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bonanova
Daedalian Member

 Posted: Fri Jul 20, 2012 2:33 am    Post subject: 1 Ned, Red, Ted and Zed are identical quadruplets, alike in every way except one: the way that they describe the children in families that have two children. 1. Ned says one is a boy, if that is a true statement; otherwise he says one is a girl. 2. Red says one is a girl, if that is a true statement; otherwise he says one is a boy. 3. If the older child is a boy, Ted says one is a boy; otherwise he says one is a girl. 4. Zed flips a coin and considers the taller (heads) or shorter (tails) child. If the coin-selected child is a boy, he says one is a boy; otherwise he mentions the sex of the other child. One of the four men, we don't know which, then tells us: "So there is this family with two kids. One of them is a girl." What is the probability that the other child is a girl? Assume the usual 50% birth numbers and a family chosen at random from a statistically neutral population._________________ Vidi, vici, veni.
Duke Gnome
Daedalian Member

 Posted: Fri Jul 20, 2012 9:28 am    Post subject: 2 [0.5714? 37.5% of instances will be someone other than Zed, saying "girl" 6.25% of instances will be Zed saying "girl" 25% of all instances will be someone saying girl and the other kid being a girl 25/43.75=57.14%]
Trojan Horse
Daedalian Member

 Posted: Fri Jul 20, 2012 4:24 pm    Post subject: 3 Bah. I should've visited this site last night. I agree with Duke Gnome.
Zag
Unintentionally offensive old coot

Posted: Fri Jul 20, 2012 6:13 pm    Post subject: 4

I disagree with DG (and TH)

Before we heard anything, there were sixteen cases with equal probability. Here's a chart of what that person would say in each of the four possible arrangements of the family in question.

Sorry, code sections don't "spoiler."

 Code: Actual case      GG GB BG BB Ned   G  B  B  B Red   G  G  G  B Ted   G  G  B  B Zed   G  B  G  B

When the man told us that one is a girl, all we know is to eliminate some of the cases. The others still have equal probability. Of the 8 where the speaker would have said girl, for half of them the other child is also a girl, so the answer is 50%.

(I'm assuming the coin flip selected the second child. By symmetry, it doesn't matter.)

_________________
 Death Mage wrote: I couldn't agree with you more, Zag.
Trojan Horse
Daedalian Member

 Posted: Fri Jul 20, 2012 7:45 pm    Post subject: 5 I disagree with your conclusions about Zed, Zag. If the family has one girl and one boy, then Zed will always say that the family has a boy. If the coin-selected child is the boy, then Zed will say that the family has a boy. If the coin-selected child is the girl, then Zed will mention the sex of the other child: namely, a boy.
Zag
Unintentionally offensive old coot

Posted: Fri Jul 20, 2012 9:46 pm    Post subject: 6

Sure enough, I misread Zed's algorithm. I'll start again and see if I now come up with the same answer DG did

In looking at it, I realize that Zed's rule is the same as Ned's. He only says 'Girl' if both children are girls, no matter what he flips.

(1) Zed's coin selection chooses the first-mentioned child
(2) Zed's coin selection chooses the second-mentioned child

 Code: Actual case         GG GB BG BB Zed(1)   G  B  B  B Zed(2)   G  B  B  B

So my new table looks like this
 Code: Actual case      GG GB BG BB Ned   G  B  B  B Red   G  G  G  B Ted   G  G  B  B Zed   G  B  B  B

These are all equal in chance of happening. Again, from the statement, we can eliminate all the cases where the speaker said Boy, and the remaining cases remain equal probability of happening. Of the 7 cases left, in 4 of them the other child is also a girl, so the chances are 4/7 or approximately 57.14%. It looks as if I agree with DG and you.
_________________
 Death Mage wrote: I couldn't agree with you more, Zag.
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