In this problem, would you still try to distinguish the two identical red poker chips using your logic?
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."
No. I'm just assuming he has two children, and randomly mentions something about one of them. The GG scenario is only eliminated after he makes his statement, because we then know he has at least one boy.
Put it this way. before he says it, we have GG, GB, BG, BB
after he says "I have a child that is [MALE OR FEMALE]" we have (where the capital letter is the child whose sex has been mentioned, and the lowercase letter is the other child):
Gg, gG, Gb, gB, Bg, bG, Bb, bB
So if he has said the sex is male, then we have four combinations left:
gB, Bg, Bb, bB.
Understand that we go to more scenarios (8) based on which child is mentioned, before we go to fewer. Actually the order of the children is something you can and should ignore, however as you are holding on to it, I show it this way....
Someone had a nice link higher up: http://mikeschiraldi.blogspot.com/2011/11/tuesday-boy-proble...