Posted on 06/29/2010 4:35:59 AM PDT by mattstat
>> If you have one boy, the probability of having two boys after your next baby is born is 1/2. <<
This is absolutely true. But that would fix that the ELDEST child is a boy. The question is whether EITHER child is a boy. When EITHER child is a boy, the odds that the OTHER child is a boy is only 1 in 3.
It doesn’t matter if the first is a boy or the second; the cases are equivalent.
The stipulated male child is either the eldest or the youngest. He can’t be both and he can’t be neither. The Briggs solution assumes he is both. You have to discount the impossible.
If he is the eldest the choice count is:
Boy-Boy
Boy-Girl, i.e. 1/2
If he is the youngest, then the choice count is:
Boy-Boy
Girl-Boy, again 1/2
This is binary: either-or. But in both cases the answer is the same: 1/2.
Case closed tight, not to be re-opened.
Before I start calculating I have to ask, you’re not Warren Beatty are you? That would complicate the problem.
Ok, so I read up on this problem. There are quite extensive discussions about the various levels of ambiguity in this and similar problems, although the kind of ambiguity I pointed out was not totally on target.
Nope. You and pelican001 have changed the problem. It does not say, as pelican wrote in his Monty Hall analogy, that the first choice is given. It says one of the choices is given. If it said the boy was born first, you’d be right. The choices would be boy or girl, 1/2.
But the boy could be the first or second child, so it could still be boy - girl, girl - boy, or boy - boy. Only one of the three outcomes is true. Probability is 1/3.
Correct (if we assume chance of having a boy is always exactly 1/2).
“Boy, girl” and “girl, boy” are just two different names for the same possibility.
No, they are not. They are two of the three different possibilities for a pair of children where there is at least one boy and no funny stuff.
Say I have two children, Alex and Sam. If I tell you that at least one is a boy, but I don't say which, then we can list three possibilities, only one of which is that they are both boys.
Now, if I tell you that Alex is a boy that eliminates one of the possibilities, and now there is a 50% chance that I have two boys.
An equivalent problem. I flip two coins. At least one of them is heads. What is the probability that the other is heads?
Answer : 1/3.
Try it.
What are the possibilities of the other? 1. Heads; or 2. Tails.
On of two things will follow the first confirmed heads. Right?
well, dang, here is an earth shaking issue i had never contemplated prior
I see (said the blind man.)
The question as posed comes down to "What is the possibility that a newborn is male?"
It's the same thing as if I flip a coin five times and it comes up heads, what is the probability it will come up heads on the sixth flip? The odds are fifty-fifty, as that's the probability each time you flip.
Given 2 sons, the probability one was born on a Tuesday is just 1 minus the probability both were born on some other day, ((6/7) squared). That comes out to .2653. Given exactly one son, the probability that son was born on a Tuesday is 1/7, or .1429. From these, we can calculate the odds favoring two sons given a son born on any Tuesday. That is
(.2653x.25)/(.1429x.50)=.9283.
Odds of .9283 implies the probability in question is .4814.
As I have explained, it does not matter a whit if the boy was born first or second. Not at all, as long as he is actually born. You are answer a problem that is entirely different than the problem posed.
As you have posed it, you are correct, but you have not posed an equivalent problem.
Here’s the equivalent problem: One coin is flipped and the result hidden from your view. The second coin is placed heads up on the table. What are the odds that the hidden coin is heads?
Or, alternatively for the proposition that it makes a difference whether the boy was born first or second:
Place a coin heads up on the table. Flip a second coin with the results hidden from your view. What are the odds that the hidden coin is heads.
THIS is the equivalent problem using coins.
Presented for your edification.
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