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Probability question

gnome said:
You misunderstand... one has been identified as a boy... does it matter if you know he's child "A" or his name is "Tom"... what does that tell you? It's arbitrary.
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I didn't misunderstand at all. Reread my answer with this very question in mind.
 
Ok Jellby, I get your point. I made a slight misinterpretation of your original question. Which brings us back to a main theme: the OP is poorly stated. Clearly one has to be very careful in how the problem is stated and in how it is interpreted...
 
I am afraid that my knowledge of statistics not brilliant.

I can understand that if you don't know which child is male then the odds are 1/3, and if you know which child (elder or younger) is male then the odds are 1/2.

If I don't know if it is the elder or youger that is male, I can guess at elder, which changes the odds to 1/2. If I guess younger, that also changes the odds to 1/2.

So I am in the position of having odds of 1/3, but either of the two alternatives available change the odds to 1/2.

What have I missed, or got wrong ?

Dave
 
You missed nothing. You got it. That's exactly what everyone has been arguing like fools for 2 weeks about. Everyone has been arguing over symantics of the original question. You in essence, boiled 2 solid weeks of bickering down into 1 post. Good Job, and everyone else, read Dave_46's post. It it eloquent and if luck, has anything to do with it, this thread is done.
 
Originally posted by Dave_46
I am afraid that my knowledge of statistics not brilliant.

I can understand that if you don't know which child is male then the odds are 1/3, and if you know which child (elder or younger) is male then the odds are 1/2.

If I don't know if it is the elder or youger that is male, I can guess at elder, which changes the odds to 1/2. If I guess younger, that also changes the odds to 1/2.

So I am in the position of having odds of 1/3, but either of the two alternatives available change the odds to 1/2.

What have I missed, or got wrong ?
Click to expand...
What you missed is that your two alternatives are not mutually exclusive: if both children are male, then the elder is male and simultaneously the younger is male. So, if you "combine" the two 1/2-probability alternatives because you don't know which holds, yielding a supposed combined probability also of 1/2, you end up counting the overlap twice. Since the probability in the overlap is very high---it's 1, in fact---you end up with a combined probability that's too high: 1/2 instead of the correct 1/3.
 
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