The daughter paradox

[Pebble]

[QUOTE=Matt;89921]Both Correct,



If however you're told that "one of her children is a girl" but not whether it's "bag-head" or "the other one" then that's a whole different ball game. QUOTE]


I'm not sure.

Unidentified girl can have an older or younger brother or an older or younger sister - probability 1:2.

[Matt]

[QUOTE=Pebble;89927]

Both Correct,



If however you're told that "one of her children is a girl" but not whether it's "bag-head" or "the other one" then that's a whole different ball game. QUOTE]


I'm not sure.

Unidentified girl can have an older or younger brother or an older or younger sister - probability 1:2.

Older/younger is irrelevant. All that matters is that you consider the children';s gender's as independant probabilites to get a binomial distribution. Older/younger is one convenient label you can use to do that.

[Pebble]

[QUOTE=Matt;89928]

Older/younger is irrelevant. All that matters is that you consider the children';s gender's as independant probabilites to get a binomial distribution. Older/younger is one convenient label you can use to do that.

Absolutely, but this is precisely wht you have done with the naming procedure.

So the argument is in these three pairs (2BG, 1GG) there are four possible girls so any girl identified or unidentified has a 50:50 chance of having a sister. However a mother has only a 33% chance of having two girls identified or unidentified.

The issue causing confusion is that if one girl is named, the mother has had double the chance of ascribing that name to a girl if she had 2 girls. But since all girls have a name, this must be true in all families. We therefore no longer need the identifier to generalise the statement. So for your argument to be true Girl-Girl combinations, where at least one of the girls has a name, would have to be equally frequent as BG combinations, and that we know not to be true.

[Matt]

I toss two coins, a 50p and a 20p but keep the results secret from you.

With no information you may correctly say that the odds that the 50p is heads is 1/2
With no information you may correctly say that the odds that the 20p is heads is 1/2

If I tell you that the 50p is heads you have information about just one coin.
With the information that the 50p is heads you may correctly say that the odds that the 50p is heads is 1
The information that the 50p is heads doesn't affect the probability of the 20p. You may correctly say that the odds that the 20p is heads is 1/2

If I don't tell you that the 50p is heads but instead tell you that (at least) one of the coins is heads then you have information that affects both coins.
With this information you may correctly say that the odds of the unidentified coin also being heads is 1/3

Don't believe me just try it. Toss two coins. If they comes up tails-tails don't note anything down.
If it comes up heads-tails or tails-heads write down "tails" (the "other" coin)
If it comes down heads-heads write down heads.
Repeat until the pattern becomes clear. Here's my results

TTHTTTTHHHTHTTTTHT

6/18 = 1/3

That's bang on the expected result. I was lucky there. There's was a out a 1 in 5 chance of getting that, a 50-50 chance of getting 5,6 or 7 (within one)

I think if you actually do it you might see what's going on.

[Pebble]

Using a 50p and 20p coin is like ascribing the coin to a younger or older position and the sex to heads or tails, the information that provides does indeed alter the calculation, since one has been given two pivotal pieces of information.

Not only do you know that the 50p is heads, you also know that tails is impossible in the 50p position, the excluded information is what changes the probability.

For exactly the same reason identifying a girl as 'this girl' (rather than a girl) would give her a 1:3 chance of having a sister rather than a 50:50 chance

[Pebble]

The 50p 20p problem has given me an idea.

A girl called Matilda in a two child family has a 50:50 chance of having a sister.

My cousin Matilda has a sibling - she has only a 33% chance of having a sister.

Why is this? Well by specifying that Matilda exists in that family, she has occupied one of the avialable slots - we may not know what slot that is, but we definitely know that there is a slot that cannot be filled by anyone but Matilda. Her sibling may be older or younger, but the sex ratio is now determined purely by genetic probability, since her position is firmly occupied in reality, as opposed to a notional position. Only if we define her as older or younger do we reduce the chance of a brother to 1:2.

So perhaps the reverse of what we have calculated is true. Since the mother and one ofspring are firmly identified and linked to each other, this excludes certain possibilities - leaving both the mother with a 1:3 chance of having two daughters and Emma Louise witha 1:3 chance of having a sister?

[Janot]

I think China has go it right. Limiting the family to just one child makes life much easier. Probably.

[Pebble]

The 50p 20p problem has given me an idea.

A girl called Matilda in a two child family has a 50:50 chance of having a sister.

My cousin Matilda has a sibling - she has only a 33% chance of having a sister.

Why is this? Well by specifying that Matilda exists in that family, she has occupied one of the avialable slots - we may not know what slot that is, but we definitely know that there is a slot that cannot be filled by anyone but Matilda. Her sibling may be older or younger, but the sex ratio is now determined purely by genetic probability, since her position is firmly occupied in reality, as opposed to a notional position. Only if we define her as older or younger do we reduce the chance of a brother to 1:2.

So perhaps the reverse of what we have calculated is true. Since the mother and one ofspring are firmly identified and linked to each other, this excludes certain possibilities - leaving both the mother with a 1:3 chance of having two daughters and Emma Louise witha 1:3 chance of having a sister?

Apologies, this is complete nonsense.

If we know her position other child 1:2, if we don't she has equal probability of occupying 4 possible positions , so 1:2.

[Pebble]

Matt. I think I can finally see why we diffir on this problem and conclude that you are right.

I have taken the genetic approach, an individual mother has a 1:3 chance of having two daughters as I have outlined.

However, you have assessed the probability of the subset of mothers with a named daughter having two daughters.

So your analysis of the six named children, does actually square this circle - though presumably it will only work for numbers of pairs divisible by both 4(or2?) & 3, on a population basis this is probably irrelevant.

[Pebble]

In the problem, is Jane an individual mother who happens to have a daughter named Emma Louise, or a representative of a subset of mothers with a daughter named Emma Louise, or both? Are there different answers depending on how you classify Jane?

[Matt]

Matt. I think I can finally see why we diffir on this problem and conclude that you are right.

I have taken the genetic approach, an individual mother has a 1:3 chance of having two daughters as I have outlined.

However, you have assessed the probability of the subset of mothers with a named daughter having two daughters.

So your analysis of the six named children, does actually square this circle - though presumably it will only work for numbers of pairs divisible by both 4(or2?) & 3, on a population basis this is probably irrelevant.

As I understand it, it should work on any sample where the probability of an individual having any particular name does not depend upon the number of their siblings.

Also that the probability that the name will be used amongst siblings will increase linearly with the number of siblings.

[Admin]

I think China has go it right. Limiting the family to just one child makes life much easier. Probably.

What about rural China?

If a woman has 7 children, what's the odds of any of them being a girl?

[Janot]

What about rural China?

If a woman has 7 children, what's the odds of any of them being a girl?

In rural China? pretty small. Elsewhere, the probability that all 7 are boys is 1 in 2^7 = 128, so I guess that the probability of at least one girl is 127 in 128. Was that the question?

[Pebble]

As I understand it, it should work on any sample where the probability of an individual having any particular name does not depend upon the number of their siblings.

Also that the probability that the name will be used amongst siblings will increase linearly with the number of siblings.

What I was trying to get at was that your population of I believe 60 odd pairs with 3 boys and 3 girls names, could only work perfectly for all childrens' names if the number evaluated was divisible by both of the 'mother's probabilities 1:2 & 1:3).

Where I got stuck is that mother's as a population have a 1:3 chance of having two grils, indiviudal mothers have a 1:3 chance of having two girls whatever their name. But within the group of mothers of any child with any single name the proportion of mothers mapped to same sex offspring is increased.

The question for me therefore was,does Jane remain an individual after naming her daughter, or simply become Emma Louise's mother? If the question is what is Jane's probability of having two daughters it remains 1:3, but if the question is what is the probability of Emma Louise's mother having two daughters the answer is 1:2.

Unclear why I found that so difficult when I happily discarded the chances of two boys in the probability analysis.

[Admin]

I thought I'd revive this thread.

Because...

The way I finally understood this was:

...(1) .(2)
A) 50 - 50 (boy-boy)
B) 50 - 50 (boy-girl)
C) 50 - 50 (girl-boy)
D) 50 - 50 (girl-girl)

If we have an initial set up like that and we ask the first question: "Jane has two children. One is a daughter. What’s the probability that she has two daughters?"

Then we are forced to choose between rows (B), (C), and (D) so we get the 2/3 - 1/3 split.

I'm going to disagree with myself (again! ).

I think choosing between rows is where the error is made. What is normally done is to change the percentages based on the number of rows to:

...(1) .(2)
A) 50 - 50 (boy-boy) 0%
B) 50 - 50 (boy-girl) 33.3%
C) 50 - 50 (girl-boy) 33.3%
D) 50 - 50 (girl-girl) 33.3%

instead of changing the percentage based on numbers (frequencies) to:

...(1) .(2)
A) 50 - 50 (boy-boy) 0%
B) 50 - 50 (boy-girl) 25%
C) 50 - 50 (girl-boy) 25%
D) 50 - 50 (girl-girl) 50%

So I'm going to go with my previous conclusion that the 'standard' answer to question A (1/3) is wrong.

The error is based on the problem of dealing with percentages.

An example being: I left the house this morning with £10 in my pocket. I then found £5 on the pavement and put it in my other pocket. I'm now 50% richer than when I left the house. Later on, I went into my pocket for the £5 only to find a hole in the pocket and the £5 missing. I'm now 33% poorer. Oh well, I can't complain. I'm still 17% richer than when I left the house!

I think a similar error of reasoning about percentages is occurring in the standard solution to question A.