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# One Girl - One Boy

## Question

Guest

Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

Assume safely that the porbability of each gender is 1/2.

Ofcourse its not 1/2 else would make it a lousy puzzle...

Ans: 1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Girl - Girl

Girl - Boy

Boy - Girl

Boy - Boy

Since we know one of the children is a girl, we will drop the Boy-Boy possibility from the sample space.

This leaves only three possibilities, one of which is two girls. Hence the probability is 1/3

## 348 answers to this question

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The probability that the second child is a girl is 50%. Just as the probability that the first kid was a girl was 50%. Just as the probability that the 6th kid they have (should they be so inspired) would also be a girl would be 50% too.

The probability of having two children that are both girls, now that's a different story. There, the probability would be 25%

The puzzle gives the gender of the first kid as a given. The probability of any one child being a girl is 50%. First, Second, Sixth, it doesn't matter.

50%. Just like a coin toss. On any given flip, the probability is (theoretically) 50% that it will come up heads. The previous flips have no bearing on the probability of the current given flip.

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The answer is 25%. Before they had children the probability for a girl first is 50%. The probability for a girl second is 50%. BUT, to have 2 girls the probability moves to half of half, or 25%.

Comparing a coin toss is a common mislead. A single coin toss is of course 50/50. But 2 consecutive tosses is 50/50 for the first, and 25/75 for the second. Taken as a set (as you must in a family), you cannot discount the total probability by looking at the single toss.

Much like this:

I have 3 boxes (a - b - c).

2 have nothing in them.

1 has a million dollars.

If you guess the correct one, you get the money.

You pick one.

Out of the remaining 2, I show you that one is of course empty.

Do you have better, worse, or the same odds if you DO change?

Better. The first guess you have a 1 in 3 chance of being right. Once I remove one, your odds move up to 2 in 3 by selecting the OTHER box, because it would have been like selecting the 2 boxes you didn't pick.

If I had offered for you to trade your ONE pick for the TWO you didn't pick, before showing one was empty would you?

Ahhh:

You are right in your "Monty Hall" puzzle, but wrong with the One boy one girl puzzle:

That the couple have 2 children is a given, One is a girl is a given, the variable is only the gender of one child, the second, which is set at 50%. The question isn't what is the chances of any couple having 2 female children, which would in fact be 25%.

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Guest
The puzzle gives the gender of the first kid as a given.

No, it doesn't. Please read the riddle again and this thread in its entirety so the solution and the reasons for it don't have to be explained all over again.

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just read my post near the end of page 5. it explains everything. i think this topic is over. really. it cant go much anywhere else...

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there is a lot of confusion between the two sides of this problem, and i've read all ur arguements, and it isnt about conditionals or conjunctionals or anything

Ok, so Teanchi and Beanchi are a married couple (dont ask me whose he and whose she)!

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

Assume safely that the porbability of each gender is 1/2.

this is the problem.

let me requote the important part (middle line)

"They have two kids, one of them is a girl, what is the probability that the other kid is also a girl."

"they have two kids." They already have them. They're not expecting kids... they already have both of them born. The kids could be 20 and 18 years old for all we care.

"One of them is a girl"

okay... so if we have two kids, already born, and we can rule out B/B:

G/B

B/G

G/G

"what is the probability that the other kid is also a girl."

not: "what is the probability that the other kid will be a girl."

In that case it would be 1/2, since the other child does not affect the probability. But it doesnt say that. It said "is also a girl". The child is already born.

it all comes down to two sides:

Side A (answer is 1/2): We're looking at the second child's birth as an independent event.

Side B (answer is 1/3): We're looking at both together.

and I am sorry to say Side B is right (as I was on Side A for quite some time)

Why?

Because the second child has already been born. We dont care what was the probability that it was boy/girl while it was in its mother's womb. We know thats 1/2. The fact is, the gender has already been determined. We're just trying to figure out the probability they are both girls.

Since the options are:

G/B

B/G

G/G

we can safely say, G/G has a 1/3 chance.

There is no fallacy. There is no bla-bla-bla. It all comes down to wording:

They have two kids, one of them is a girl, what is the probability that the other kid is also a girl.

the "other kid" is already born. The solution doesnt call for the probability that the other kid will be a girl. It calls for the probability that they are (as in right now) BOTH girls.

We're not looking at one girl's chances. In that case there would be two options:

B

G

but we're not. BECAUSE OF THE WORDING, we are looking at both children at once:

G/B

B/G

G/G

It is 1/3.

It pains me to say this (I defended 1/2 for a while until i realized i was wrong) but I'm not being bitter and I'm trying to prove what I now see is right.

its a fine line, based on skale's wording. but the answer is 1/3. I rest my case. I think this topic should be done. The discussion cant go much elsewhere.

Thanks Unreality. And Martini.

It does come down to the wording.

"One of them is a girl"

versus

"One of which is a girl"

The statement in the riddle is not "One of which" though. Nor is it "At least one of them is a girl"

It's a matter of assigning. As soon as you say "one of them" you've assigned, say, the first born, or the second born to be a girl. If one of them, say the first born, is in fact a girl, the chances of the other being a girl are 50%.

in:

G/B

B/G

G/G

it does look to be 1/3. But, by assigning, by saying "one of them is a girl" I submit that you can, in a way, add a second G/G set.

Or, of those 6 kids, 4 are girls. The chances of "one of them" girls having a sister is 50%.

Semantics maybe, but that's how I see the puzzle.

But then again, I may just be "one of them" guys.

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Guest

It does come down to the wording.

"One of them is a girl"

versus

"One of which is a girl"

The statement in the riddle is not "One of which" though. Nor is it "At least one of them is a girl"

It's a matter of assigning. As soon as you say "one of them" you've assigned, say, the first born, or the second born to be a girl. If one of them, say the first born, is in fact a girl, the chances of the other being a girl are 50%.

You have not assigned the first or second as a girl by saying "One of them is a girl". The fact that one is first or second hasn't even been mentioned. Saying "one of them" means simply that, "one of the two". "One of them is a girl" is no different than "one of which is a girl".

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I think it might be a not-so-sneaky way of saying "I read too much into the puzzle"

I thought the puzzle was working with an additional, or deeper twist:

like:

"you have two coins in your pocket totalling 30 cents in value. One of them is not a quarter"

That classic works as a puzzle, specifically because it does assign. So when I looked over the words "one of them" I took that to be a crucial part of the puzzle.

I've been wrong before and I'll be wrong again Dammit.

Sneaky?... not so much.

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It didn't say "one of them is NOT a girl"

it said "one of them is a girl"

which could also mean "one of them is NOT a guy" but that just means that one that isnt a guy, is a girl, which is the same conclusion

no, it was just a probability problem, which, because of wording, has the answer of 1/3

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this is why i have a problem with this aspect of statistical analysis. let's see what we have here.

1] if you know the oldest child is a girl, the probability the other is also a girl is 1/2.

2] if you know the youngest child is a girl, the the probability the other is also a girl is 1/2.

3] if you do not know which child is a girl, the probability changes to 1/3?

sorry, but this defies any logic i understand.

i encountered a problem similar to this called the 3 siblings which gave an even more ridiculous reason for one answer over another. no sense confusing this issue more though.

but ask yourself logically. why does assigning change the probability? and don't just regurgitate all the past answers. i know the statistical processes pretty well and as i said before, this is not just a debate you guys are having. there are many websites dedicated to both sides of this question.

another similar problem is the 2 bears.

2 bears are standing in the woods. one is brown and one is black. you are told that one of them is a male. what is the probability the other is also a male.

virtually the same as this problem. as i said earlier, most mathematicians will agree with the 1/3 answer. but there are many that challenge it and contend it is 1/2. they're having the same debate over it that you're having. in fact marilyn van zant has found out the many people disagree with her 1/3 answer. and it's not just a matter of semantics either.

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this is why i have a problem with this aspect of statistical analysis. let's see what we have here.

1] if you know the oldest child is a girl, the probability the other is also a girl is 1/2.

2] if you know the youngest child is a girl, the the probability the other is also a girl is 1/2.

3] if you do not know which child is a girl, the probability changes to 1/3?

It's not that the probability changes to 1/3, it's that the problem is altogether different.

It becomes easier to understand if you take the time to do the experiment I mentioned earlier using 100 pairs of pennies.

If you randomly lay out 100 pairs of pennies, about 50 of the pairs will have the first penny being heads up. Of those 50 pairs, about half of those will also have the second penny heads up. In other words, following your correct logic in scenario 1, "if you know the oldest child (first penny) is a girl (heads up), the probability the other is also a girl (heads up) is 1/2".

However, about how many of the pairs will have either penny being heads up? About 75 of the pairs will have at least one heads up penny. So, now we have 75 pairs of pennies with at least one heads up penny. Of those 75 pairs, 1/3 of the pairs will be HH, 1/3 of the pairs will be HT, and one third will be TH. So, out of 75 pairs where at least one of the pair is heads up, about 25 0f the pairs will be both heads up. 25 is one third of 75.

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Martini, Unreality - you guys are good logicians. But when good logicians develop blind spots, their ego prevents them from listening to anyone else!

One point I and others have repeatedly tried to point out:

In your example of 100 pair coins, 75 pairs have either one head. Ok fine. BUT WHY? Why 75??? Because the sum total is 100, you are getting 75 out of 100. If it were 200, you would have got 150 out of 200. This 75/150 is dependant on the fact the total is 100/200! But then you remove 25, and out of the 75 you start divisions. You cannot do that! As soon as the total becomes 75, all other numbers change! You cannot first use probability, and then select a portion and start doing statistics on that! As soon as you select, the probability changes! The numbers in a probability is always related to the total.

I tried to state this in various ways - do not mix permutation and combination, etc. Maybe my command of the language is not good; but the point is you cannot declare a probability, and remove some of them, and take the same numbers in another reckoning! As soon as you eliminate some, the probability changes! This is one of the paradoxes you have been stating again and again, so much so, that you firmly believe in it like the holy grail - even if your common sense tells you otherwise!

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Martini, Unreality - you guys are good logicians. But when good logicians develop blind spots, their ego prevents them from listening to anyone else!

Uh huh.

One point I and others have repeatedly tried to point out:

In your example of 100 pair coins, 75 pairs have either one head. Ok fine. BUT WHY? Why 75??? Because the sum total is 100, you are getting 75 out of 100.

Are you saying that the original 100 should still be relevant? It's not. The TT pairs don't matter because the riddle states there is at least one girl. It is then impossible to have a boy/boy (TT) pair. Out of all of the possible penny pairs (after all the TT pairs are removed), 1/3 will be HH, 1/3 will be TH, and 1/3 will be HT. PLEASE try the experiment yourself and then get back to us!

Speaking of blind spots, you have said in this thread that "We are confusing sequential events with simultaneous events" and then went on to state nonsense about how that makes a difference.

You then went on to state "If we toss a coin, the probability of getting a head is 1/2. If we toss two coins together, the probability of getting 2 heads is 1/3 (3 possibilities, head-head, tail-tail and head-tail)" with the illogical premise that HT = TH/HT. If you don't believe that the probability of getting two heads in a row are 1/4, again, PLEASE do the experiment yourself!

If you try the experiment and you are still convinced the pennies are lying, then take a look at the following links:

<!-- m --><a href="http://www.learninghaven.com/articles/r" target="_blank">http://www.learninghaven.com/articles/r ... asers.html</a><!-- m -->

If the above reputable sites aren't enough to convince you that you're wrong, don't tell us that we are the ones having our egos preventing us from listening to someone else.

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Guest

Ok, I finally admit I was wrong. I tried to search for those relevant sites for arguments for or against,, without results.

My sincerest apologies to everyone

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Guest

No apologies necessary. It takes a big man to admit he's wrong. Just glad this one's finally over. Or am I speaking too soon?

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Guest

as i stated, i know statistical analysis pretty well and i understand all the examples that you guys have given. for those of you who have changed sides, you are following the standard view. after you read the argument proposed at this website, tell me again how it's not ambiguous. here's a quote to get you interested. note the source. also note the date. this debate has been going on a long time.

Scientific American Declared this Problem Ambiguous in the 1950's

Nick Ratti Jr. of Bristol, RI, sent me a photocopy of a clipping from "Scientific American, circa 1950's." (There is no page number, date, or other information on the clipping itself.)

Another example of ambiguity arising from a failure to specify the randomizing procedure appeared in this department last May. Readers were told that Mr. Smith had two children, at least one of whom was a boy, and were asked to calculate the probability that both were boys. Many readers correctly pointed out that the answer depends on the procedure by which the information "at least one is a boy" is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says "at least one is a boy." If both are girls, he says "at least one is a girl." And if both sexes are represented, he picks a child at random and says "at least one is a ..." naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. (This is easy to see because the informant makes a statement in each of the four cases -- BB, BG, GB, GG -- and in half of these case both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.

Eldon Moritz reports that he also received a copy of this column, and was told that appeared in the October 1959 issue, and was written by Martin Gardner.

here is the site. it has to do with an 'ask marilyn' question. i'm sure you can guess what the question was but check the link.

this is why you guys have spent so long discussing it. it is ambiguous.

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this is why you guys have spent so long discussing it. it is ambiguous.

No, it's not why we've been spending al ot of time discussing this. We've been spending a lot of time discussing it because of others misunderstanding probability.

Look again at what your article says:

"Many readers correctly pointed out that the answer depends on the procedure by which the information "at least one is a boy" is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3."

Then look at what it specifies about how the answer of 1/2 is obtained. It says the same thing we've been saying all along:

"And if both sexes are represented, he picks a child at random and says "at least one is a ..." naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2."

We have already pointed out that if the child is named or if the riddle stated "the first child" then the answer would have been 1/2. The riddle as given in the OP did not name the child, hence, the riddle and the answer are not ambiguous and has the answer that your cite says it would under the given circumstances: 1/3.

The very article you quoted confirms that the logic you don't understand is the correct logic:

this is why i have a problem with this aspect of statistical analysis. let's see what we have here.

1] if you know the oldest child is a girl, the probability the other is also a girl is 1/2.

2] if you know the youngest child is a girl, the the probability the other is also a girl is 1/2.

3] if you do not know which child is a girl, the probability changes to 1/3?

sorry, but this defies any logic i understand.

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mmkay, another proof of 1/3:

you have a bag. it is not see through. You have two marbles in the bag. Marbles can be red or yellow.

The marbles are already there. They are inside the bag. There are two of them. There is nothing else in the bag.

One of them is red. What is the probability that both marbles are red?

these are the options of what is inside the bag:

Red and Yellow

Yellow and Red

Red and Red

Yellow and Yellow

but at least one has to be red... so we can say goodbye to "Yellow and Yellow". Now lets look at it again:

Red and Yellow

Yellow and Red

Red and Red

tada! there is a 1/3 chance they are both red.

I FOUND THE SOURCE OF THE ARGUEMENT!

everyone on both sides agree to remove (Yellow - Yellow)... leaving however many things are left. Thus, people who believe 1/3 believe in this set:

Red and Yellow

Yellow and Red

Red and Red

Yellow and Yellow

When you take one away, you have three to pull one out of. 1/3.

The people who believe in one-half answer believe this is the set:

Yellow and Red

Red and Red

Yellow and Yellow

when you remove 1, there are two left to pull one out of.

Basically, the people who think it is 1/2, they dont think (Red/Yellow) and (Yellow/Red) are both equal chances. The fact is, having one of each is twice the probability of having both. It all comes down to this:

People who think it is 1/3:

Mixed (yellow and red, or red and yellow): 1/2

Both yellow: 1/4

Both red: 1/4

While people who think it is one half:

Mixed: equal (1/3)

both yellow: equal (1/3)

both red: equal (1/3)

thus when you take away yellow, they think the answer is 1/2.

but it isnt

The probability for Mixed is twice as much... the sample space is this:

Red and Yellow

Yellow and Red

Yellow and Yellow

Red and Red

NOT THIS:

Mixed

Yellow and Yellow

Red and Red

therein lies the rift between the two sides. Hopefully the people who think it is 1/2 will finally understand now... that the answer is most definitely 1/3

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Guest

so I've read all the responses, and still believe that 1/2 is the correct answer. Here's one way of looking at it that I don't think anyone has brought up:

We know there are 4 possibilities:

GG

BG

GB

BB

we can throw out the last one - BB, leaves 3 combinations, so the thinking is 1/3 is what's left. The fallacy as I see it, is we don't know which one is the girl. Since we know one of them is a girl, we have 2 possibilities - one where the first child is the girl we're told about, and one where the second child is the girl.

So if the girl is the first child, the possibilities are:

GG

GB

or 1/2 the time the second child is also a girl

If the girl is the second child, the possibilities are:

GG

BG

or 1/2 the time the first child is also a girl

since the first scenario results in 1/2 and the second scenario results in 1/2, we have:

(50% * 1/2) + (50% * 1/2) = 1/4 + 1/4 = 1/2

Thus, like many have said - the odds are clearly 1/2 that the other child is also a girl.

another way to look at it - if we were told that the first child was a girl - what are the odds the second is a girl - again 1/2 (we throw out BB & BG, and are left with GG or GB). If instead we were told the second child was a girl, what are the odds the first one is also a girl - again 1/2 (we throw out BB & GB, and are left with GG or BG). Simply knowing or not knowing which one is the girl does not change the odds for the other one - which is a random event not dependent on the sex of the other child.

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You are assuming we are told which girl is first and second...

WE ARE NOT TOLD IF THE GIRL IS FIRST OR SECOND!

We are just told that ONE OF THEM is A GIRL...

we agree on this:

GG

GB

BG

we dont know if the girl is first, or the girl is second. I agree. If we KNEW the girl was first:

GG

GB

it would be 1/2

if we KNEW the girl was second:

GG

BG

it would be 1/2

BUT WE DO NOT KNOW!

GG

BG

GB

It is 1/3

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Guest
So if the girl is the first child, the possibilities are:

GG

GB

or 1/2 the time the second child is also a girl

If the girl is the second child, the possibilities are:

GG

BG

or 1/2 the time the first child is also a girl

since the first scenario results in 1/2 and the second scenario results in 1/2, we have:

(50% * 1/2) + (50% * 1/2) = 1/4 + 1/4 = 1/2

Thus, like many have said - the odds are clearly 1/2 that the other child is also a girl.

Doesn't work that way. Why on Earth are you multiplying 50% by 1/2 twice and adding them together?

If you think your answer is logical and correct, PLEASE do the experiment with pennies that I laid out several times. When all TT combinations are removed, about 1/3 of the pennies will be HH, 1/3 will be HT, and 1/3 will be TH.

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yeah i already shot him down too with a different explanation. no way he can even say its 1/2 now... muhahahahahahahahahahaaaaaa

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What we need to resolve the issue for all beyond any doubt is to resolve the various apparent paradoxes that arise, instead of giving different analogies or more cases.

Q. If the first child is a girl, the probability of the other child being a girl/boy is 1/2.

If the second child is a girl, the probability of the other child being a girl/boy is 1/2.

If any of the children is a girl, how can the probability of the other being a girl be less(i.e., 1/3)? Aren't we merging both cases?

A. Yes we are merging two cases, creating duplicate members of the set.

In the first we have 1 in 2 - GB, GG; in the second we have 1 in 2 - BG, GG.

Merging them we have a duplicate member GG which is occuring twice. So 1in2 and 1in2 do not form 2in4 but rather 1in3 due to overlapping members.

Q. Does already having a girl decreases the chance of having another girl?

A. Having a girl increases the chance of having a subsequent boy, thus decreasing the chance of having another girl. Just like if we get several heads in succession on tossing a coin, the chance of getting a subsequent tail increases.

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Q. Does already having a girl decreases the chance of having another girl?

A. Having a girl increases the chance of having a subsequent boy, thus decreasing the chance of having another girl. Just like if we get several heads in succession on tossing a coin, the chance of getting a subsequent tail increases.

I guess you think if you are at a roulette table and have just seen red come up 6 times in a row you should bet on black? Please, enough inaccurate information has been given in this thread!

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I was with the 1/2 arguement for a while then the penny arguement made me think. in a family of two the ratio would be like this 1:2:1 (BB:BG:GG) and I was leaning towards the1/3 arguement but not quite convinced so I googled it and came up with this site. [url:ba5d7]http://www.cut-the-knot.org/Curriculum/Probability/FamilyStats.shtml although the problem there isn't the same as the debate here the statitistics and analysis does show that the 1/3 arguement is correct. You can play with the applet and see that a family with two children there has a statistical chance that two out of four families will have a boy and a girl one out of four will have two girls and one out of four will have two boys. take the two boy families out of there and two out of three families will have both sexes and one out of three will have both girls.

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The reason assigning which member of the group meets the condition changes the probability is because it changes the possibilities. If you know the oldest child is a girl or the brown bear is a male, then there are only two possible results with one meeting the criteria, which results in the 50% answer. Without assigning either member specifically, then there are three possible results with only one meeting the criteria.

The brown bear/black bear scenario might actually make the whole thing easier to understand, since we get away from the sequencing issue that many people are hung up on. Without any conditions, there are four possible results; brown male/black male, brown male/black female, brown female/black male and brown female/black female. Once you stipulate the criteria that one of the bears be male, you eliminate only the instance where both are female. Leaving three possibilities with only one meeting the criteria.

I hope that adds something to the discussion, but it is really just rewording of what has already been said many times.

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