go on, give me an answer to some of the questions put to you.
I'm particularly interested to see how you answer the "I didn't throw two tails" scenario or flatboys 2p and 10p.

Ever heard the saying "holding the cat's arse to the sun?"

No, but I'm guessing this is where you claim you were trolling all along and you were simply pretending to be a moron.

Instead of probability use logic. 2 possible outcomes 50:50 chance.

We're nearly there!

If you are stood in front of this hypothetical woman who you know has one girl and another child and think to yourself 'what are the odds she has a girl and a boy' you can work through the probability logic already laid out nicely and get 66% or you can consider that as determining the gender of the other child determines whether she has a girl and a boy and as the probability of the other child being a boy is 50% that means the probability of her having a girl and a boy is also 50%. Two different ways to view the problem and two different answers. 🙂

Andy.

Instead of probability use logic.

That is rather the point of this entire exercise. The maths reveals that the apparently obvious and intuitive logic is incorrect. In mathematics there is truth, and all that.

This isn't some kind of stupid trick. I'm not using maths to show that the sky is green.
The actual result [i]is[/i] genuinely 66% as my spreadsheet and the numerous other empirical tests show.
The maths leads us to the correct logical reasoning and explains why this is so.

"Logic" tells you that the plane won't take off and that switching doors doesn't make you less likely to get a goat.

Rona/Andy... but that's the subtlety and why the question is worded in such a way, and if you do any of the various experiments the results will bare that out.

The problem with your second approach is that you are using information that you do not have. You know there are 2 children, and you know that one of them is a girl, but you don't know which of the two children that actually is. As soon as you specify one child as being one sex or the other you are using information that you don't have.

For example, say you have a cat and a dog, then I ask you 'Is at least one of your pets female?' and you answer 'Yes".

Now, you can either have a female cat and a male dog, a female dog and a male cat, or you have a female cat and female dog, but you don't know which of those three it could be, and they are all distinct choices.

This is the limit of the information that you know in the question above, it's just that you assume that you know more.

As soon as you know the sex of one of the pets for certain, the sex of the other then becomes 50:50, and that is where the mistake lies, you just don't know the sex of either child in the original puzzle, all you know that one of them is a girl. The girl could be child A, or it could be child B, or it could be both of them, but you just don't know.

"As soon as you know the sex of one of the pets for certain, the sex of the other then becomes 50:50, and that is where the mistake lies, you just don't know the sex of either child in the original puzzle, all you know that one of them is a girl. The girl could be child A, or it could be child B, or it could be both of them, but you just don't know."

That's irrelevant and that's the problem, the original question requires you to know that they aren't both boys but at the same time not to know that one of them is a girl - something which can't be done when you say one of them is a girl. As soon as you know one of them is a girl you can determine whether or not the women has and boy and a girl or a girl and a girl by knowing the answer to 'what is the gender of the other child' as this is a question with a 50:50 outcome the probability of her having a boy as the other child becomes a 50:50 coutcome etc. . .

It isn't possible to word the original question in a subtle enough way for the above logic not to work, i.e. as the answer to my question above will correctly tell you whether the woman has two girls or a boy and a girl the probability of the possible outcomes is the probability that she has a girl and a boy. If you have a stituation where you know that the kids weren't boy-boy but were none the wiser beyond that the probability of boy-girl (or girl-boy) becomes 66%.

Andy.

Apart from the fact that there are twice as many families with a boy and a girl as there are families with a girl and a girl...

So, in order to get the answer of 50%, you need to know more information than is in the original question.

I'll point you again at my example of the cat and the dog. If you know at least one of the pets is female then you have:-

female cat/male dog
male cat/female dog
female cat/female dog

What are the odds that the other pet is a male?

How is this any different to the question with the children?

I just checked the front page. This was answered correctly with the second reply.

How did it get to 8 pages?

"How is this any different to the question with the children?"

It's different because the order of the children is not an issue in the original question whereas the species is in yours, you don't need to determine the order to know whether she has boy-girl or girl-girl. All you need to do in this example is determine the gender of the unknown child and this tells you whether she has boy-girl or girl-girl.

The problem is the information in the question, not that the probability of boy-girl vs either same gender combo isn't 2:1 a point you seem to be missing.

Andy.

because its good fun!

Why is order not important with the children? Why does age not allow us to differentiate the two children, as species does for the pets?

With your thinking, does older boy equal younger boy?

The problem is that you are reading too much information into the original question which isn't there. You are trying to determine the sex of the unknown child, but you don't know which child you know the sex of, just as you don't know what species of pet the female is.

Okay, if, as you are saying, the chance is 50%, then that means the chances of a girl/girl are equal to that of boy/girl. Is that correct?

Rona: incredibly you still seem to be missing the point.
This isn't a fanciful whimsy. If you do this for real (or any of the equivalent experiments suggested) then you will find that 66% of the women who say they have at least one girl will have a girl and a boy.

At the risk of repeating myself: take a population of 100 women who all have two kids. The distribution will be 25 have two boys, 25 have two girls and 50 have mixed gender. Yes?

Ask every woman in that population the question and 75 of them will say yes. That 75, consists of the 50 with mixed gender and the 25 with all girls. A ratio of 50:25, or 66%

That is the actual answer by manually performing the experiment.

So either you need to prove that 2:1 is 50%, as Smee attempted to do, or you need to explain how 50 out of 75 is 50%

Neither of you appear to be able to read an entire answer.

The probability of mixed sex out of boy-girl, girl-boy and girl-girl is clearly 66%, the problem is the wording of the original question.

Which statement do you think is incorrect:
In the original question there is a child of unknown gender.
The probability of an unknown child being a boy or a girl is 50%.
The gender of the unknown child determines whether the woman has mixed or same sex children.

If it's the middle one then that means the gender of the female child influences the gender of the unknown child.

Andy.

Getting so close to 300 now.

indeed!

Not long now.

Chance of Smee being honest and saying "Damn! It is 66%" and not claim he was a troll?

I have read your posts, and just don't agree with you.

It is the middle one that is not correct in this circumstance, and that's the subtlety.

You have said, the probability of an unknown child being boy or girl is 50%.

This is true for a specific child.

But the question, as originally asked, does not allow you to specify either child as being the one you know the sex of. As soon as you tie that down you get the 50% answer, but you are unable to tie that down.

If you are correct, then you should be able to answer my pet question as being 50% without any more information, as it gives you exactly the same information as you have in this question.

You know these things:-

There are 2 children = there are 2 pets
One child is older than the other = one is a cat and one is a dog
At least one is a girl = at least one is a female

They are directly analogous to one another.

[img] [/img]

[img] [/img]

So how long have you been hanging around refreshing waiting to post that then Drac? 😉

I was lucky was just about to disappear when I finished another thread, typed 300 to reserve it then found a pic.

funknick, GrahamS, let's play the Race card.

Seeing as the cat/dog situation is not fully understood, let's say the 2 children had different fathers, of different Race to each other, but we don't know which was which in terms of order or sex.

This way we don't know whether the girl that we know exists, was first or 2nd out of the womb, or whether her Dad was Afro-Caribbean or Caucasian either. And the child that we don't know whether is male or female, well we don't know whether that one was 1st or 2nd out either (still), and now we don't know which Race its father was either.

So, in this case, Girl/Boy cannot, in any way, even by those so stubbornly stuck in their wildest trolling dreams, be confused with Boy/Girl, as potentially they are not only 1st/2nd or 2nd/1st out of the womb as well, but also potentially Afro-Caribbean/Caucasian origin, or Caucasian/Afro-Caribbean too! Stick that in your pipe and smoke it Smee.

And well done people, 300 posts. Who says maths is boring! 😉

Well done mboy your question has sufficient info to work, Graham the problem with your question is that it leaves a single unknown which you need to be a double unknown thus allowing a logical approach to determine the outcome with a (incorrect in reality but correct for your question) probability of 50%. Your question works if worded as 'what are the odds she had a daughter first'. That way there isn't a logical approach that can correctly get the incorrect answer. Anyway that helped get it over 300 so that'll do!

🙂

Andy.

Andy... but it doesn't leave a single unknown, you are just making that assumption.

The question put by mboy, gives you three unknowns, but that third one is irrelevant. The original difference, birth order, is still there even without the race issue. You could just as well do it where one child has ginger and one child has blonde.

If you actually do the experiment as GrahamS describes, which involves asking each family the question 'Is at least one of your children a girl?' then you get the 66% result as described.

If what you are saying is correct, that there is only a single unknown in the original question, then you can tell me which of the childern, older or younger, is definitely a girl.

Two kids, you know the sex of one. How many does that leave whose sex you don't know?

Aaaah, but Smee, you don't know which of the two children is that girl.

If what you and Rona/Andy are saying is correct, then can you please demonstrate by use of an experiment, that can be performed, using only the information given in the first question, and not using any assumed information.

Smee - Member

Two kids, you know the sex of one. How many does that leave whose sex you don't know?

at first i thought you were understandably taken in by the problem. then i thought you were taking the piss. now it's clear you're just not that bright...

The wording in my question is pretty much the standard way to ask this question in a unambiguous manner. It matches very closely with the wording on the Dr Math website that someone linked to earlier.

Rona: there are a million things you don't know about the children. Hair colour, skin tone, favourite ice cream, shoe size... it doesn't make any difference at all.

You are both still dodging the fact that the actual result comes out as 66%

I think your issue may be that you are selecting a woman from a population of women who have one daughter.
That is pre-screening, it is not what the question says and it will alter your result.

I find it very hard to believe this isn't trolling now. This answer has been answered so well, by so many people, in so many interesting and intuitive ways that if the same amount of mental effort and inventiveness had been applied to the palestine/israel conflict the entire gaza strip would now be filled with people eternally holding hands and singing 'we are the world'. It ain't ever gonna be resolved now, the people who could be convinced either way have already been convinced.

[img] [/img]

😆

It's like a pictoral representation of this thread! In that it's being repeated several times on one page.

Okay, here is a related but even simpler example that may help you understand:

[i]Suppose you have a deck of 60 cards, consisting of:

• 20 black cards that are black on both sides,
• 20 white cards that are white on both sides, and
• 20 mixed card that are black on one side and white on the other.

The cards are shuffled. One is pulled out of the deck at random, and placed on a table so that you can only see one side (at random). The side facing up is black. What is the probability that the other side is also black?[/i]

I will go 50 / 50 ...what are the odds i mean probability I mean ratio of us getting to 10 pages now then

Impossible, for a start I'm refusing to submit any more posts.

Me too...

Bugger..

😉

That's all very well funky, but when I say it, I mean it.

The cards are, of course, 66% too.

The problem is different, but the reasoning is similar in a lot of ways. Information that you gain (the black side of the card) lets you establish which subset of the population it belongs to.

Two kids, you know the sex of one. How many does that leave whose sex you don't know?

You know the sex of one.

There is only one child whose sex you do not know, but that child is one of several [i]possible[/i] children.

The cards example hopefully also demonstrates that having two possible outcomes doesn't mean that the odds are 50/50.

Otherwise there is no point to statistics at all:
"Will it rain tomorrow? Well either it will or it won't, so it must be 50/50."
"Either I win the lottery or I don't. 50/50"

You know one of the children is a girl, [u]but you don't know which one[/u].