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Boy or girl paradox

All in all, do we have the correct answer? I really wonder... If I understood the question correctly, it seems the easiest question in stats. If not then what is correct?
 
Just google it.
Again, I am not trying to be smart about it. I put all the details in my last post (and those are not even mine). This problem is a Paradox for which there are 2 diff answers depending on how it was explained/understood.
At least now, if I am asked this question, I can post a full explanation on how it can be understood/solved.
Tstone, look at the wiki link (http://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Second_question). It has a lot of details on how to solve it.
The way it was written in this question makes it understood as having a prob of 1/2.
A small change (making it "that both children are boys"), will lead to a prob of 1/3.
This thread could drag for years (because we are going to keep arguing it)...
 
peterruse said:
read the earlier posts
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If I rephrase the question as: "You met your friend as a shop, he was with one of his child, you known that he had 2 children, what's the probability that the other child is a boy?"
Given the same assumptions as in the first post, what's the probability that the other child is a boy?
 
Jacksc said:
If I rephrase the question as: "You met your friend as a shop, he was with one of his child, you known that he had 2 children, what's the probability that the other child is a boy?"
Given the same assumptions as in the first post, what's the probability that the other child is a boy?
Click to expand...

the answer to this question is exactly the same as the first. seeing the child doesn't give you any new information; it only confirms that one child is a boy.
 
peterruse said:
the answer to this question is exactly the same as the first. seeing the child doesn't give you any new information; it only confirms that one child is a boy.
Click to expand...

So now if I toss two coins, I reveal one coin to you, the coin is head for example, so the probability of the other coin is also head is 1/3. This doesn't make sense to me, given that the two coins is independent and H/T chance is 50:50.
 
Jacksc said:
So now if I toss two coins, I reveal one coin to you, the coin is head for example, so the probability of the other coin is also head is 1/3. This doesn't make sense to me, given that the two coins is independent and H/T chance is 50:50.
Click to expand...

yes, the answer to your coin question is also 1/3, assuming i don't know which coin you showed me... you just showed me a coin and it was heads. because, conditional on this information, i know the possibilities are HT, TH, HH.

on the other hand, if you tell me that the first coin came up heads, then it is 1/2.
 
peterruse said:
yes, the answer to your coin question is also 1/3, assuming i don't know which coin you showed me... you just showed me a coin and it was heads. because, conditional on this information, i know the possibilities are HT, TH, HH.

on the other hand, if you tell me that the first coin came up heads, then it is 1/2.
Click to expand...

Put it this way, person A tossed one coin, person B tossed one coin. B showed you the coin, the coin is tail (T) for example, so how in the world the coin tossed by person A is affected by the result showed by B?
 
So now if I toss two coins, I reveal one coin to you, the coin is head for example, so the probability of the other coin is also head is 1/3. This doesn't make sense to me, given that the two coins is independent and H/T chance is 50:50.
Click to expand...

Jacjsc. Look. Do you know why in binomial distribution there appears the combinations formula? Because to adjust for the order of outcome. Imagine it this way and Ill make clear why it is 1/3. I just understood the first part of ambigious question that's why I initially answered 1/2.

Now. You flip a coin n times. Let x be the number of success (H or T). Then we have x successes and n-x failures. Let's mark successes as "+" symbol and failures as "-". If for example there are 5 independent trials, each with probability of success P=0.6 and the probability of exactly 3 successes is required. So the outcomes are:

+++-- or +-+-+- or ++--+ ....

The probability of this specific outcome is (0.6^3) *(0.4^2) = 0.03456

But the thing is that we are not concerned with the exact orderings of outcomes, we just need exactly 3 successes regardless of order. So we have to include all number of combinations that can give us 3 successes and 2 failures which is 10. So the true probability is:

C(5,3)*0.03456 = 0.3456 .

This is the case in girl-boy problem. We don't know the ordering of boy-girl. We don't know which one is older. So we have to include all the combination of girl-boy satisfying the condition- "at least one boy". That's why the probability is 1/3, since there are 3 outcomes satisfying the condition and the probability of one of them being chosen is 1/3.
 
Jacksc said:
Put it this way, person A tossed one coin, person B tossed one coin. B showed you the coin, the coin is tail (T) for example, so how in the world the coin tossed by person A is affected by the result showed by B?
Click to expand...

this is different. in this scenario you know *who* tossed a tail. the other guy then has a tail with probability 1/2. this is different from just knowing that one of them got a tail, but not knowing *who*. probability is all about information. the information you have affects your probabilities.
 
So the way I see it is: In the case when I see the child, I don't know whether the child is a younger or older kid, so I have 50% chance of the child being the older and 50% chance of the child being the younger, for each of these cases, the chance for the other child is boy/girl is 50/50. So I can conclude that the probability of the other child being a boy is 50%.

So now if I toss two coins, I reveal one coin to you, the coin is head for example, so the probability of the other coin is also head is 1/3. This doesn't make sense to me, given that the two coins is independent and H/T chance is 50:50.
Click to expand...

For this coin tossing experiment, after revealing you the coin, the probability space is:
HT, HH. There's no TH, so the probability that the other coin is head is 50%
Discuss
 
Jacksc said:
So the way I see it is: In the case when I see the child, I don't know whether the child is a younger or older kid, so I have 50% chance of the child being the older and 50% chance of the child being the younger, for each of these cases, the chance for the other child is boy/girl is 50/50. So I can conclude that the probability of the other child being a boy is 50%.



For this coin tossing experiment, after revealing you the coin, the probability space is:
HT, HH. There's no TH, so the probability that the other coin is head is 50%
Discuss
Click to expand...

Again, you have to think about what is being revealed exactly. That is what you're conditioning on. In your argument, you're not conditioning on the proper information.
 
peterruse said:
Again, you have to think about what is being revealed exactly. That is what you're conditioning on. In your argument, you're not conditioning on the proper information.
Click to expand...

Can you be more clear by what do you mean "you're not conditioning on the proper information" ?
 
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