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#1
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a probability brainteaser from OOT
help me y'all,
I'm either doing something wrong or explaining something wrong. answer in this thread |
#2
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Re: a probability brainteaser from OOT
After quickly reading through the thread it seems that most people converged to the correct answer of 1/3. The argument you provide it the standard one given for this rather famous problem.
There is usually a second part. Now if the mother instead answers that her eldest child is a girl, does change the probability? Yes, the probability of the younger child being a girl is 1/2 not 1/3 as in the previous case. This is generally referred to as the census takers problem. |
#3
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Re: a probability brainteaser from OOT
did you see the 1% stipulation?
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#4
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Re: a probability brainteaser from OOT
The 1% stipulation comes from assuming a finite population size, n, with n/2 men and n/2 women. If n is small then the non-replacement effect you mentioned is significant. However, as n grows large this effect becomes insignificant, i.e. much less than 1%.
One should note that the above model for the populations is very inaccurate. A better model is to assume that at every birth there is a 1/2 probability of a boy and a 1/2 probability of a girl. Under this model, the replacement effect is wiped out since we are drawing from a stationary distribution. |
#5
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How\'s this for an approach
Take an average hand (say Q8u) and figure the probability for having the nuts for that hand. It may not extrapolate perfectly to having the nuts for any hand, but it may be close. So, what's the probability of ending up with a flop that is the nuts for q8u?
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#6
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Huh? Tell me what I\'m missing...
How could it possibly matter what the girl's name is?
The probability is 1/3... right? |
#7
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Re: Huh? Tell me what I\'m missing...
[ QUOTE ]
How could it possibly matter what the girl's name is? The probability is 1/3... right? [/ QUOTE ] A women is more likely to have a girl named Sarah if she has two girls. Conversely, if he she has a child named Sarah.... |
#8
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Re: a probability brainteaser from OOT
Someone please explain how the first answer is 1/3. The woman has two children, on of them IS a girl. This only leaves one other child, which is either a boy or girl. The chances of having a boy or a girl are equally likely. Therefore shoudn't the answer be 1/2.
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#9
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Re: a probability brainteaser from OOT
[ QUOTE ]
Someone please explain how the first answer is 1/3. The woman has two children, on of them IS a girl. This only leaves one other child, which is either a boy or girl. The chances of having a boy or a girl are equally likely. Therefore shoudn't the answer be 1/2. [/ QUOTE ] There are three possibilities, all of which are distinct and equally likely. G,B B,G G,G. So the answer is 1/3. |
#10
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Re: a probability brainteaser from OOT
Ok, but if she has at least one girl B,G and G,B are exactly the same. The one that isn't a girl, is a boy.
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