51–60 of 152 posts
"If you surveyed all such families, you would find that roughly 13/27 of them have two boys." roughly 13/27? How rough are we talking, maybe it's roughly 13/26 instead.
Earlier quoted context omitted.
It is about specifying a child, and then saying it is a boy. The one extreme is: one of my children is a boy. in this case the other is a boy with 1/3 probability. The other extreme is: One of my children has a national unique id=... He is a boy. The other is a boy with 1/2 probability. And other cases are in between. The more information you provide on the first child (the less chance there is that the other can hav…
"the less chance there is that the other can have the same property" No! That is not true! The two events are independent, specifying information on one has zero impact on the other.
If you say "the older is a boy" then you have a statement about one son and no information whatsoever about the younger one, therefore p=1/2.
Earlier quoted context omitted.
No, the 13 was arrived at by excluding the possibility that the second child was a boy on Tuesday. (Read the article again and see.) Once you include that possibility it's back to 14/28.
When considering the case that the older son was the one born on a Tuesday, that gives 14/28 possibilities. One of those 14 is the case that both were born on Tuesday. When considering the case that the younger son was the one born on a Tuesday, that gives 14/28 possibilities. One of those 14 is the case that both were born on a Tuesday. But woops, we've already covered the case that both were born on a Tuesday in ou…
If you have two kids there are 4 possibilities, not 3: BG, BG, BB, GG
BG seems to be the same as GB, except that it's not. And it's not in this case either.
Earlier quoted context omitted.
"the less chance there is that the other can have the same property" No! That is not true! The two events are independent, specifying information on one has zero impact on the other.
They are independent, but you get information about both ("One of them is a boy."), therefore p=1/3. If you say "the older is a boy" then you have a statement about one son and no information whatsoever about the younger one, therefore p=1/2.
In this case we are asking if the other is a boy NOT if the other was born on Tuesday.
I hate this one because while it says: "I have two children and one is a son born on a Tuesday." It actually means: "I have two children and only one of them is a son born on a Tuesday." You are supposed to just assume this modification.
There are 14 possible permutations for a child
Mon-Sun Girl
Mon-Sun Boy
which for two children gives a total of 28:
Child 1 is Boy, born Mon-Sun
Child 1 is Girl, born Mon-Sun
Child 2 is Boy, born Mon-Sun
Child 2 is Girl, born Mon-Sun
Note that we haven't said which child is 1 and which is 2, and in fact we don't know because we haven't been told - this is the important bit.
So, for our unknown child to be a boy, it must be one of the following 14 permutations:
unknown child is Child 1, boy, born Mon-Sun
unknown child is Child 2, boy, born Mon-Sun
but one of those permutations is already taken by our known Tuesday boy, so we have to remove one of the Tuesday permutations leaving 13 possible outcomes out of 27. Note that one of those 13 outcomes is 'unknown child is son born on Tuesday' - it's perfectly valid to have both sons born on a Tuesday.
Note also that we can't say which permutation we're removing until we know which is Child 1 and which is Child 2, just that one of them is taken.
If the original statement were phrased as '... the older child is a son born on a Tuesday' then you would have a constraint on which was Child 1 and which was Child 2 and then the probability would be 1/2 as expected because you would know up front that you were entirely discounting, say, Child 1, AND that the removed Tuesday permutation also belonged to Child 1.
But I stand to be corrected!
I hate this one because while it says: "I have two children and one is a son born on a Tuesday." It actually means: "I have two children and only one of them is a son born on a Tuesday." You are supposed to just assume this modification.
No I don't think that's true. Bear with me :-) There are 14 possible permutations for a child Mon-Sun Girl Mon-Sun Boy which for two children gives a total of 28: Child 1 is Boy, born Mon-Sun Child 1 is Girl, born Mon-Sun Child 2 is Boy, born Mon-Sun Child 2 is Girl, born Mon-Sun Note that we haven't said which child is 1 and which is 2, and in fact we don't know because we haven't been told - this is the important b…
In any case, as I replied here http://news.ycombinator.com/item?id=3290118 you can not remove the duplication! It's two different situations, even though they may appear the same.
Let's simplify the question:
If a man says "I have two children, one was born on a Tuesday," what is the probability that they are both born on a Tuesday?
Is the answer to this 0 or 1/7 in your opinion? (Or something different).
The difference is between the information being offered, and determined independantly. I'll do this with the boy/girl problem, for simplicities sake.
If we ask a man if he has at least one boy, and he says yes, we can work out the chance the other child is a boy like so:
Assume all four possibilities are equally likely:
BB
BG
GB
GG
BB
BG
GB
However! Because the father offered the information on his own, it effectively turns it into the author's other problem, where the older child is a boy, find out the gender of the younger child.
The trick is that he's equally likely to give information about either of his children, therefore there are eight possibilities:
He gives information about child A:
BB - B
BG - B
GB - G
GG - G
BB - B
BG - G
GB - B
GG - G
BB - 50%
GB - 25%
BG - 25%
You meet a man on the street and he says, “I have two children and one is a son born on a Tuesday.” What is the probability that the other child is also a son?
Is 1/2
The answer to this:
A man has two children, and one is a son born on a Tuesday. What is the probability that the other child is also a son?
Is 13/27
It's nitpicky, but I think the author should be very exact about this kind of thing, since he's trying to clear things up.
Earlier quoted context omitted.
If we are determined to look at order of birth (ie. have a BG and GB) then we should consider all cases by order of birth, so where 'B' represents the boy we know and 'b' or 'g' represents a child we dont, and the first character represents the first child, and the second the second childe - we have: Bb, bB, Bg, gB 50%.
Imagine a similar problem but with red and blue poker chips. Say, for example, that I have a bag and I pull out two chips, one at a time. In this problem, would you still try to distinguish the two identical red poker chips using your logic?
Under your inference, the man wouldnt have mentioned anything unless he had at least one male child. (in which case you can say the GG scenario is gone, but GB BG and BB are equally probable)
Under my inference, the man just told me the sex of one child at random.... (in which case BB is twice as probable as GB or BG - where he could have equally said 'i have at least one girl')
That sound reasonable to you? I have it in ruby form if you are interested :)
The blogpost linked to by someone above (http://blog.tanyakhovanova.com/?p=221) uses this explanation, which is very different (to me) than the one in the main link, and I can see why this gets to 1/3:
"A father of two children is picked at random. If he has two daughters he is sent home and another one picked at random until a father is found who has at least one son."
The more sons you have, the more likely that one of them is born on a tuesday. Thus, if you take all the two-child families with at least one son, and eliminate the families without a son born on tuesday, the two-boy families are more likely to remain than the one-boy families, and you will end up with a higher proportion of two-boy families than before. Simple. (edited)
Simple. And wrong. Note that the correct answer is 13/27 probability of a boy, which is less, not more, than 50%.