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What are the odds that the other child is a boy?

 
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Re: Probability Question

Postby mightyal on Sat Sep 16, 2006 2:00 pm

I agree that the wording is important.
sully800 wrote:A woman has 2 children. One of them is a boy.

Clear as day, if one is a boy, then one is a girl.
0%
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Re: Probability Question

Postby sully800 on Sat Sep 16, 2006 6:54 pm

mightyal wrote:I agree that the wording is important.
sully800 wrote:A woman has 2 children. One of them is a boy.

Clear as day, if one is a boy, then one is a girl.
0%


If both are boys, then by default one is also a boy. It certainly doesn't say ONLY one is a boy.
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Postby sully800 on Sat Sep 16, 2006 7:00 pm

Nappy Bone Apart wrote:This is a totally different situation. The only way your logic would work in your original question is if both children were born at the same time.


The thing is though, you have no frame of reference for time. There is nothing that specifies which child was born first, the known boy or the unknown child. So it doesn't matter that they aren't twins (actually if they were twins and one was a boy the chances of another boy would be MUCH higher....but thats a whole different story.)



And Stopper, I like how you provide that link from wiki, and it basically proves the 1/3 allegation.

wikipedia article wrote:Second question

* A two-child family has a boy. What is the probability that it has a girl? Or that the other is a girl?

When we know that one child is a boy; there cannot be two girls, so {GG} is to be eliminated and the sample space shrinks to:

{BB, BG, GB}.

Since there are two combinations out of three that include girls, the probability that the family has a girl is 2/3.


This question is worded the same way, except instead of looking for the other child to be a boy, they ask what are the chances that its a girl. Simple logic tells you that if its a 2/3 chance for a girl (as the article says) its a 1/3 chance for a boy.
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Postby sully800 on Sat Sep 16, 2006 7:01 pm

And if you think my wording is different than their wording (other than the final question) please explain how. Because I view them as exactly equivalent.
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Postby Stopper on Sat Sep 16, 2006 8:38 pm

Well, I suppose the link I provided sort of proved the 1/3 allegation, but like I say, your wording is confusing...

Your question was:

A woman has 2 children. One of them is a boy. What is the probability the other is a boy?

And as Wikipedia says:

"A two-child family has at least one boy. What is the probability that it has a girl?

There are variations in the exact wording; often the second question confusingly asks about the "other child"."

You also said "one of them is a boy", instead of "at least one is a boy", which confused things further. After that, it's 50/50 what the other child is. Unless you're a hyperliteralist, and you assume the other MUST be a girl, in which case the probability of a boy is 0%. :wink:
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Postby sully800 on Sat Sep 16, 2006 9:36 pm

Yeah, I understand how its confusing. It really should say "at least one" but I thought that was implied (since it wouldn't be much of a question without it.

As for "the other child" being confusing, I don't find that part confusing at all. But whatever I trust wiki :P

The answer is the same either way. If you think "one is a boy" means one and only one then the answer is 0%. Otherwise it would be 33%. No 50%!
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Postby mightyal on Sun Sep 17, 2006 3:36 am

I learned this as the gameshow question.

On a gameshow, you are shown three boxes and asked to choose one. Two are empty; the other contains the grand prize. After makiing your choice, the host opens one of the other 2 boxes and shows you that it is empty. He offers you the option of sticking with your original choice or switching. Well?
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Postby Stopper on Sun Sep 17, 2006 6:05 am

What, are there no goats involved?
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Postby Pacifist on Sun Sep 17, 2006 6:38 am

We know that one of the children is a boy, and despite the fact that this leaves three options: BB, BG and GG, the probability is not 33.3%. We know that one child is a boy, so we have B?. This can either be BB or BG, thus the odds are 50:50, so we get 50%.

For those of you who have decided to say that we have three options, I also have another explanation:

There are four options (each time we know the bold child):
BB, BG, BB and GB!!!

The genetics actually show that you are more likely to have boys than girls, and you are more likely to have two children of the same gender than one of each.
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Postby sully800 on Sun Sep 17, 2006 7:15 pm

Pacifist wrote:We know that one of the children is a boy, and despite the fact that this leaves three options: BB, BG and GG, the probability is not 33.3%. We know that one child is a boy, so we have B?. This can either be BB or BG, thus the odds are 50:50, so we get 50%.

For those of you who have decided to say that we have three options, I also have another explanation:

There are four options (each time we know the bold child):
BB, BG, BB and GB!!!

The genetics actually show that you are more likely to have boys than girls, and you are more likely to have two children of the same gender than one of each.


Now that's some faulty logic :roll:
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Postby AndyDufresne on Sun Sep 17, 2006 7:18 pm

mightyal wrote:I learned this as the gameshow question.

On a gameshow, you are shown three boxes and asked to choose one. Two are empty; the other contains the grand prize. After makiing your choice, the host opens one of the other 2 boxes and shows you that it is empty. He offers you the option of sticking with your original choice or switching. Well?


I've heard that similar thing. I remember hearing it's better to change your selection, due to the fact that you had a 2/3 chance of being wrong when you made your first selection. It improves your odds to change your selection...or something similar to that?

Let's Make a Deal was the gameshow.


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Postby Nappy Bone Apart on Sun Sep 17, 2006 10:41 pm

If you're relying on wikipedia to prove your theory, remember that wikipedia entries can, and are, changed by random people who think they know what they're talking about. So wikipedia proves zilch.

And if you're going to do logic problems, NOTHING can be implied. Logic depends on every fact you give, and you can't base anything on what you don't give. Thus the confusion here. As you said, you're given no reference for time, so every answer given in this thread can be correct. So what does that prove?
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Postby owheelj on Sun Sep 17, 2006 11:14 pm

Carefully read:

A woman has 2 children. One of them is a boy. What is the probability the other is a boy?


-----

First we need to assume that if a person has 1 child the chance of it being a boy or a girl is 50%. In fact it appears that in todays society it is slightly more likely for it be a girl but not statistically relevent in this question I think.


So a woman has 2 children, we know 1 is a boy. What is the probability that the other is a boy? Well there are two genders to choose from - boy or girl. We're asking what the probability of a boy is - that's 1 particular event.

Therefore the probability is 1 in 2 or 50%.

The fact that the woman has 2 children is irrelevent. You could word the question as

A woman has 200 children. 199 of them is a boy. What is the probability the other is a boy?

and the answer is still 50%.

A womans body does not remember what she gave birth to last time so it's a 50% chance of every baby being boy or girl. Of course the chance of having 2 boys is 1/2 x 1/2 = 1/4 or 25% but the question isn't asking us what the chance of both children being boys, just what the chance of the gender of one being a boy is.

People have put foward the arguement that there are four options - b,b b,g g,b and g,g and since we know that one is a boy we can discount the g,g option.

In fact we also have to discount either b,g or g,b. We are specifically told that one child is a boy and asked what the gender of the other is. If we know that one is a boy then we know that either the first child or the second child is a boy. If the first child is a boy - we have the options of b,b or b,g. g,b or g,g aren't possible because we know that the first child is a boy. Hence there is a 50% chance that the other child is a boy. If instead the second child is a boy then our options are b,b or g,b. b,g and g,g are not options since we know that the second child is a boy. Again 50% chance. Either way it is impossible for both b,g and g,b to be options.
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Postby vtmarik on Mon Sep 18, 2006 1:06 am

Nappy Bone Apart wrote:If you're relying on wikipedia to prove your theory, remember that wikipedia entries can, and are, changed by random people who think they know what they're talking about. So wikipedia proves zilch.


Wikipedia has a very critical core group that maintains the site and stops unverfied or incorrect information from burrowing too deeply. Whenever a page gets vandalized or attacked and altered radically, the page is reset to an earlier version and locked.

Either check your facts, or stop spouting useless rhetoric.
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Postby Nappy Bone Apart on Mon Sep 18, 2006 9:56 am

vtmarik wrote:
Nappy Bone Apart wrote:If you're relying on wikipedia to prove your theory, remember that wikipedia entries can, and are, changed by random people who think they know what they're talking about. So wikipedia proves zilch.


Wikipedia has a very critical core group that maintains the site and stops unverfied or incorrect information from burrowing too deeply. Whenever a page gets vandalized or attacked and altered radically, the page is reset to an earlier version and locked.

Either check your facts, or stop spouting useless rhetoric.


I'm not. Using wikipedia as your only source to prove a theory leaves much to be desired. That "core" can't check every page once a year, much less a week or whatever. wikipedia is a great resource, but I wouldn't accept it as law.
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Postby sully800 on Mon Sep 18, 2006 5:57 pm

I wasn't using the wiki article as a source to prove my case. Someone else posted it saying it shows that 50% is the correct answer. I say it verifies that the correct answer is 33%
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Postby sully800 on Mon Sep 18, 2006 5:58 pm

owheelj wrote:In fact we also have to discount either b,g or g,b. We are specifically told that one child is a boy and asked what the gender of the other is. If we know that one is a boy then we know that either the first child or the second child is a boy. If the first child is a boy - we have the options of b,b or b,g. g,b or g,g aren't possible because we know that the first child is a boy. Hence there is a 50% chance that the other child is a boy. If instead the second child is a boy then our options are b,b or g,b. b,g and g,g are not options since we know that the second child is a boy. Again 50% chance. Either way it is impossible for both b,g and g,b to be options.


So you're making 2 options with 2 options each? And you don't think that affects your results?
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Postby sully800 on Mon Sep 18, 2006 6:03 pm

Nappy Bone Apart wrote:And if you're going to do logic problems, NOTHING can be implied. Logic depends on every fact you give, and you can't base anything on what you don't give. Thus the confusion here. As you said, you're given no reference for time, so every answer given in this thread can be correct. So what does that prove?


The whole point is that there is no time reference. If it said the first child is a boy, what are the chances that she has another boy....the puzzle wouldn't fool anyone. Because as owheelj just took a very long time to repeat, the incidents would be unrelated.

I agree that the logic of the puzzle depends on every fact given. That's why saying "one child is a son" is sufficient. That statement does NOT say ONLY one child is a son. I agree that its better worded as "at least one child is a son" but the statements are equivalent if you only use the information given to you.
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Postby foutballfreek15 on Mon Sep 18, 2006 6:05 pm

It matters on the father. Some fathers can only have boys. While others can only have girls. then others can have both.
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Postby sully800 on Mon Sep 18, 2006 10:05 pm

:D
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Postby maniacmath17 on Mon Sep 18, 2006 11:53 pm

Sully is right, its 33% that the woman has two boys. If you can't figure out why, imagine flipping 2 coins 100 times and make a list of the results.

Statisticaly you should get 25 times where they came up heads heads, 25 tails tails, and 50 one of each.

Now you are told to only keep the results with atleast one heads. That means you take the 25 out that are tails tails. From this information, you are asked how many in this list have two heads. The answer is 25 out of 75, which is 33%

I think the question should read: Given that a woman has two children and atleast one is a boy, what are the chances that both are boys?
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Postby owheelj on Tue Sep 19, 2006 11:01 am

sully800 wrote:
owheelj wrote:In fact we also have to discount either b,g or g,b. We are specifically told that one child is a boy and asked what the gender of the other is. If we know that one is a boy then we know that either the first child or the second child is a boy. If the first child is a boy - we have the options of b,b or b,g. g,b or g,g aren't possible because we know that the first child is a boy. Hence there is a 50% chance that the other child is a boy. If instead the second child is a boy then our options are b,b or g,b. b,g and g,g are not options since we know that the second child is a boy. Again 50% chance. Either way it is impossible for both b,g and g,b to be options.


So you're making 2 options with 2 options each? And you don't think that affects your results?


No it's 2 options of 1 option each - that's to say if you choose either option there is only one choice.
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Postby OwlLawyer on Tue Sep 19, 2006 12:51 pm

This is the classic gamblers fallacy.

If you are playing roulette and the wheel is spun 100 times, and someone says, "What are the chances that 100 spins in a row will be red?" Well, the answer is ASTRONOMICAL. BUT... if the wheel has already been spun 99 times, and it has been read every time (assuming that this is random chance and the wheel is not tilted or broken or something), then what is the chance that the next spin will be red?

Well, it's approximately 50% (remember 0 and 00 are not red). It's not astronomical, because you are asking about one event.

If, before either child was born, you asked, what are the chances that a woman has two boys, then you would be right, but when asking about ONE event, it's still 50%.
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Postby Nappy Bone Apart on Tue Sep 19, 2006 12:56 pm

maniacmath17 wrote:Sully is right, its 33% that the woman has two boys. If you can't figure out why, imagine flipping 2 coins 100 times and make a list of the results.

Statisticaly you should get 25 times where they came up heads heads, 25 tails tails, and 50 one of each.

Now you are told to only keep the results with atleast one heads. That means you take the 25 out that are tails tails. From this information, you are asked how many in this list have two heads. The answer is 25 out of 75, which is 33%


Same fallacy we've been discussing all thread. However....

maniacmath17 wrote:I think the question should read: Given that a woman has two children and atleast one is a boy, what are the chances that both are boys?


There we go. This I can get behind.
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Postby Nappy Bone Apart on Tue Sep 19, 2006 12:56 pm

OwlLawyer wrote:This is the classic gamblers fallacy.

If you are playing roulette and the wheel is spun 100 times, and someone says, "What are the chances that 100 spins in a row will be red?" Well, the answer is ASTRONOMICAL. BUT... if the wheel has already been spun 99 times, and it has been read every time (assuming that this is random chance and the wheel is not tilted or broken or something), then what is the chance that the next spin will be red?

Well, it's approximately 50% (remember 0 and 00 are not red). It's not astronomical, because you are asking about one event.

If, before either child was born, you asked, what are the chances that a woman has two boys, then you would be right, but when asking about ONE event, it's still 50%.


*applause*
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