ChatterBank32 mins ago
On A Similar Theme - Probability Again
I remember reading about a problem which went something like this:
Mr Smith has a daughter named Tallulah. What is the probability of his other child being a girl?
The answer was very surprising and counter-intuitive and not the same as the answer in Factor-Fiction's related thread, but the author went through every step and it all made sense. Unfortunately, I can't find the book or remember its title. Does anyone recognise this, and tell me the title or author?
Mr Smith has a daughter named Tallulah. What is the probability of his other child being a girl?
The answer was very surprising and counter-intuitive and not the same as the answer in Factor-Fiction's related thread, but the author went through every step and it all made sense. Unfortunately, I can't find the book or remember its title. Does anyone recognise this, and tell me the title or author?
Answers
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No best answer has yet been selected by Cloverjo. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.I don't know the book, but the argument goes something like this.
Mr. Smith has two children (since he has a daughter and "another child"). A priori, he could have:
Girl+Girl
Girl+Boy
Boy+Girl
Boy+Boy
Each of these has a probability of 1/2 x 1/2 = 1/4 (25%). But the middle two (Boy+Girl and Girl+Boy) are the same, so we have:
2 girls = 25%
1 of each = 50%
2 boys = 25%
We know that the "2 boys" option is ruled out, because he has a daughter called Tallulah. So, were are left with "1 of each" or "2 girls". The probability of "1 of each" is twice as high as "2 girls". So, *given* that he has a daughter called Tallulah, there's only a 1/3 chance (33.3%) that his other child is a girl.
Mr. Smith has two children (since he has a daughter and "another child"). A priori, he could have:
Girl+Girl
Girl+Boy
Boy+Girl
Boy+Boy
Each of these has a probability of 1/2 x 1/2 = 1/4 (25%). But the middle two (Boy+Girl and Girl+Boy) are the same, so we have:
2 girls = 25%
1 of each = 50%
2 boys = 25%
We know that the "2 boys" option is ruled out, because he has a daughter called Tallulah. So, were are left with "1 of each" or "2 girls". The probability of "1 of each" is twice as high as "2 girls". So, *given* that he has a daughter called Tallulah, there's only a 1/3 chance (33.3%) that his other child is a girl.
Paul, that argument works when we don't know the name of the girl. There is a thread on that subject elsewhere on AB. The point is that knowing the unusual name of one of the children changes the probability of the other child being a girl to 1/2. See the link I posted at 3.40 on Thursday 30th Nov.
Incidentally, for others interested, I remember where I read the argument for the first time. It was a book called The Drunkard's Walk, and the girl in the problem is called Florida (as in my other link) not Tallulah. (My memory!)
The author explains it in a simpler way. You can see here by going to page 112 of Amazon's Look Inside function. (Hopefully)
http:// www.ama zon.com /The-Dr unkards -Walk-R andomne ss-Rule s/dp/03 0727517 5
Incidentally, for others interested, I remember where I read the argument for the first time. It was a book called The Drunkard's Walk, and the girl in the problem is called Florida (as in my other link) not Tallulah. (My memory!)
The author explains it in a simpler way. You can see here by going to page 112 of Amazon's Look Inside function. (Hopefully)
http://
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