The daughter paradox

[Matt]

1. A woman has two children. One is a daughter. The chance of the second child being female is 1 in 2

Correction.

I see here you say second daughter rather than other daughter. The question in the OP is "Other Daughter"

BB
GB
BG
GG

Strangely however you've said "One daughter" rather than "first child is a daughter" so you still don't get the same answer as the OP

The bottom three options above are where there's one daughter. The bottom two are where the second child is a daughter. This includes option three where the one (and only) daughter is the second child. This gives a two in three chance.

Mental or what?

[Janot]

I see here you say second daughter rather than other daughter.
Mental or what?

We are having a heat wave, or I'm having a hot flush. Whatever, I can't think clearly. I should not have said 'second' daughter, but rather 'other daughter'

A. A woman has two children. I'm looking at one of them right now, and it's a girl. What is the probability that the other is a girl?

B. A woman has two coins placed on the table. I'm looking at one right now, and it is showing tails. What is the probability that the other is tails?

I claim that A and B are the identical statistical problem (with obvious assumptions). The answer to the second one is clear.

[Matt]

We are having a heat wave, or I'm having a hot flush. Whatever, I can't think clearly. I should not have said 'second' daughter, but rather 'other daughter'

A. A woman has two children. I'm looking at one of them right now, and it's a girl. What is the probability that the other is a girl?

B. A woman has two coins placed on the table. I'm looking at one right now, and it is showing tails. What is the probability that the other is tails?

I claim that A and B are the identical statistical problem (with obvious assumptions). The answer to the second one is clear.

Yes there are identical to each other. But no identical to the question in the OP.

IN the OP you can have

BB - discounted
BG and you're looking at the Girl
GB and you're looking at the Girl
GG and you're looking at one of the Girls.

The way you've phrased the question you can have BG and still discount that pair because you happen to be looking at the Boy.

[Admin]

The answer to A, "if a woman has two children at least one of which is a girl, what's the probability of the other one being a girl?" is 1/3

The answer to B "choosing a girl at random and seeing what she's paired with." is 1/2

The reason they're different is that in A you're picking pairs, and in B you're picking girls. As such in B you're twice as likely to pick from a pair with two girls.

Yes, I agree. This is where the 'paradox' appears to come from.

You get different answers depending on how the question is conceptualised. Even when the target population is identical.

In question A we're picking from pairs, in question B we're picking from individuals so you get different answers (thus the appearance of a paradox); but if question A is framed in the same way as question B and you choose a girl individually then the answer is the same (apart from when the name/attribute is of very high frequency).

[Harryprice]

Let's hope this calculation of odds never occurs in a real life scientific experiment. Lots of highly intelligent, logical people appear to get completely thrown by it, which could easily include your average scientist.

[polomint38]

On a tangent
http://snipr.com/zlczr

[Matt]

Let's hope this calculation of odds never occurs in a real life scientific experiment. Lots of highly intelligent, logical people appear to get completely thrown by it, which could easily include your average scientist.

I like how we've managed to disagree vehemently without calling each other Dung-Fer-Brains or becoming so entrenched in our views that we're afraid to change our minds in the face of compelling evidence.

Think it speaks well for intelligent, logical people.

[Matt]

On a tangent
http://snipr.com/zlczr

I have three boys.

[Harryprice]

On a tangent
http://snipr.com/zlczr

Suppose an attractive couple have a daughter, Emma-Louise. What’s the probability that they have two daughters?

[Croydon Bob]

Think it speaks well for intelligent, logical people.

And polomint.

[Admin]

Let's hope this calculation of odds never occurs in a real life scientific experiment. Lots of highly intelligent, logical people appear to get completely thrown by it, which could easily include your average scientist.

Calculating odds is a doddle. It's correctly conceptualising the problem that is the difficult thing. And make no mistake, this is the sort of problem that vexes maths professors and such like - think of the responses to the Monty Hall problem!

I've thoroughly enjoyed doing this and looking at how others have approached it (particularly Matt) has helped me eventually get to the bottom of it (I think.... ).

[Pebble]

OK I want to lay Pebble's worries to rest so I come up with a sample containing 6 names.

Each gender, name and allowed permutation of name combination are equally represented.

Accepted.

It appears that any unique identifier can work, because it brings the total number of a given sex into play. It follows, I think, that if the origional question was "if a woman has two children, one of whom is a girl, what is the probability that this girl has a sister?" then one must accept that "this girl" could be D1 or D2 in a girl girl combination and therefore the chances must immediately increase to 1:2.

[Pebble]

This is an old one but it was published in the ASKE newsletter with a twist I hadn't come across before:

1. The first part of the puzzle is a classic problem: ‘Jane has two children. One is a daughter. What’s the probability that she has two daughters?’
   .
2. The second part is: ‘Jane has two children. One is a daughter, Emma-Louise. What’s the probability that she has two daughters?’

.

I feel like I am worrying a bone here but.....

If the final part of the question is whether the mother has two daughters, then the name is irrelevant - XX chromasomes is what counts.

Only if the question can be changed to look at the liklihood of Emma Louise having two sisters do we get to 1:2.

So there must be an error in the process that forces us to look at the number of sisters, instead of looking at the problem from the mother's perspective - where there cannot be an increase in the proportion of daughters, whatever their names.

My suspicion is that we are approaching the problem from the child's name, which means that the name is the first event we consider - hence the relationship has to be 1:2. But from the mother's perspective the name follows conception & delivery - first and foremost she has only a 1:4 chance of having two daughters. Excluding two sons she has a 1:3 chance of having two daughters. She may then name those daughters as she wishes.

So all the work on the proportion of offspring with given names misses the point - we should only look at the mothers perspective - we need to be able to calculate her probability - not the offspring's probabilities.

[Pebble]

One more then I'll stop.

Mother's probability

2/3 chance of son + Emma Louise (older or younger)
1/3 chance of 2 daughters (Emma Louise older or younger)

Offspring's probability

1/2 Brother (Emma Louise older or younger)
1/2 Sister (Emma Louise older or younger)

If and only if the mother has two daughters do her chances of naming a daughter Emma Lousie double, but not her chances of having two daughters to name.

[Croydon Bob]

take two coins. Toss them both. The are three possible outcomes.

both heads,
both tails
one of each.

If you repeat this over and over recording the results you will see that "one of each" comes up twice as often (half the time) as either both heads (quarter of the time) or both tails (also quarter of the time)

You'll also see that out of all the times that there's at least one head, (three quarters of the time) in approximately one third of those instances (one quarter of the time) there will also be a second head.

For me it was this post by Matt that nailed it.

Re-word the original Q to say: A woman tossed a coin, got "heads", called that coin "Emma-Louise". What are the chances that when she tossed a second coin she got a second "heads"? And you'll see that it is a very simple puzzle (excluding hermaphrodites and slight extra chance of having a boy, etc).