Poll: A women has two kids, one is a boy, what are the odds the other is also a boy?

[NickIsCool]

Snip

That's a very elaborate way of completely missing the point. It's all language.

Most people would have no difficulty with solving the following:
"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?" (33%)

this question can be paraphrased like the OP

"A woman has two kids, (atleast) one is a boy, what are the odds the other is also a boy?"

but not as

"A woman has two kids, the oldest a boy, what are the odds #2 is also a boy?" This is because it now contains more information: which of the 2 kids the question is about is now defined.

i dont know what part of the language your looking at, but when most people are asked the odds of something, it stops being a language problem, and starts to be math... you know that thing with odds aka probabilities

also, thanks for reading the whole post

i cant tell that you got the part about me saying that the order of the children WAS NOT part of the question

i guess what i was trying to get across was that people where pulling information out of nowhere instead of using what little information the problem gave us

Most people would have no difficulty with solving the following:
"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?" (33%)

apparently you have a problem figuring that one out

based on the information supplied, she either has

two boys
or
a boy and a girl

i dont know where people are getting the order of the children from

would someone please show me in the original posters question where it stated the order of the children?

 

[veloper]

Snip

That's a very elaborate way of completely missing the point. It's all language.

Most people would have no difficulty with solving the following:
"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?" (33%)

this question can be paraphrased like the OP

i dont know what part of the language your looking at, but when most people are asked the odds of something, it stops being a language problem, and starts to be math... you know that thing with odds aka probabilities

also, thanks for reading the whole post

i cant tell that you got the part about me saying that the order of the children WAS NOT part of the question

i guess what i was trying to get across was that people where pulling information out of nowhere instead of using what little information the problem gave us

Most people would have no difficulty with solving the following:
"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?" (33%)

apparently you have a problem figuring that one out

based on the information supplied, she either has

two boys
or
a boy and a girl

The odds:
2 girls - 0%, because we're told.
2 boys - 33%
boy AND girl - 66%.

 

[NickIsCool]

snip

The odds:
2 girls - 0%, because we're told.
2 boys - 33%
boy AND girl - 66%.[/quote]

where does it tell you that either child came first?

... oh wait

i just realized my mistake

so you where right about your question

but the answer to the OPs question is still 50%

 

[veloper]

The odds:
2 girls - 0%, because we're told.
2 boys - 33%
boy AND girl - 66%.

where does it tell you that either child came first?

... oh wait

i just realized my mistake

so you where right about your question

but the answer to the OPs question is still 50%

Both are the same question


"A woman has two kids, (atleast) one is a boy, what are the odds the other is also a boy?"

EQUALS

"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?"

The odds of getting the same uncertain thing twice in a row are lower.
Lower than the odds of getting [first A and then B OR first B and then A] combined.

 

[TZer0]

Interesting. However it is flawed. She has a boy. So you do not select 'one of the families with any number of boys', you select -A BOY-. So you do not choose 'one of the pairs that succeeds the criteria' you select 'one of the heads.' The thing is, the double pairs of heads have two heads, and thusly you're more likely to choose one of those pairs than a pair of tails/heads or a pair of heads/tails.
(...)
John and Fred being of equal probability, means that outcome has twice as many occurances as Bill/Tracy or Jill/Don.

First of all.. never, ever split someone's post up in pieces when quoting. Delete as much as you want of it, remove parts which you don't want to comment, but don't split it up, keep it in one quote. It makes things so very unreadable when you don't.

Obviously, you haven't understood the program. while c1 == "girl" and c2 == "girl": causes the loop underneath to run while we're getting two girls, an impossible choice, the loop runs down to the line c2 = "girl". When we get something that isn't two girls, we go out of the loop and then if we have two boys, we add one to the counter using the if-test. Thus, we're picking two tails. In other words, the counter increases only if we get two boys and the experiment is re-run when we get two girls. (The only error I can see in my program is the fact that there are two more occurrences for boy than girl in the random.random()<0.5-test. However, the random-samples are so big that adding two on one side isn't going to change the numbers in any significant way.)

Next thing, don't exemplify such statistic-boards with different names. This isn't needed. You've set up the exact same board as me, however, you interpreted it incorrectly.
Let me illustrate using a spreadsheet:

There is no way B-B is going to occur more often or just as often B-G and G-B together. Wikipedia is against you, my numerical solution is against you, my spreadsheet is against you.

 

[NickIsCool]

The odds:
2 girls - 0%, because we're told.
2 boys - 33%
boy AND girl - 66%.

where does it tell you that either child came first?

... oh wait

i just realized my mistake

so you where right about your question

but the answer to the OPs question is still 50%

Both are the same question


"A woman has two kids, (atleast) one is a boy, what are the odds the other is also a boy?"

EQUALS

"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?"

The odds of getting the same uncertain thing twice in a row are lower.
Lower than the odds of getting [first A and then B OR first B and then A] combined.

wrong

they are not the same question

with the original question we KNOW that one of them is a boy

we need the probability of the other one being a boy

and with us knowing that, regardless, one is a boy, we have two options for the other one

 

[DracoSuave]

Obviously, you haven't understood the program.

No I understand the algorithm perfectly, but it is based on a flawed premise, that being, that you can reduce the weight of outcomes without increasing the weight of other outcomes for the -exact same reason.-

A failure to agree with the basis of your algorithm isn't a failure to understand it. In fact, I believe I pointed out the problem.

So I will repeat it again so you can understand the counter argument.

You do not select a random family. You select a random -boy-. Then you see if he has a brother or a sister.

If you do a sampling of all -boys- then you arrive at the correct answer to the question.

Also this is flawed:

'If we remove 0.25 we must distribute evenly.'

Why? You are removing it because of zero boys being in the G-G case. But why must you distribute it evenly? Answer that. 'Just because' isn't an answer.

But regardless, in your algorithm, as much as you hate to admit it, 50% of your boys have a brother, and 50% of your boys have a sister. Do the math. Really -do the math.-

You'll find it works out that way.


---------

You know you have a boy. So, statisticly, you look at all boys with a single sibling, and you take a sampling of them. The problem with your algorithm is that it excludes, systematicly, any boy who is a younger sibling of a boy. Why? Why do you not include them? The question does not. So you should not.

 

[Socken]

Something similar. [http://en.wikipedia.org/wiki/Monty_Hall_problem]

 

[TZer0]

The odds:
2 girls - 0%, because we're told.
2 boys - 33%
boy AND girl - 66%.

where does it tell you that either child came first?

... oh wait

i just realized my mistake

so you where right about your question

but the answer to the OPs question is still 50%

Both are the same question


"A woman has two kids, (atleast) one is a boy, what are the odds the other is also a boy?"

EQUALS

"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?"

The odds of getting the same uncertain thing twice in a row are lower.
Lower than the odds of getting [first A and then B OR first B and then A] combined.

wrong

they are not the same question

with the original question we KNOW that one of them is a boy

we need the probability of the other one being a boy

and with us knowing that, regardless, one is a boy, we have to options for the other one

NickIsCool, you're wrong.
Without any information, two children:
B-G
B-B
G-B
G-G
Both questions narrow it down to
B-G
B-B
G-B
Because they only exclude the G-G-option.

 

[Allan53]

The birth of each is an independent event. For simplicity's sake, it's the same as flipping a coin, and we're being asked the chance that we're getting two of a kind. 50% per result for each of 2 results = .5 * .5 = 25% chance that the results will be the same.

Nope. The odds of the results being the same is 50%.

1. The result of the first is irrelevant, as long as it matches the second.

2. The second (assuming individual events) has a 50% of being any particular result. One of these matches the first, the other doesn't.

Therefore, 50% chance of matching. There is a 25% of being both boys or both girls (25% each, that is, for 50% total of a match).

 

[The_root_of_all_evil]

If you're using statistics to solve a real world problem with laboratory guidelines, you either fail math or are doing junior homework. Or, as in the above case, you're using wordplay to go n'yeh, which is rude.

 

[Zac_Dai]

Already been asked, and spawned a 31 page, Norman Conquest's worth of comments (1066 at last count) thread:

http://www.escapistmagazine.com/forums/read/18.73797#1519525

Yeah I checked the wiki article on it, interesting read. Not that it will stop our elite escapist intelligentsia from arguing about it for another 31 pages again though lol.

Also just noticed your profile pic. Your debating style makes more sense now.

 

[DracoSuave]

Something similar. [http://en.wikipedia.org/wiki/Monty_Hall_problem]

The problem with that:

Let's say you choose door 1, door 3 has a goat. If it is advantageous to switch to door 2, then that means it is more likely to have a car.

Let's say you have chosen door 2, door 3 has a goat. The same argument says it is advantageous to switch, as it is more likely to have a car.

Both doors are therefore more likely to have a car.

That doesn't make any sense, mathematically, or logically.

Choosing to switch means that you are re-choosing door 1 or door 2. That mathematical proof falls under its own weight under reductio ad absurdum.

Hmmm...

18 different outcomes--

1/3 of the time you guess the correct spot right...

Oh. I see where I made the mistake. I weighed it wrong.

See, if you have a 1/3 chance of guessing the correct spot right, and under -every- possibility you see a goat, then you have a 2/3 chance of the other spot being a car under all possibilities. The governing random chance is not in the goat spot, but in the spot you picked, and the goat spot will not affect that one bit...

...unless the situation exists where they open up a spot that has a car, in which case, you lose then and there, but you're no longer at 1/3 if you see a goat.

Wow. Interesting.

 

[veloper]

*snipping double quote*

The odds:
2 girls - 0%, because we're told.
2 boys - 33%
boy AND girl - 66%.

where does it tell you that either child came first?

... oh wait

i just realized my mistake

so you where right about your question

but the answer to the OPs question is still 50%

Both are the same question


"A woman has two kids, (atleast) one is a boy, what are the odds the other is also a boy?"

EQUALS

"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?"

The odds of getting the same uncertain thing twice in a row are lower.
Lower than the odds of getting [first A and then B OR first B and then A] combined.

wrong

they are not the same question

with the original question we KNOW that one of them is a boy

You mean to say, that you would not be able to figure out that:
(atleast) "one of them is a boy"

when you are given the following information:

"A woman has two kids, she does NOT have TWO girls"?

 

[Amnestic]

If you're using statistics to solve a real world problem with laboratory guidelines, you either fail math or are doing junior homework. Or, as in the above case, you're using wordplay to go n'yeh, which is rude.

Grats on 10k posts by the way Mr. Evil.


Who on earth voted for 100%? There's three of you out there.

 

[The_root_of_all_evil]

That doesn't make any sense, mathematically, or logically.

It does, but there's an extra piece of information involved.

The door chosen to remove is chosen by the host (who is aware of the statistics), so he is manipulating your choice.

The other door IS more likely to have a car mainly because the host HAS to pick the goat, removing that choice from your switch.

If the host doesn't know though, it drops back down to 50/50.

Grats on 10k posts by the way Mr. Evil.

Cheers

 

[jantunen]

50% the first is a boy, and one the second occasion there is a 50% chance it is a boy
but when both need to happen, its 0,5x0,5=0,25=25%

 

[NickIsCool]

The odds:
2 girls - 0%, because we're told.
2 boys - 33%
boy AND girl - 66%.

where does it tell you that either child came first?

... oh wait

i just realized my mistake

so you where right about your question

but the answer to the OPs question is still 50%

Both are the same question


"A woman has two kids, (atleast) one is a boy, what are the odds the other is also a boy?"

EQUALS

"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?"

The odds of getting the same uncertain thing twice in a row are lower.
Lower than the odds of getting [first A and then B OR first B and then A] combined.

wrong

they are not the same question

with the original question we KNOW that one of them is a boy

we need the probability of the other one being a boy

and with us knowing that, regardless, one is a boy, we have to options for the other one

NickIsCool, you're wrong.
Without any information, two children:
B-G
B-B
G-B
G-G
Both questions narrow it down to
B-G
B-B
G-B
Because they only exclude the G-G-option.

yes both questions narrow it down to that, but the original question narrows it down even more

"A women has two kids, ONE IS A BOY, what are the odds the other is also a boy?"

now we know for certain one is a boy

"A women has two kids, one is a boy, what are the odds THE OTHER is also a boy?"

that leaves us with either
B-B
or
B-G

because we KNOW that one is a boy and we are only concerned with the gender of the other one

either way one is a boy, so he is no longer a factor

i have made a diagram to illustrate my point

the B on the left represents the boy we KNOW the gender of

the B and G on the right represent the possible outcomes for THE OTHER childs gender

 

[DracoSuave]

That doesn't make any sense, mathematically, or logically.

It does, but there's an extra piece of information involved.

The door chosen to remove is chosen by the host (who is aware of the statistics), so he is manipulating your choice.

The other door IS more likely to have a car mainly because the host HAS to pick the goat, removing that choice from your switch.

If the host doesn't know though, it drops back down to 50/50.

Grats on 10k posts by the way Mr. Evil.

Cheers

Yeah I realized that under rigorous mathematical analysis.

However, that only bolsters the 50/50 argument for the mom with a boy... as choosing by family statistically excludes one of the boys from a double-boy pairing each and every time.

 

[gmacarthur81]

I love how many people are reading so much into a simple probability question.

Unrelated events are just that, unrelated. There is no way that they can have an impact on one another. Case in point.

A woman has 1 child that is a boy. Her father's third cousin's niece also has 1 child. What are the chances that her father's third cousin's niece is a boy?

There is no difference between the OP's question and my own, both are asking the probability of an independant event. By the way you are working out the math with your statistical table then the chances of her having a boy would be 33% when in REALITY we know that the chance would be 50% (taking out any genetic predispos).

The premise you are going on is "cum hoc ergo propter hoc" or correlation proves causation. You are saying since she already had one boy, there is only a 33% chance of a second boy because of the correlation of data inputed to the table says so. Therefore the causation of her other child being a boy is directly related to the other child being a boy when there is no logical premise for this outside of a flawed statistical table.

While a statement like that could be made with a certain statisticaly likelyhood, it is a logical fallacy.