The question seems to be asking:

A family has 2 children. What is the probability that one of them is a girl, given one is a boy?

This is equivalent to:

What fraction of families with exactly 2 children and at least 1 boy (B), have exactly 1 boy (A)?
This is very revealing. Applying what we know about conditional probability:

(A) The probability that a family with 2 children has exactly 1 boy is 2/4 (BG, GB, but not BB, GG)
(B) The probability that a family with 2 children has AT LEAST 1 boy is 3/4 (BB, BG, GB, but not GG)
(A intersect B) This is equivalent to event A (At least 1 boy AND exactly 1 boy)

P(A|B) = P(A intersect B) / P(B) = (2/4)/(3/4) = 2/3



Look at it this way:
In a family of 4, the expected distribution of children would be 2 girls and 2 boys (regardless of the age distribution). From the perspective of one of the boys 2/3 of his siblings are girls. The converse is true from the perspective of a girl.

9/13/2007 3:08:45 PM

This shouldn't be conditional probability, though. The outcome of A does not influence or have any bearing on the outcome of B. So one child is a boy (the king). The probability that the second child is a girl is still 50%. It doesn't magically go up to 66% because the first child was a boy.

[Edited on September 13, 2007 at 6:56 PM. Reason : .]

9/13/2007 6:54:32 PM

2/3. Take it from someone who has done this very question before.

Quote :

"you KNOW the king is male. he can have a younger brother, older sister, or younger sister. 2/3."

Or equivalently:

Quote :

"All possible 2 sibling families: BB, BG, GB, GG. GG is eliminated b/c you know you have a boy. So now your possibilities are BB, BG, or GB. 2/3 of those have a G (sister)."

*****************************************************

WHY ARE

SOME

PEOPLE

ASSUMING

FIRST

CHILD

WAS THE

KING ???


[Edited on September 13, 2007 at 7:44 PM. Reason : ]

9/13/2007 7:41:35 PM

^ read it again.

IF the king is the elder sibling, THEN there is a 50% chance his younger sibling will be G.

IF the king is the younger sibling, THEN there is a 50% chance his younger sibling will be G.

...

now tell me how you get 66.6% from that.

9/13/2007 8:04:41 PM

your reasoning is nice, but the fatal flaw in it is that BB gets counted twice.

so essentially you are looking at:

BB
BB (again)
BG
GB

and so there is a girl 2/4 times = 50%

but the correct way is to write out the sample space:

BB
BG
GB
GG

GG is out, and the current situation could be either of the first three.

of those first three (BB, BG, GB), we have a girl 2/3 of the times.

that's the answer that any math prof would give. i am certain of it enough to bet large amouts of money on it.

(you can also do it as a conditional probability question: P(one child is G | one child is B) = 0.5/0.75 = 2/3)

9/13/2007 8:19:10 PM

^^ You're missing the point entirely. You don't know if the king is the younger or older sibling, so you have to take into account both the probability he has an older sister and the probability he has a younger sister. Because you know there are only two siblings, it doesn't matter if he has a brother who is younger or older. The fourth possible outcome of GG isn't possible in this situation isn't possible because you know the king is male. Therefore the correct probability that his sibling is his sister is P(sister) = 2 outcomes where sibling is sister / 3 possible outcomes. Hence 2/3.

9/13/2007 10:19:14 PM

^, ^^

bah.

9/13/2007 11:17:46 PM

shit's retarded. poorly worded question. are we asking about two children in a family or a single sibling?

9/13/2007 11:22:29 PM

Quote :

"Look at it this way: In a family of 4, the expected distribution of children would be 2 girls and 2 boys (regardless of the age distribution). From the perspective of one of the boys 2/3 of his siblings are girls. The converse is true from the perspective of a girl."

Seriously. To illustrate this point I made a spreadsheet in OpenOffice. There are 5 columns:

Column 1: =IF(RAND()>0.5;"B";"G") - Randomly choses boy or girl with 50% probability
Column 2: same as column 1 - Represents second child

Column 3: =IF(OR(A2="B";B2="B");1;"") - is 1 if there is a boy, blank otherwise
Column 4: likewise, only determines the presence of a girl

Column 5: =IF(D2=C2;1;"") - is 1 if there is a boy AND a girl, blank otherwise

(Sum of column 5) / (Sum of column 3) is the fraction of boys that have sisters in a 2-child family.

First 10 lines of spreadsheet (simulating 384 families):

1st 2nd has boy? has girl? has both? G G 1 # with a m child: 298 G G 1 # with a m+f child: 197 B G 1 1 1 G G 1 Fraction of males that 0.66 G B 1 1 1 have sisters: B B 1 B B 1 B G 1 1 1 G G 1

You can recreate this spreadsheet yourself or I can send it to you. Maybe some of you will understand what's going on better.

9/14/2007 3:36:21 AM

i find your ideas intriguing and i would like to subscribe to your newsletter.

9/14/2007 11:07:26 AM

^^ GG on simulating this dude.

Yup, any time there is a probability problem that causes problems, the best way to find the solution is by simulation in Excel. Like that Monty Python problem with the 3 doors, and a prize behind one of them. The one that caused massive worldwide controversy a decade or so ago.

I got this pm from someone. Posting it here so that everybody can benefit:

*******************************************************************
Hey,

why are you taking both

Quote :
"BG, GB"


into account?

Birth order is not part of the problem, so they are equivalent.

Also, this problem can be proven false empirically. In all families with one boy, the next child born always has .5 chance. If that werent true, the world's pop wouldnt be 50/50 boy girl, which it is.
*******************************************************************

And the answer is BECAUSE birth order is not part of the problem. We don't know if the King is older or the younger sibling, so we have to count both possibilities.

Or as someone just recently said:

Quote :

"^^ You're missing the point entirely. You don't know if the king is the younger or older sibling, so you have to take into account both the probability he has an older sister and the probability he has a younger sister."

Everybody knows this, but is not applying it here:

When you toss a coin, what is the chance of getting 2 heads? Or 2 tails?

Shouldn't it be 1/3 because some of you should be looking only at

HH
TT
HT (or TH as they are "equal")

And the chance of one H and one T is also 1/3?

But we know from our 1st prob course that the sample space is

HH
TT
HT
TH

and so, prob of 2 heads/tails is 1/4, and prob of one of each is 1/2.


THE PROBLEM

IN THIS THREAD

IS NOT DIFFERENT

AT ALL

!!!!!!!!!!!!!!!!!!!!!!!!

9/14/2007 12:35:24 PM

Indeed, the fraction males that were the oldest that have a sister in the above simulation is half. The reason the probability changes is you put another condition on the space that eliminates all doubt about one of the children.

Now let's say you want to consider birth order because the first son is heir to the throne. The answer is NO DIFFERENT. He comes from a family with 2 boys with 1/3 probability, or BG with 2/3 probability.


^ The problem you're referring to is the MONY HALL problem (not Monty Python) http://en.wikipedia.org/wiki/Monty_Hall_problem

9/14/2007 2:15:53 PM

Haha yeah, Monty Hall, not Monty Python

BTW, the answer will be 0.5 ONLY IF you are told that the King was the first child, or the second child, i.e., you are told which number child he is.

9/14/2007 5:35:15 PM

double post

[Edited on September 14, 2007 at 5:36 PM. Reason : ]

9/14/2007 5:35:41 PM

^^ Yes, that is, indeed, correct. Only if you eliminate all doubt about the the king himself does the probability become 50%.

9/14/2007 5:59:23 PM

Quote :

"You're missing the point entirely. You don't know if the king is the younger or older sibling, so you have to take into account both the probability he has an older sister and the probability he has a younger sister.""

goddammit this is gonna drive me nuts.

if you "take into account both the probability he has an older sister and the probability he has a younger sister", then WHY don't you likewise take into account the probability he has an older brother and the probability he has a younger brother

:grr:

i mean i see the argument youre making with the spreadsheet example, but I'm inclined to call that trickery.

i still see it like this:

you have the King. the King is a boy, represented as "B". he has one other sibling, which will be represented as "B" (boy) for a brother, or "G" (girl) for a sister.

there are four combinational possibilities here.

BB
BG
BB
GB

in two of the four possiblities, the king has a brother "B". That's 50%
in the other two of four possibilities he has a sister "G". That's 50%

Im calling statistical shenanigans on the 66% chance.



[Edited on September 14, 2007 at 8:53 PM. Reason : ]

9/14/2007 8:45:34 PM

no. 2/3 is right. if you got up to say 100 guys that are in a two sibling family and be like "hey do you got a sister?", tally up the numbers. about 2/3 of them will say yes. simple survey. or you can just figure it out mathematically and say 2/3.

9/14/2007 9:04:11 PM

you know what?

i dont like your statistics.

I'm going to invent a new branch of statistics.

I'll call it Joestistics.

9/15/2007 12:28:42 AM

Quote :

"there are four combinational possibilities here. BB BG BB GB"

You can look at it this way if you want.

Quote :

"in two of the four possiblities, the king has a brother "B". That's 50% in the other two of four possibilities he has a sister "G". That's 50% Im calling statistical shenanigans on the 66% chance."

The flaw in your logic is the OUTCOMES YOU LISTED ARE NOT OF EQUAL PROBABILITY.

25% of families are GB
25% of families are BG
25% of families are GG
25% of families are BB

Therefore 50% of families have 1 boy and one girl. 25% have 2 boys. The sample space in this problem is 75% of the above sample space, because there is no possible GG. Normalize the probabilities by dividing by the sum over the mass function (0.75) and you find that 2/3 of the families in the sample space have 1 girl and 1 boy, while one third has 2 boys.

In your description you count BB as 50% of your sample space. A family with 2 boys is only half as likely as a family with a boy and a girl. This is the point you're overlooking. Your description is akin to this:

Consider a random male, who is known to have exactly one sibling. The probability mass function of the sibling looks like this:

- Older brother (BB): 16.67%
- Younger brother (BB): 16.67%
- Older sister (GB): 33.33%
- Younger sister (BG): 33.33%



Don't call the spreadsheet trickery. It simulates a random group of families with 2 children and then asks each of the males "Do you have a sister" and 2/3 of them say yes. This is not statistical shenanigans. It is a matter of FACT.


I don't understand why people have so much trouble comprehending statistics. I never took statistics until recently and I knew how to do it all. It's all commons sense as far as I can tell. I don't even know why they give it names like "Bayes' Theorem" and shit. I've been using this shit since elementary school.

[Edited on September 15, 2007 at 4:54 AM. Reason : ]

9/15/2007 4:48:16 AM

i have a confession to make.

i had one ST class required for my major. (ST 371, i think)

i went the first week, then didnt show up again til final exam.

i accepted the C- and moved along.

9/16/2007 3:29:20 PM

so your grade was only based on 1 final exam? no tests? good job.

[Edited on September 16, 2007 at 7:39 PM. Reason : final]

9/16/2007 7:39:14 PM

i know i missed a few tests, which forced me to have the final weighed heavier. and there was no graded HW. but yeah, maybe i did show up for one test during the semester.

thanks for keeping score.

9/16/2007 10:34:08 PM

np, its what i do

9/16/2007 10:44:05 PM

^^^^ j00 R PWNT

9/16/2007 11:53:36 PM

pwnt? haha no way.

this just gives me more credibility when I submit my "Joestistics" to the journals.

9/17/2007 9:15:49 AM

I agree - the plane will not take off. :-)

[Edited on September 17, 2007 at 7:03 PM. Reason : Edited to remove all probability discussion, since this question is so screwy and requires way too m]

9/17/2007 6:58:18 PM

^nope, got the homework back today. we were right, 2/3.

9/17/2007 7:03:42 PM

you should dispute the grade, and submit this thread to the office of Academic Integrity as evidence.

9/17/2007 8:25:34 PM

why dispute? i got it right

9/17/2007 11:35:01 PM

i say you got it wrong.

9/18/2007 2:03:01 AM

1/2

9/18/2007 3:49:18 AM

anyone who STILL believes the answer is 1/2 after i repeatedly and clearly showed why the answer is 2/3, should promptly kill himself (or herself as the case may be) for failing at life.

9/18/2007 12:57:25 PM

YOUR statistics shows it's 2/3, yes.

JOESTISTICS shows it's 1/2.

i'll be submitting a formal paper to the journals shortly. stay tuned.

9/18/2007 1:00:57 PM

anybody who still thinks the plane will not take off is a retard.

9/18/2007 5:21:21 PM

^ The plane obviously did take off a few days ago, and rosschilen obviously missed it.

9/18/2007 9:18:03 PM

monty python problem

9/18/2007 10:01:48 PM

mount my python problem

9/18/2007 10:07:06 PM

vpython problem

9/19/2007 11:34:05 AM

you guys can twist and turn this problem as much as you want, but you have to look at it how its worded. How its worded, the answer is 100%. If you think about it in real terms, the answer in 50%. Order is never mentioned so its not 67%

9/19/2007 12:04:52 PM

To anybody still confused by the 2/3 answer:

Do you agree that out of all families with two children, having a boy and a girl is twice is likely as having two boys?

25% of all such families have two boys.

25% of all such families have two girls.

50% of all such families have a boy and a girl.

Therefore, if there's at least one boy in such a family, his sibling is twice as likely to be a sister than a brother.

9/19/2007 1:03:10 PM

^ i was going to write some smartass comment about 'welcome to last week' ...

but actually thats a REALLY good, and concise, explanation.








[Edited on September 19, 2007 at 1:30 PM. Reason : ]

9/19/2007 1:24:10 PM

Quote :

"you guys can twist and turn this problem as much as you want, but you have to look at it how its worded. How its worded, the answer is 100%. If you think about it in real terms, the answer in 50%."

I can see why you and some others are sticking with the 50%, but it is very stupid to say the 100% thing. How can you say 100%... I don't see it.

Quote :

"Order is never mentioned so its not 67%"

BECAUSE order is never mentioned, the answer is 2/3. IF ORDER WAS MENTIONED, the answer would be SMALLER.

IF YOU ARE TOLD for a fact which order the King was born (1st or 2nd), then the answer becomes 1/2. BECAUSE the order is not told, the answer is BIGGER.

The same reason permutations are always more than combinations.

If I gave you 5 pictures to put on the wall, and I said you could choose any 3 to put up, the number of ways you could choose 3 pics out of 5 is 10 (combinations). But if you counted each order separately (for eg., 123 is different from 213), then the answer is 60. (permutations).

FIXING the order DECREASES the possible choices. NOT FIXING the order INCREASES the possible choices.

9/19/2007 4:30:41 PM

i haven't read this thread, but i have a feeling that whatever 0EPII1 suggests is off base

http://www.brentroad.com/message_topic.aspx?topic=489649

9/19/2007 4:57:44 PM

^no, what he posted is, in fact, entirely correct. also, his statement that for the set 1,1,2,2,3,3,4,4, the mode is a meaningless quantity (whether it exists by definition or not) is also correct.

Quote :

"^ i was going to write some smartass comment about 'welcome to last week' ... but actually thats a REALLY good, and concise, explanation. "

I gave the exact same explanation:

Quote :

"The flaw in your logic is the OUTCOMES YOU LISTED ARE NOT OF EQUAL PROBABILITY. 25% of families are GB 25% of families are BG 25% of families are GG 25% of families are BB Therefore 50% of families have 1 boy and one girl. 25% have 2 boys. "

[Edited on September 19, 2007 at 5:02 PM. Reason : ]

9/19/2007 4:59:17 PM

^^ Yea, except that everybody (including all professors) agree with what I am saying (and vice versa), save for a couple of molasses-brained folks.

So maybe you should read the thread before you make yourself appear foolish/vindictive.



[Edited on September 19, 2007 at 5:02 PM. Reason : ]

9/19/2007 5:01:04 PM

in the set {1,1,2,2,3,3,4,4} there are 4 modes {1,2,3,4} however, in this case (and many others), while the mode exists by definition, it is an entirely meaningless quantity, which is why the confusion arises. the only time the mode(s) can be interpreted as having significance is when the number of modes is small compared to the cardinality of the set

9/19/2007 5:04:17 PM

^^ i heard you still piss the bed.






[Edited on September 19, 2007 at 5:05 PM. Reason : damn, three seconds ]

9/19/2007 5:04:20 PM

1 - Umm, ok? What does that have to do with this thread?
2 - You heard wrong. I don't still piss the bed.
3 - Shut up or I will piss in your butt.


^^

1st sentence: exactly

2nd sentence: i would change it to "much smaller than the cardinality of the set". for example, for 1,1,2,2,3,3,4,4,5 the modes are 1,2,3,4. and still, quite meaningless.



[Edited on September 19, 2007 at 5:15 PM. Reason : ]

9/19/2007 5:11:06 PM

[Edited on September 19, 2007 at 5:15 PM. Reason : ]

9/19/2007 5:14:42 PM

i know i started this thread, but it needs to die. people who think it is 50% are going to be wrong and stick with it, but just let this thread die.

9/19/2007 6:59:06 PM