Pop quiz.
A new family has moved in next door and I’ve been told that they have two children. I have spotted a girl.

What is the chance that the other child is also a girl?

______________________________________________________________________________________________________________________________________________

btm suggested:

Suppose I toss a fair coin (probability of heads == probability of tails = 50%) twice. The probability of getting heads twice is 25%. If I know that one of the results was a head, what’s the probability that the other was a head? What’s the probability it was a tail?
This is the one sibeen posted earlier, but with coins instead of children (because coins are always either heads or tails, and if they’re fair coins, it’s 50% for either, whereas children are… complicated.)

Possible outcome probability
head, head 25%
head, tail 25%
tail, head 25%
tail, tail 25%
__________ _____
total: 100%
Now suppose that I know that one of tosses resulted in heads. What’s the probability that the other was a head? I make it 1/3 (because the “tail, tail” line of the above table is no longer a possible outcome, so there are three possibilities.) The probability of a head and a tail is then 2/3.

deevs suggested that the answer was 50%, with the caveat that I’d only spotted one child.

The Rev is suggesting that there is not enough information.

Party Pants thinks that btm is onto something but that the answer is 50% anyway.

My daughter has used a decision tree and come to the conclusion that the answer is 25%.

Me, I told her immediately that the answer was 50%.

sibeen said:

Pop quiz.
A new family has moved in next door and I’ve been told that they have two children. I have spotted a girl.

What is the chance that the other child is also a girl?

______________________________________________________________________________________________________________________________________________

btm suggested:

Suppose I toss a fair coin (probability of heads == probability of tails = 50%) twice. The probability of getting heads twice is 25%. If I know that one of the results was a head, what’s the probability that the other was a head? What’s the probability it was a tail?
This is the one sibeen posted earlier, but with coins instead of children (because coins are always either heads or tails, and if they’re fair coins, it’s 50% for either, whereas children are… complicated.)

Possible outcome probability
head, head 25%
head, tail 25%
tail, head 25%
tail, tail 25%
__________ _____
total: 100%
Now suppose that I know that one of tosses resulted in heads. What’s the probability that the other was a head? I make it 1/3 (because the “tail, tail” line of the above table is no longer a possible outcome, so there are three possibilities.) The probability of a head and a tail is then 2/3.

deevs suggested that the answer was 50%, with the caveat that I’d only spotted one child.

The Rev is suggesting that there is not enough information.

Party Pants thinks that btm is onto something but that the answer is 50% anyway.

My daughter has used a decision tree and come to the conclusion that the answer is 25%.

Me, I told her immediately that the answer was 50%.

but 50 % of which side of the argument?

The elder sprogs reasoning via diagram. Each branch is a 50/50 split:

1/2 ——G———-1/2 G

1/2 ——B

sibeen said:

The elder sprogs reasoning via diagram. Each branch is a 50/50 split:

1/2 ——G———-1/2 G

1/2 ——B

And ignore all that as I accidentally hit submit.

I’ll put this here.
sibeen said:

I’m with btm on this, knowing there is one head only takes out the tail/tail option. My daughter has been using similar logic but using a decision tree instead. If you follow her logic the answer is a 1/4.

How does your daughter get around the fact that knowing that there’s one head necessarily reduces the event space to 3 (or 2, following p_p’s argument)?

As an aside, I considered using the conditional probability formula, P(A|B) = P(A&B)/P(B), but, since tossing two heads (P(A&B) means two heads AND one head, in two tosses) is only possible if the first toss was a head (so the two events are not independent), it doesn’t apply.

This is elder sprogs decision tree using heads (girls) and tails (boys).

At the first toss there is 1/2 chance in each leg.

If a tail is flipped (lowest branch), as we know there is already a head then the next decision branch is a 1, and the total of that branch is 1 × 1/2 = 1/2.

On the upper segment we have the original 1/2 and then this branches out again with each fork producing a 1/2. The total end of each branch is 1/2 × 1/2 = 1/4.

There are two branches at the top each with a 1/4 chance, so a total of a 1/2 chance for the total branches and there is a 1/2 chance for the bottom branch.

The chance of a Head – Head is 1/4.

sibeen said:

Party Pants thinks that btm is onto something but that the answer is 50% anyway.

No, btm and I are in disagreement.

I think it is 50/50.

I was trying to use the decision tree. Start with all 4 options mapped out, and then look at the results known so far and cross out those ones that didn’t happen. Decision tree will look shitty, but here goes.

First Child —— Second Child —— Outcome
girl ——————- girl ——————- girl, girl
———————— boy —————— girl, boy

boy —————— girl ——————- boy, girl
———————— boy —————— boy, boy

So, if the girl spotted was the first child we can cross off the bottom two options and we are left only with “girl,girl or girl,boy – so 50%
If the girl spotted was the second child then cross out 2 and 4, and we are left with girl,girl or boy,girl – once again 50%

boy,girl and girl,boy are different outcomes, they don’t both come into play when we find out that one of the children is a girl, one of the two will be crossed out.

86!

Arts said:

86!

da dum di doo Pah. Maxwell Smart.

A similar question has turned up on stackexchange; see https://math.stackexchange.com/questions/86797/whats-the-probability-of-2-head-given-at-least-1-head. One of the respondents refers to another question on the site, asking whether knowing that a man has two children, and seeing him with one of his sons, makes the probability that the other is a son 1/3.

See also boy or girl paradox on TATE.

sibeen said:

“/uploads/f0b5b967-aba0-4b40-b0ea-f06f9d3a29ed.jpe”

This is elder sprogs decision tree using heads (girls) and tails (boys).

At the first toss there is 1/2 chance in each leg.

If a tail is flipped (lowest branch), as we know there is already a head then the next decision branch is a 1, and the total of that branch is 1 × 1/2 = 1/2.

On the upper segment we have the original 1/2 and then this branches out again with each fork producing a 1/2. The total end of each branch is 1/2 × 1/2 = 1/4.

There are two branches at the top each with a 1/4 chance, so a total of a 1/2 chance for the total branches and there is a 1/2 chance for the bottom branch.

The chance of a Head – Head is 1/4.

I can see what she’s trying to do, and I can see why it’s wrong. The second decision branch in the bottom branch is still only 1/2 for head and 1/2 for tail, since the outcome of the toss is undecided until after the toss — we don’t know that there’s at least one head until after the experiment has been conducted.

btm said:

A similar question has turned up on stackexchange; see https://math.stackexchange.com/questions/86797/whats-the-probability-of-2-head-given-at-least-1-head. One of the respondents refers to another question on the site, asking whether knowing that a man has two children, and seeing him with one of his sons, makes the probability that the other is a son 1/3.

See also boy or girl paradox on TATE.

There ya go, it has a wiki page :)

I suspect the maths faculty will now throw the question in the bin :)

btm said:

sibeen said:

“/uploads/f0b5b967-aba0-4b40-b0ea-f06f9d3a29ed.jpe”

This is elder sprogs decision tree using heads (girls) and tails (boys).

At the first toss there is 1/2 chance in each leg.

If a tail is flipped (lowest branch), as we know there is already a head then the next decision branch is a 1, and the total of that branch is 1 × 1/2 = 1/2.

On the upper segment we have the original 1/2 and then this branches out again with each fork producing a 1/2. The total end of each branch is 1/2 × 1/2 = 1/4.

There are two branches at the top each with a 1/4 chance, so a total of a 1/2 chance for the total branches and there is a 1/2 chance for the bottom branch.

The chance of a Head – Head is 1/4.

I can see what she’s trying to do, and I can see why it’s wrong. The second decision branch in the bottom branch is still only 1/2 for head and 1/2 for tail, since the outcome of the toss is undecided until after the toss — we don’t know that there’s at least one head until after the experiment has been conducted.

which brings us back to…

btm said:

sibeen said:

“/uploads/f0b5b967-aba0-4b40-b0ea-f06f9d3a29ed.jpe”

This is elder sprogs decision tree using heads (girls) and tails (boys).

At the first toss there is 1/2 chance in each leg.

If a tail is flipped (lowest branch), as we know there is already a head then the next decision branch is a 1, and the total of that branch is 1 × 1/2 = 1/2.

On the upper segment we have the original 1/2 and then this branches out again with each fork producing a 1/2. The total end of each branch is 1/2 × 1/2 = 1/4.

There are two branches at the top each with a 1/4 chance, so a total of a 1/2 chance for the total branches and there is a 1/2 chance for the bottom branch.

The chance of a Head – Head is 1/4.

I can see what she’s trying to do, and I can see why it’s wrong. The second decision branch in the bottom branch is still only 1/2 for head and 1/2 for tail, since the outcome of the toss is undecided until after the toss — we don’t know that there’s at least one head until after the experiment has been conducted.

I actually said to her earlier this evening that I bet it’s just a weird variation on teh Monty Hall problem. I then explained the MH problem to her and tried to explain why it is a bit counter intuitive.

sibeen said:

btm said:

sibeen said:

“/uploads/f0b5b967-aba0-4b40-b0ea-f06f9d3a29ed.jpe”

This is elder sprogs decision tree using heads (girls) and tails (boys).

At the first toss there is 1/2 chance in each leg.

If a tail is flipped (lowest branch), as we know there is already a head then the next decision branch is a 1, and the total of that branch is 1 × 1/2 = 1/2.

On the upper segment we have the original 1/2 and then this branches out again with each fork producing a 1/2. The total end of each branch is 1/2 × 1/2 = 1/4.

There are two branches at the top each with a 1/4 chance, so a total of a 1/2 chance for the total branches and there is a 1/2 chance for the bottom branch.

The chance of a Head – Head is 1/4.

I can see what she’s trying to do, and I can see why it’s wrong. The second decision branch in the bottom branch is still only 1/2 for head and 1/2 for tail, since the outcome of the toss is undecided until after the toss — we don’t know that there’s at least one head until after the experiment has been conducted.

I actually said to her earlier this evening that I bet it’s just a weird variation on teh Monty Hall problem. I then explained the MH problem to her and tried to explain why it is a bit counter intuitive.

The real point is that it isn’t worth spending money on.

sibeen said:

btm said:

sibeen said:

“/uploads/f0b5b967-aba0-4b40-b0ea-f06f9d3a29ed.jpe”

This is elder sprogs decision tree using heads (girls) and tails (boys).

At the first toss there is 1/2 chance in each leg.

If a tail is flipped (lowest branch), as we know there is already a head then the next decision branch is a 1, and the total of that branch is 1 × 1/2 = 1/2.

On the upper segment we have the original 1/2 and then this branches out again with each fork producing a 1/2. The total end of each branch is 1/2 × 1/2 = 1/4.

There are two branches at the top each with a 1/4 chance, so a total of a 1/2 chance for the total branches and there is a 1/2 chance for the bottom branch.

The chance of a Head – Head is 1/4.

I can see what she’s trying to do, and I can see why it’s wrong. The second decision branch in the bottom branch is still only 1/2 for head and 1/2 for tail, since the outcome of the toss is undecided until after the toss — we don’t know that there’s at least one head until after the experiment has been conducted.

I actually said to her earlier this evening that I bet it’s just a weird variation on teh Monty Hall problem. I then explained the MH problem to her and tried to explain why it is a bit counter intuitive.

The MH problem is counterintuitive until you realise that he’s giving you more information by opening the first door.

btm said:

sibeen said:

btm said:

I can see what she’s trying to do, and I can see why it’s wrong. The second decision branch in the bottom branch is still only 1/2 for head and 1/2 for tail, since the outcome of the toss is undecided until after the toss — we don’t know that there’s at least one head until after the experiment has been conducted.

I actually said to her earlier this evening that I bet it’s just a weird variation on teh Monty Hall problem. I then explained the MH problem to her and tried to explain why it is a bit counter intuitive.

The MH problem is counterintuitive until you realise that he’s giving you more information by opening the first door.

So maths has drama as well?

btm said:

sibeen said:

btm said:

I can see what she’s trying to do, and I can see why it’s wrong. The second decision branch in the bottom branch is still only 1/2 for head and 1/2 for tail, since the outcome of the toss is undecided until after the toss — we don’t know that there’s at least one head until after the experiment has been conducted.

I actually said to her earlier this evening that I bet it’s just a weird variation on teh Monty Hall problem. I then explained the MH problem to her and tried to explain why it is a bit counter intuitive.

The MH problem is counterintuitive until you realise that he’s giving you more information by opening the first door.

Exactly what I explained to elder sprog.

i think the question cannot be answered as a single probability.

My answer is in two parts and goes like this:
If the girl observed is the first-born child, then the probability that the second-born child is also a girl is 50%
If the girl observed is the second-born child, then the probability that the first-born child is also a girl is 50%

I cannot just crossout boy.boy and add the three possible outcomes together and give the overall probability at 1/3 because that would mean making it possible for the observed child to be both the first and second born at the same time. Since we don’t know if she is the first or second born, but cannot be both, I split the answer into two parts.

party_pants said:

i think the question cannot be answered as a single probability.

My answer is in two parts and goes like this:
If the girl observed is the first-born child, then the probability that the second-born child is also a girl is 50%
If the girl observed is the second-born child, then the probability that the first-born child is also a girl is 50%

I cannot just crossout boy.boy and add the three possible outcomes together and give the overall probability at 1/3 because that would mean making it possible for the observed child to be both the first and second born at the same time. Since we don’t know if she is the first or second born, but cannot be both, I split the answer into two parts.

does it matter whether the spotted child is first or second born? there is no mention in the OP of any order, just that a child was spotted and it was a girl.

ChrispenEvan said:

party_pants said:

i think the question cannot be answered as a single probability.

My answer is in two parts and goes like this:
If the girl observed is the first-born child, then the probability that the second-born child is also a girl is 50%
If the girl observed is the second-born child, then the probability that the first-born child is also a girl is 50%

I cannot just crossout boy.boy and add the three possible outcomes together and give the overall probability at 1/3 because that would mean making it possible for the observed child to be both the first and second born at the same time. Since we don’t know if she is the first or second born, but cannot be both, I split the answer into two parts.

does it matter whether the spotted child is first or second born? there is no mention in the OP of any order, just that a child was spotted and it was a girl.

That’s all the information available. or is it?

ChrispenEvan said:

party_pants said:

i think the question cannot be answered as a single probability.

My answer is in two parts and goes like this:
If the girl observed is the first-born child, then the probability that the second-born child is also a girl is 50%
If the girl observed is the second-born child, then the probability that the first-born child is also a girl is 50%

I cannot just crossout boy.boy and add the three possible outcomes together and give the overall probability at 1/3 because that would mean making it possible for the observed child to be both the first and second born at the same time. Since we don’t know if she is the first or second born, but cannot be both, I split the answer into two parts.

does it matter whether the spotted child is first or second born? there is no mention in the OP of any order, just that a child was spotted and it was a girl.

Yes, that information is important because the order gives us our 4 possible starting outcomes.

A Bayesean analysis is interesting.
Definitions:
P(GG) prob. that both are girls (25%)
P(BB) prob. that both are boys (25%)
P(GB) prob. of girl then boy (25%)
P(BG) prob. of boy then girl (25%)
P(G|GG) prob. at least one girl given that both are girls (=100%)
P(GG|G) prob. both girls given that at least one is a girl
P(G) prob. at least one girl.

P(GG|G) = P(G|GG) x P(GG)/P(G) = 1 × (.25/.75) = ¼×¾ = 1/3.

This is a naive probability, in that it’s theoretical. Suppose now that we examine one of the children to determine whether it’s a girl. Then,
P(g) is prob. that a sampled child is a girl (=50%), and

P(GG|g) = P(g|GG) x P(GG)/P(g) =1 × (.25/.5) = ¼×½ = (2/4) = 1/2 = 50%.

So, if we know the neighbours have two children and at least one is a girl, the probability that they’re both girls is 1/3.
If we know the neighbours have two children and we see that one of them is a girl, the probability that they’re both girls is 1/2.

Have any of you mentioned that the child might be transgender or intersex?

kii said:

Have any of you mentioned that the child might be transgender or intersex?

Yes. That’s one of the reasons I used the coin tosses rather than children.

Curiously, an American columnist in the magazine Parade asked families with exactly two children, at least one of which was a boy, to write and give details of the two. Of 17946 responses, 35.9% had two boys. See Making Babies by the Flip of a Coin? (subscription required for more then the abstract,) in The American Statistician, 59: 180-182, 2005, for details and analysis.

btm said:

Curiously, an American columnist in the magazine Parade asked families with exactly two children, at least one of which was a boy, to write and give details of the two. Of 17946 responses, 35.9% had two boys. See Making Babies by the Flip of a Coin? (subscription required for more then the abstract,) in The American Statistician, 59: 180-182, 2005, for details and analysis.

That illustrates the point.

It’s not knowing “at least one” (random gender) that is important.

It is only if you know that there is “at least one of specified gender”, and that you will always know if there is at least one.

In the kid next door context an example where the answer might be 1/3 is if you were discussing the new neighbours with your wife and she knows they have two kids, of unknown gender. You say “I hope they have at least one girl”, and she then looks out of the window and says “actually I can see them both, and yes there is at least one girl”.

Even then the probability of 2 girls depends on unknowns:
Can she see the gender of both children?
Is she a literal sort of woman, who will give you the information you asked for, and no more, or if it had been two girls, would she have said so?
If it had been two boys, would she have said so?

btm said:

So, if we know the neighbours have two children and at least one is a girl, the probability that they’re both girls is 1/3.
If we know the neighbours have two children and we see that one of them is a girl, the probability that they’re both girls is 1/2.

This part confuses the dickens out of me, but then again I am not mathematically inclined.

Divine Angel said:

btm said:

So, if we know the neighbours have two children and at least one is a girl, the probability that they’re both girls is 1/3.
If we know the neighbours have two children and we see that one of them is a girl, the probability that they’re both girls is 1/2.

This part confuses the dickens out of me, but then again I am not mathematically inclined.

That could be because the statement “at least one is a girl” is ambiguous, rather than lack of mathematical inclinations.

If you see that one child is a girl, then you know that at least one is a girl, so is the probability of two girls 1/3 or 1/2?

It’s 1/2 (assuming equal gender distribution, and equal probability of seeing a child of either gender).

It only becomes 1/3 if you know that you will always be informed if there is at least one girl.

So if you ask “is there at least one girl?”, and get answer yes, or “are there two boys?”, and get answer no, then the probability of two girls is 1/3, but if you are told “there is at least one girl” you don’t know if the person telling you thinks girls are special, or if she only knows the gender of one of the children, so you don’t know what the probability of two girls is.