Poll of the Day > Help me understand your reasoning in solving the following easy problem

(+) Yeah! (+)

iwantmyoldid posted...

DarknessLink7 posted...

GG
GB

Which makes the odds of having two girls 50% since the two scenarios above are equally likely. However, if the "left kid" isn't guaranteed to be a girl, we instead have the situation where the "right kid" is the girl. Then it's obvious the only possibilities are:

GG
BG

Which again makes the odds of having two girls 50% since the two scenarios above are equally likely. Since either the "left kid" or the "right kid" must be a girl, one of the two situations above must be true. In either situation, the odds seem to be 50% that both kids are girls.

By not guaranteeing "left kid" is a girl, gives you GG,BG,GB. So 1/3. The way you did it here ls guaranteeing right kid a a girl which is not the same as nt guaranteeing left.

THANK YOU! I think this is what I was looking for! I'll check back again after getting some sleep and see if I still get it.
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Lightning Bolt posted...

At least one is a girl, because you know that for reasons, but you don't know which is which. So the possibilities are:

GG
GB
BG

Giving us a 1/3 chance of them both being girls.

GB and BG cancel eachother out as they are the same outcome.

You could write it as:

g1g2
g2g1
(b-g1
b-g2
or
g1b-
g2b-)

But it's still 2/4 or 50%.

Only 1 boy can exist so it is unnumbered and it can't be both sides so it's an interchangeable not simultaneously present. This is quantum at its core, numbers of definitive value in undefinitive location.
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Kyuubi4269 posted...

Lightning Bolt posted...

At least one is a girl, because you know that for reasons, but you don't know which is which. So the possibilities are:

GG
GB
BG

Giving us a 1/3 chance of them both being girls.

GB and BG cancel eachother out as they are the same outcome.

You could write it as:

g1g2
g2g1
(b-g1
b-g2
or
g1b-
g2b-)

But it's still 2/4 or 50%.

Only 1 boy can exist so it is unnumbered and it can't be both sides so it's an interchangeable not simultaneously present. This is quantum at its core, numbers of definitive value in undefinitive location.

They are not the same outcome

Kyuubi4269 posted...

GB and BG cancel eachother out as they are the same outcome.

They aren't the same outcome.
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Lightning Bolt posted...

They aren't the same outcome.

Firstborn girl and firstborn boy occurs, same event.
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And even if they are the same outcome, it would not cancel out.

Canceling out means just GG and BB posible (when having a kid) but since one is a girl, BB cannot be possible that means GG is 100% likely which is absurd.

Kyuubi4269 posted...

Lightning Bolt posted...

They aren't the same outcome.

Firstborn girl and firstborn boy occurs, same event.

I don't really get what you're trying to say. Could you explain it in more detail?
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DarknessLink7 posted...

Kyuubi4269 posted...

Lightning Bolt posted...

They aren't the same outcome.

Firstborn girl and firstborn boy occurs, same event.

I don't really get what you're trying to say. Could you explain it in more detail?

a girl and a boy being born is the same as a boy and a girl being born.

You can also argue 25% by first round being 50% of being possibily GG, and in the successful round you have a 50% chance of GG in the 50% chance. 50%*50%=25%.
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Kyuubi4269 posted...

a girl and a boy being born is the same as a boy and a girl being born.

So, by your logic, a family with two random kids has a 33% chance of having two girls?

GG
GB
BB
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Lightning Bolt posted...

Golden Road posted...

Let's assume you have a bag with four discs inside. -snip-

Once you know which side is which (by literally looking at it), you have more information than the OP said we have.

A more accurate parallel would be picking one out randomly without looking, and having the person next to you confirm Y/N whether at least one side is black. If he says Yes, then you take a look.
When phrased that way, it's super super clear why it's 1/3, right?

OK, it's not a great way to explain myself after all =P

I'll go back to this:

Ann, Courtney
Heidi, Bradley
Gary, Emily

Yes, only one of these three families has 2 girls, while two of them have a boy and a girl. But they're not equally likely to be chosen. If you choose Ann or Courtney, then you have the family with two sisters. If you choose Heidi or Emily, you have a family with a brother and sister. You're twice as likely to choose the girl-girl family, which offsets having only half as many girl-girl families.
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Golden Road posted...

I'll go back to this:

Ann, Courtney
Heidi, Bradley
Gary, Emily

Yes, only one of these three families has 2 girls, while two of them have a boy and a girl. But they're not equally likely to be chosen. If you choose Ann or Courtney, then you have the family with two sisters. If you choose Heidi or Emily, you have a family with a brother and sister. You're twice as likely to choose the girl-girl family, which offsets having only half as many girl-girl families.

You're not choosing the individual kids, you're choosing the family. Your choices are family A (Ann and Courtney), family B (Heidi and Bradley), or family C (Gary, Emily).
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piccolo1104 posted...

Golden Road posted...

I'll go back to this:

Ann, Courtney
Heidi, Bradley
Gary, Emily

Yes, only one of these three families has 2 girls, while two of them have a boy and a girl. But they're not equally likely to be chosen. If you choose Ann or Courtney, then you have the family with two sisters. If you choose Heidi or Emily, you have a family with a brother and sister. You're twice as likely to choose the girl-girl family, which offsets having only half as many girl-girl families.

You're not choosing the individual kids, you're choosing the family. Your choices are family A (Ann and Courtney), family B (Heidi and Bradley), or family C (Gary, Emily).

You're choosing Ann's family, Courtney's family, Heidi's family, or Emily's family. We don't care about Bradley's or Gary's families. While there are only three families, one of those families is twice as likely to be chosen than the other two.
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Golden Road posted...

piccolo1104 posted...

Golden Road posted...

I'll go back to this:

Ann, Courtney
Heidi, Bradley
Gary, Emily

Yes, only one of these three families has 2 girls, while two of them have a boy and a girl. But they're not equally likely to be chosen. If you choose Ann or Courtney, then you have the family with two sisters. If you choose Heidi or Emily, you have a family with a brother and sister. You're twice as likely to choose the girl-girl family, which offsets having only half as many girl-girl families.

You're not choosing the individual kids, you're choosing the family. Your choices are family A (Ann and Courtney), family B (Heidi and Bradley), or family C (Gary, Emily).

You're choosing Ann's family, Courtney's family, Heidi's family, or Emily's family. We don't care about Bradley's or Gary's families. While there are only three families, one of those families is twice as likely to be chosen than the other two.

True. We dont care about the boys' fam but that does not make thr girl's family twice as likely.

Golden Road posted...

piccolo1104 posted...

Golden Road posted...

I'll go back to this:

Ann, Courtney
Heidi, Bradley
Gary, Emily

Yes, only one of these three families has 2 girls, while two of them have a boy and a girl. But they're not equally likely to be chosen. If you choose Ann or Courtney, then you have the family with two sisters. If you choose Heidi or Emily, you have a family with a brother and sister. You're twice as likely to choose the girl-girl family, which offsets having only half as many girl-girl families.

You're not choosing the individual kids, you're choosing the family. Your choices are family A (Ann and Courtney), family B (Heidi and Bradley), or family C (Gary, Emily).

You're choosing Ann's family, Courtney's family, Heidi's family, or Emily's family. We don't care about Bradley's or Gary's families. While there are only three families, one of those families is twice as likely to be chosen than the other two.

Based on the original question, you are choosing a family that has at least one girl. Ann's family is not considered a separate choice from Courtney's family.
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the number of possibilities of having two children (or flipping two coins) is 4.

the question is really: how many possiblities does the phrase "at least one of them is a girl" exclude?

"at least one of them is a girl" translates into "they are not both boys"

"they are not both boys" excludes exactly 1 of the 4 possibilities

the chance is 1/3

people that say 1/2 are mistaking "at least one of them is a girl" to be as much information as knowing "the first child is a girl" (or "the second child is a girl"; the ambiguity is an indication of the lack of information)
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Which child is 100% female is irrelevant. Say you rolled 2 dice at the same time, whatever number is on the first has a 1/6 chance of being on the second. Now if i told you that one of the die had a 1 on it, does this change the number on the other?
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valid_is_noob posted...

Which child is 100% female is irrelevant. Say you rolled 2 dice at the same time, whatever number is on the first has a 1/6 chance of being on the second. Now if i told you that one of the die had a 1 on it, does this change the number on the other?

Prior/partial knowledge of a sample space affects probability.

The rolled number wont change. Individually they have a 1/6 change but as a group, the probability is different (it would be some multiple of 1/36), now, knowing 1 of the dice restricts the space to a smaller subset of the original 36.

I believe it's 50% but I'm very curious if it's 33% what the logic is behind that.

A family has two children, and at least one of them is a girl. What is the probability that they have two girls?

I don't follow why any of the conversations regarding first or second child or right or left child etc. are relevant.

It is guaranteed that one of the children is a girl. That is without question, and I feel like at that point the question is no longer asking about two girls. It's asking about the sex of the unknown child, which is independent of the guaranteed girl. So why wouldn't it be 50%? How is this any different than asking what the probability of a family with one child is of it being a girl?

I'm looking forward to the solution
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Sahuagin posted...

the number of possibilities of having two children (or flipping two coins) is 4.

the question is really: how many possiblities does the phrase "at least one of them is a girl" exclude?

"at least one of them is a girl" translates into "they are not both boys"

"they are not both boys" excludes exactly 1 of the 4 possibilities

the chance is 1/3

people that say 1/2 are mistaking "at least one of them is a girl" to be as much information as knowing "the first child is a girl" (or "the second child is a girl"; the ambiguity is an indication of the lack of information)

But that is already cut down because we know one is a girl. GB is the exact same as BG, you guys are adding to much complications on to this. Who ever is the first born does not matter. Why are you guys bringing that into this? it is actually

GG
BB
GB

Having two GB is ridiculous. The BB gets cut down and we are left with GG or GB. Now if the question was what is the probability of the girl being born first or something then yeah. But this is such a easy question being torn apart by mathmaticians lol.
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yutterh posted...

GB is the exact same as BG

They are NOT the same thing. At all.

Look at it this way, having a boy and and a girl (regarless of order) is twice as likely as having two of one type, which shows that GB is not the same as BG (They add up to make boy+girl(ignoring order) twice as likely)

iwantmyoldid posted...

yutterh posted...

GB is the exact same as BG

They are NOT the same thing. At all.

Look at it this way, having a boy and and a girl (regarless of order) is twice as likely as having two of one type, which shows that GB is not the same as BG (They add up to make boy+girl(ignoring order) twice as likely)

Having a boy and a girl is as likely as having two of the same gender.

RoboT_Ripper posted...

I believe it's 50% but I'm very curious if it's 33% what the logic is behind that.

A family has two children, and at least one of them is a girl. What is the probability that they have two girls?

I don't follow why any of the conversations regarding first or second child or right or left child etc. are relevant.

It is guaranteed that one of the children is a girl. That is without question, and I feel like at that point the question is no longer asking about two girls. It's asking about the sex of the unknown child, which is independent of the guaranteed girl. So why wouldn't it be 50%? How is this any different than asking what the probability of a family with one child is of it being a girl?

I'm looking forward to the solution

It is all about the order of the events. No matter how you guys dont accept it, AB = / = BA and so since we are not told the order of the girls, we have to take all orders into account.

If no kids are born yet, all three combinations have the same likelihood of occuring, yes? Gg, bb, bg/gb. It's not bg & gb, it's one of each gender. Now, we know one combination can't happen in this case: bb. Two are left, with an equal chance. Gg & one of each. 50%.

nurlen posted...

iwantmyoldid posted...

yutterh posted...

GB is the exact same as BG

They are NOT the same thing. At all.

Look at it this way, having a boy and and a girl (regarless of order) is twice as likely as having two of one type, which shows that GB is not the same as BG (They add up to make boy+girl(ignoring order) twice as likely)

Having a boy and a girl is as likely as having two of the same gender.

Lets say you have two kids. The first born is a boy then the second born can be boy or girl. Lets say it is a Boy, then you have BoyBoy. Let's say it is a girl, then you have BoyGirl. Lets say the First was a girl, then you can Have GirlGirl or Girl Boy. The sample space of two childrens contain 4 possibilities., two of which have a boy and a girl, so boy+girl is twice as likely.

Just look at the real world, how often do you see a family with two of the same sex compared to one of each?

The topic says assume chances are 50% for each gender. Real world stats do not apply here, and even if they did, do not help your case as there are more females than males.

nurlen posted...

The topic says assume chances are 50% for each gender. Real world stats do not apply here, and even if they did, do not help your case as there are more females than males.

Chances are 50 for each gender in the real world too.

iwantmyoldid posted...

nurlen posted...

The topic says assume chances are 50% for each gender. Real world stats do not apply here, and even if they did, do not help your case as there are more females than males.

Chances are 50 for each gender in the real world too.

If that were true then why did you ask me to "look at the real world"? If it's the same as the example it doesn't matter and all three combinations are equally likely, and there would not be twice as many bg as gg or bb. Counting bg and gb as separate is like counting g1g2 and g2g1 separate. Same with b1b2 and b2b1.

iwantmyoldid posted...

valid_is_noob posted...

Which child is 100% female is irrelevant. Say you rolled 2 dice at the same time, whatever number is on the first has a 1/6 chance of being on the second. Now if i told you that one of the die had a 1 on it, does this change the number on the other?

Prior/partial knowledge of a sample space affects probability.

The rolled number wont change. Individually they have a 1/6 change but as a group, the probability is different (it would be some multiple of 1/36), now, knowing 1 of the dice restricts the space to a smaller subset of the original 36.

As a group you are still only told the outcome of 1 die. The answer is still 1/6, because as you said it doesnt change the rolled number. Being told anything doesnt change the outcome of the dice.
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nurlen posted...

iwantmyoldid posted...

nurlen posted...

The topic says assume chances are 50% for each gender. Real world stats do not apply here, and even if they did, do not help your case as there are more females than males.

Chances are 50 for each gender in the real world too.

If that were true then why did you ask me to "look at the real world"? If it's the same as the example it doesn't matter and all three combinations are equally likely, and there would not be twice as many bg as gg or bb. Counting bg and gb as separate is like counting g1g2 and g2g1 separate. Same with b1b2 and b2b1.

g1g2 is the same as g2g1 because once girl 2 is born firs then she is girl one. basically, there is no such thing as g2g1 here, when the numbers are counting order of being born.

I said real world chances of a single boy or girl is 50 because it is the same in the problem. Two choices, equally likely.

Idk if this link (or others I have found) help me or not: http://mathforum.org/library/drmath/view/52186.html

They basically say it is both, 1/3 or 1/2 depending how you define it.

Which is going back to square one.

Now you're talking hypocritically. 1 & 2 aren't referring to birth order. Name them if you want. Emma and Jasmine.

iwantmyoldid posted...

Idk if this link (or others I have found) help me or not: http://mathforum.org/library/drmath/view/52186.html

They basically say it is both, 1/3 or 1/2 depending how you define it.

Which is going back to square one.

But this question is after both are born, so according to that the answer is 1/3.
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piccolo1104 posted...

Golden Road posted...

piccolo1104 posted...

Golden Road posted...

I'll go back to this:

Ann, Courtney
Heidi, Bradley
Gary, Emily

Yes, only one of these three families has 2 girls, while two of them have a boy and a girl. But they're not equally likely to be chosen. If you choose Ann or Courtney, then you have the family with two sisters. If you choose Heidi or Emily, you have a family with a brother and sister. You're twice as likely to choose the girl-girl family, which offsets having only half as many girl-girl families.

You're not choosing the individual kids, you're choosing the family. Your choices are family A (Ann and Courtney), family B (Heidi and Bradley), or family C (Gary, Emily).

You're choosing Ann's family, Courtney's family, Heidi's family, or Emily's family. We don't care about Bradley's or Gary's families. While there are only three families, one of those families is twice as likely to be chosen than the other two.

Based on the original question, you are choosing a family that has at least one girl. Ann's family is not considered a separate choice from Courtney's family.

This. 1/3.

Also because I know my mathematics and this is a standard learn to count/don't trust your gut for probability exercise.

It's pretty simple, 50% chance that the other child is a Girl and that they have two girls.

If the question was what is the chance of a family with two children to have two girls, then that would be 33% (since you have 3 options, B/B, B/G, G/G).

If you ask what is the chance any one child would be a girl, that is (in theory), 50% (not sure it really is a 50% chance as there may actually be more girls born then boys etc).

In this case you are specifically asking about 1 child (the one that has not had it's gender identified yet), which means there is a set 50% chance that it will be a boy or girl.
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2 Children, 4 possibilities.
BB
BG
GB
GG

The order of the children doesn't matter. BG and GB are the same outcome. 3 possibilities.
BB
GB
GG

At least one of them is a girl. Eliminates BB. 2 possibilities.
GB
GG

1 out of 2 is GG.
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A single quantum child that is 50% male and 50% female.

BB
BG
GG
GB
4 choices, each equal, hence 25% for each choice.

We know BB can't exist. So we have:
BG
GG
GB
3 choices, each equal. So 33.3% for each choice. There's only ONE of these choices that fulfills the requirement of "two girls", hence its 1/3.

Now. Alternatively, it's also 1/2.
Why? Because if we look at it as "there's only one child left to pick", we're instead looking solely for "G".
So our choices are:
B
G
Each equally as likely, hence 50% each.
We only want G, and there's only one of these outcomes that is G. So it's 1/2.


Basically, it changes on how you look at things.
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Bayes' Theorem: P(A|B) = (P(B|A) * P(A)) / P(B)

Possible combinations:
BB
BG
GB
GG

Probabilities for each of those events:
P(BB) = 0.25
P(BG) = 0.25
P(GB) = 0.25
P(GG) = 0.25

P(Count(G) = 0) = P(BB) = 0.25
P(Count(G)>=1) = P(BG) + P(GB) + P(GG) = 0.75
P(Count(G) = 2) = P(GG) = 0.25

So, using Bayes' Theorem:

P(Count(G) = 2 | Count(G)>=1) = (P(Count(G)>=1 | Count(G) = 2) * P(Count(G) = 2)) / P(Count(G)>=1)

= (P(Count(G)>=1 | Count(G) = 2) * 0.25) / 0.75

So the only thing left to figure out is what is P(Count(G)>=1 | Count(G) = 2)?

Well, in words, that means the probability that the number of girls is greater than or equal to 1, given that the count of girls is equal to 2.

Obviously, given that the count is 2, the probability of the count being greater than or equal to 1 is 100%, or 1.0.

So putting that back in our equation above:

P(Count(G) = 2 | Count(G)>=1) = (1.0 * 0.25) / 0.75 = 0.3~, or 33%.
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I think that it's 1/3.

I suspect that the people who argue for 1/2 are interpreting the given information as "the first-born is a girl". In that case, you would have only GB or GG as viable outcomes. The outcomes thrown out would be BG and BB.

Notice also that the probability of having two girls should be similar to the probability of flipping two H in a row with a coin. If we didn't have the given info, we would answer 1/4. But we do have given info and we would only answer 1/2 if we forgot the first child/coin toss.
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Golden Road posted...

OK, it's not a great way to explain myself after all =P

I'll go back to this:

Ann, Courtney
Heidi, Bradley
Gary, Emily

Yes, only one of these three families has 2 girls, while two of them have a boy and a girl. But they're not equally likely to be chosen. If you choose Ann or Courtney, then you have the family with two sisters. If you choose Heidi or Emily, you have a family with a brother and sister. You're twice as likely to choose the girl-girl family, which offsets having only half as many girl-girl families.

It's very interesting how you interpret the problem. I'm trying to restate what you're thinking in terms of "cheating in a gameshow." The kids are behind a door and you peek and see one of them is a girl. I guess if there are actually two girls in the family, then it's more likely for you to realize one is a girl. But if you argue that, and even give them names, you gotta consider all outcomes before you "peek" and see one of them is a girl.

This is a list of how the kids could be standing behind the door. GG GB BG BB
Alice Clara (GG)
Clara Alice (GG)
Alice Bobby (GB)
Bobby Alice (BG)
Alex Bobby (BB)
Bobby Alex (BB)

So now when you "cheat" and realize one of them is a girl, it seems like GG outcome is 50% (2/4) likely, but that's not the answer to the problem in my opinion, because the outcome Alice and Clara is identical to Clara and Alice, no?
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Someone has probably posted the answer already, but I can't be assed to read through the entire topic.

The correct answer is 1/3, which seems counter-intuitive, but it's true based on the information we are given (namely that one of the children is a girl - the problem is that we don't know "which one").

Consider that the kids had to be born in a specific order (yes, even if they were twins). In this case, there's four possible combinations of birth order for two kids:

Boy, then boy.
Boy, then girl.
Girl, then boy.
Girl, then girl.

Because we know one of the children is a girl, the first combination is impossible, leaving just the other three. Two of those feature a boy and a girl, one of those features two girls. Thus, the odds of having two girls is one-in-three (although if you provided more information - like specifying whether the girl was the older or younger sibling - you could change the odds to an even 50%).

Sorry, ~80% of PotD, you're wrong.
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jramirez23 posted...

Golden Road posted...

OK, it's not a great way to explain myself after all =P

I'll go back to this:

Ann, Courtney
Heidi, Bradley
Gary, Emily

Yes, only one of these three families has 2 girls, while two of them have a boy and a girl. But they're not equally likely to be chosen. If you choose Ann or Courtney, then you have the family with two sisters. If you choose Heidi or Emily, you have a family with a brother and sister. You're twice as likely to choose the girl-girl family, which offsets having only half as many girl-girl families.

It's very interesting how you interpret the problem. I'm trying to restate what you're thinking in terms of "cheating in a gameshow." The kids are behind a door and you peek and see one of them is a girl. I guess if there are actually two girls in the family, then it's more likely for you to realize one is a girl. But if you argue that, and even give them names, you gotta consider all outcomes before you "peek" and see one of them is a girl.

This is a list of how the kids could be standing behind the door. GG GB BG BB
Alice Clara (GG)
Clara Alice (GG)
Alice Bobby (GB)
Bobby Alice (BG)
Alex Bobby (BB)
Bobby Alex (BB)

So now when you "cheat" and realize one of them is a girl, it seems like GG outcome is 50% (2/4) likely, but that's not the answer to the problem in my opinion, because the outcome Alice and Clara is identical to Clara and Alice, no?

But, you know that one of them is a girl.

So:
GG
BG or GB but only one, not both, because one of those is already eliminated
BB- eliminated

So,
GG or one only of the GB/BG combinations= .50

This topic does make a nice illustration of the Gambler's Fallacy.
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The problem here is one of semantics more than anything. People don't realize that the question is asking the probability of a specific combination after being given one of the outcomes. Which isn't surprising because probability questions require a particular awareness of what the question is supposed to mean. You can't just throw out a question like this without much context and expect everyone to get it right. It's sort of misleading and I would not call it a "simple" question.

Basically the question asks among all the possible combinations what are the odds that both children will be girls assuming we know one child is a girl. . However, people are interpreting it to mean, what is the probability the "next" child will be a girl?
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dancer62 posted...

GG
BG or GB but only one, not both, because one of those is already eliminated
BB- eliminated

The bolded line is incorrect. You can't eliminate one of these because both are feasible outcomes based on the information given. If you knew for certain whether the girl was older or younger you could eliminate one possibility and make the problem a 50/50 coin toss, but that's not the information given.

dancer62 posted...

This topic does make a nice illustration of the Gambler's Fallacy.

Gambler's Fallacy only applies if the two events are unrelated. You're not being asked about a sequence of events (aka, "What are the odds the next child will be female?"), you're being asked about two events that have already happened, with limited information about one of those events (and you're not told WHICH event the information applies to).

That, in essence, means that the two births are - in the context of the problem given - a single paired-event, rather than two unrelated events (since the information given applies to one of the two, but you have no way of knowing which). Gambler's Fallacy does not apply.
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Guys where were you all day long? It was like 4 against army. I gave up when I got called a hypocrite.

jramirez23 posted...

Golden Road posted...

OK, it's not a great way to explain myself after all =P

I'll go back to this:

Ann, Courtney
Heidi, Bradley
Gary, Emily

Yes, only one of these three families has 2 girls, while two of them have a boy and a girl. But they're not equally likely to be chosen. If you choose Ann or Courtney, then you have the family with two sisters. If you choose Heidi or Emily, you have a family with a brother and sister. You're twice as likely to choose the girl-girl family, which offsets having only half as many girl-girl families.

It's very interesting how you interpret the problem. I'm trying to restate what you're thinking in terms of "cheating in a gameshow." The kids are behind a door and you peek and see one of them is a girl. I guess if there are actually two girls in the family, then it's more likely for you to realize one is a girl. But if you argue that, and even give them names, you gotta consider all outcomes before you "peek" and see one of them is a girl.

This is a list of how the kids could be standing behind the door. GG GB BG BB
Alice Clara (GG)
Clara Alice (GG)
Alice Bobby (GB)
Bobby Alice (BG)
Alex Bobby (BB)
Bobby Alex (BB)

So now when you "cheat" and realize one of them is a girl, it seems like GG outcome is 50% (2/4) likely, but that's not the answer to the problem in my opinion, because the outcome Alice and Clara is identical to Clara and Alice, no?

The annoying thing is, if you use the cheating in a game show story, the probability actually becomes 50%. Since by seeing it's one girl behind the door, the odds of a boy being there is lowered. So in that case you having seen one of the girls in a GG pair is more likely. (And if you specifically saw that the left kid was a girl, then the probability of GG becomes 66%? Not sure about this, but it would make sense given that the only configurations could be Alice-Clara, Clara-Alice or Alice-Bobby.)

Though in the original questions I posted, the answer is 33%, since you got the information from me and not by looking.
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Let me ask you a new question:

I flip 100 coins behind a screen. I then chose to cover up one of them as I removed the screen to revealed that I flipped 99 Heads. You don't know what result I got on the coin I covered up. I ask you: "What are the odds that this single coin I'm covering is also Heads?"

Would your answer be different if I had instead said: "The coin I'm covering was the last coin I flipped. What are the odds that it's also Heads?"
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According to the braniacs here it'll be .0736% or some shit.

DarknessLink7 posted...

Let me ask you a new question:

I flip 100 coins behind a screen. I then chose to cover up one of them as I removed the screen to revealed that I flipped 99 Heads. You don't know what result I got on the coin I covered up. I ask you: "What are the odds that this single coin I'm covering is also Heads?"

The answer is slightly less than 1% (0.99%, specifically)

This is actually probably one of the more intuitive ways to demonstrate the concept driving the original problem. The only way that the coin your covering is heads is if ALL the coin tosses came up heads. On the other hand, the coin you're covering could be tails if the first toss came up tails, or the second, or the third, or the fourth, so on and so forth all the way to the hundredth toss.

In essence, there are 100 possibilities of having a single coin come up tails, but only one possibility for all the coins to come up heads. Hence the answer is 1/101 = 0.99%.

DarknessLink7 posted...

Would your answer be different if I had instead said: "The coin I'm covering was the last coin I flipped. What are the odds that it's also Heads?"

Yes it would - then the answer becomes 50% because the previous results no longer matter. In this case there's only one possibility that satisfies a heads result (99 heads followed by another heads) and one that satisfies a tails result (99 heads followed by tails), so the odds of either result is an even 50%.
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It said a "family has two children" it did not say if they all they have is two children. Which is why I put in 33%.
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