Reply With QuoteAren't more boys born than girls? I thought there was a slight leaning to make up for the genetic bits males have missing causing deadness.
The daughter paradox
This is an old one but it was published in the ASKE newsletter with a twist I hadn't come across before:
- 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?’
.- The second part is: ‘Jane has two children. One is a daughter, Emma-Louise. What’s the probability that she has two daughters?’
The answers given are:
Part 1: 1 in 3
It's the standard way of working it out. There are 4 ways of having 2 children:
Boy - Boy
Boy - Girl
Girl - Boy
Girl - Girl
If at least one is a girl then the boy-boy option is ruled out so the girl is twice as likely to have a brother than a sister.
Part 2: 1 in 2
This is the answer given:
I think that's wrong.In the second part of the puzzle the child who is a daughter is uniquely identified by her name, Emma Louise (we make the reasonable assumption that Jane does not give both her children the same name). The ‘other child’ now is the child who is not Emma Louise and can be a boy or a girl. Hence the probability now that Jane has 2 daughters is 1 in 2!
Another way of uniquely identifying the daughter in the question could be to say that she is the older of the two or the taller, and so on. Or by being introduced to her. Again the answer would be 1 in 2.
It doesn't matter what the girl's name (or any other attribute for that matter) is, the answer is 1 in 3 unless the birth order is known - in which case it becomes 1 in 2.
Any thoughts on this puzzle?
I can expand on my reasoning but I'll wait for now.
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Aren't more boys born than girls? I thought there was a slight leaning to make up for the genetic bits males have missing causing deadness.
Last edited by Floppit; 14th July 2010 at 06:49 PM.
This completely stumps me - how would the act of naming a daughter affect the statistical probability of he having a sibling of a given sex?
Are they really arguing that if you don't name your offspring till the family is complete that you will end up with different sex ratios! Perhaps they can provide evidence that only first daughters are called Emma Louise.
The art of medicine consists in amusing the patient while nature cures the disease. Voltaire
These counter-intuitive puzzles are great for getting your cognitive knickers in a twist.
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Yes you are correct. The probability of two daughters given that one is a daughter is 1 in 3 whether you know the name of the one daughter or not. As you say, it would be 1 in 2 if you wanted the probability of two daughters given that the first is a daughter.Originally Posted by John Jackson
Never interrupt your enemy when he is making a mistake. - Napoleon Bonaparte (1769-1821)
In the first example either of the two children can be a girl, in the second which child is the female is determined therefore 50/50 for the second child. In other words the second queries the gender of only one of the two children where as the first queries both but states that one (either) is female.
And the above is almost certainly complete bollox!
How do you conclude that "in the second which child is the female is determined?" Emma Louise could be the first or second child.
Never interrupt your enemy when he is making a mistake. - Napoleon Bonaparte (1769-1821)
Because if they were both daughters it would only refer to one of them - hence the comment about not giving two the same name. Therefore it's a specific daughter referred to not either daughter (if they had 2!).
Ah, I wonder?
The possibilities with two children, one of them Emma Louise are
Boy - Emma Louise
Emma Louise - Boy
Emma Louise - Girl
Girl - Emma Louise
Probability of 2 girls 2:4 ie 1:2
Never interrupt your enemy when he is making a mistake. - Napoleon Bonaparte (1769-1821)
Are they actually two different options?
This is the crux of the problem.
Yes they are 2 cases but surely their combined probability adds up to 1/3 - i.e. each choice has a probability of 1/6
The point is by stating Emma-Louise is born first or second, you end up (falsely) introducing birth order.
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I think so. How can they not be? There has to be a birth order even if we don't know it and the girls don't share a name.Are they actually two different options?
Excellent point - I withdraw the suggestion :-) And we're back to the probability of 1:3 - there's absolutely no getting away from it.
Never interrupt your enemy when he is making a mistake. - Napoleon Bonaparte (1769-1821)
Looks a bit like the billiard-ball problem in which you are shown a bag containing either two reds or, equally likely, a red and a white. You are shown a ball at random, which happens to be red. It is now twice as likely that the other is red. The point being that you have seen either a red from a mixed pair or a red from a red pair or the other red from a red pair.
I don't know why, but people who don't see this immediately have great difficulty in seeing it at all.
The style as we like is the humdrum.
Yes, it took some time but I see it clearly now. Can I explain it clearly though.
Lets have a stab.
First lets mention the assumptions.
Probability that a child is a boy is equal to the probability that a child is a girl
The probability of a second child's gender is independent of the gender of the first child.
The nit picker in me insists on mentioning that these are not 100% accurate but hey, it's close enough for jazz and I'm sure the assumptions are implicit in the hypothetical
Firstly I must confess that as is often the case my intuition was wrong. I thought that case 1 would result in a 0.5 probability that the other child was a girl.
What I had failed to consider was that a girl and a boy is twice as likely as two girls as it can be boy-girl or girl-boy.
The possible options for two children are.
boy-boy
girl-boy
boy-girl
girl-girl
This gives us a marginal probability for two girls of 0.25 and a probability for at least one girl of 0.75.
Therefore the conditional probability of two girls, given that there's at least one girl is 0.25/0.75 = 0.333...
Or one in three as we like to say.
Naming one of the girls gives us an additional option.
boy-boy
emma-boy
boy-emma
emma-girl
girl-emma
We might then make the say that that the probability of two girls is 2/5 = 0.4, the probability at least one girl is 4/5 = 0.8 therefore the conditional probability of two girls given that there is at least one girl called Emma is 0.4/0.8 = 0.5.
Or one in two as we like to say.
This may be incorrect. It assumes that all these options are equally probable. I suspect that it is in this assumption that the paradox reveals itself. I just tried to prove the assumption incorrect and got in a frightful muddle so I'm not as convinced as I was that it's a mistake.
The route I took if anyone wants to try again is by spelling out further options.
boy-boy
boy-girl
boy-emma
girl-boy
emma-boy
boy-girl
boy-emma
girl-girl
emma-girl
girl-emma
emma-emma
This is where the error creeps in. You don't actually get an extra option, you're actually splitting a single option in two.
What you actually have is:
boy-boy
emma-boy
boy-emma
Girl - Girl (one of whom is called Emma).
The probability of each being 1/3
Now, you can split the girl-girl option into 'Emma-girl and girl-Emma' but by doing so you're making the probability 1/6 in each case.
So for problem 2, the actual probability table can be represented by:
boy-boy
emma-boy 1/3
boy-emma 1/3
Girl - Girl (one of whom is called Emma). 1/3
or
boy-boy
emma-boy 1/3
boy-emma 1/3
Emma-girl 1/6
Girl-Emma 1/6
In other words, it's the splitting of an option in two but without halving the probabilities of the resulting options that leads to the error.
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