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

(+) Yeah! (+)

This is all assuming the probability of having a boy is 50% and the probability of having a girl is also 50%.

I won't influence your answer by writing what I think here, but please vote and then post your reasoning below. I hope I get a few different opinions, so don't be afraid to post your reasoning even if you pick the less popular option!
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One of those typical examples why obvious probabilities might not be obvious at all right?

Older child might be girl, younger child might be girl, both are girl. Are the three situations covered by the requirement, so 1 out of 3?

Something about set observations altering probabilities, as in, this is not the same as asking what the chance the other child is a girl is.
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Two independent events, 50% probability for each, we know the results of one event but not the other.

50%. Knowing the results of one coin toss does not affect the probability of another coin toss.
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Better way of saying what I said is that the information is not as complete as it seems. It would be 50% if it were given which child was a girl, the information is not that complete.
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Nichtcrawler X posted...

Better way of saying what I said is that the information is not as complete as it seems. It would be 50% if it were given which child was a girl, the information is not that complete.

I see your point. However, what would happen if they were twins?
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dancer62 posted...

Two independent events, 50% probability for each, we know the results of one event but not the other.

50%. Knowing the results of one coin toss does not affect the probability of another coin toss.

Interesting reasoning.
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DarknessLink7 posted...

I see your point. However, what would happen if they were twins?

Wouldn't change anything. One of the two would still be older. Or better put, they are still distinguishable individuals.

Unless you were to say they are identical twins, which would alter the probabilities...
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Nichtcrawler X posted...

DarknessLink7 posted...

I see your point. However, what would happen if they were twins?

Wouldn't change anything. One of the two would still be older. Or better put, they are still distinguishable individuals.

Unless you were to say they are identical twins, which would alter the probabilities...

Ah, all right. I didn't mean identical twins, don't worry. Could you explain what you mean by distinguishable individuals? Or rather, why is that important?
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dancer62 posted...

Two independent events, 50% probability for each, we know the results of one event but not the other.

50%. Knowing the results of one coin toss does not affect the probability of another coin toss.

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DarknessLink7 posted...

Could you explain what you mean by distinguishable individuals? Or rather, why is that important?

We know there are two children, we know one of them is a girl, just not which one. Either one being a girl with the other being a boy are 2 separate possibilities.
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Just had a flash of insight how other to explain it, no guarantees that it makes sense though...

A couple has two children. Those might be two sons, two daughters or a son and a daughter. with 25%, 25% and 50% odds. (boy-boy, girl-girl, boy-girl and girl-boy)

The statement of atleast 1 girl only excludes the 2 sons possibility. The probabilities are not altered in any other way, just readjusted to be out of 100% again.
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Well if one HAS to be a girl, and they only have two children, your only two options for the second child are a boy or a girl, so 50/50 chance of it being one or the other.
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Nichtcrawler X posted...

Just had a flash of insight how other to explain it, no guarantees that it makes sense though...

A couple has two children. Those might be two sons, two daughters or a son and a daughter. with 25%, 25% and 50% odds. (boy-boy, girl-girl, boy-girl and girl-boy)

The statement of atleast 1 girl only excludes the 2 sons possibility. The probabilities are not altered in any other way, just readjusted to be out of 100% again.

A very good explanation. Thank you.
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It doesnt matter how many children they have or the probability for the known genders.
Only the unknown genders matters, which in this case is one with a 50/50 chance.
It is the same if they had five kids and it was known that at least four of them are females.
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What if the unknown child is a girl that identifies as a boy?

It's 50%. The possibilities are:

GG
GB
BG
BB

The family obviously can't be BB, so it's either GG, GB, or BG. What some people miss, though, is that GG is twice as likely as GB or BG, since it has two possible girls to choose from, while GB and BG only have one. There's a 50% chance the girl is chosen from GG, a 25% chance GB, and a 25% BG.

Ann
Bradley
Courtney
Derek
Emily
Fred
Gary
Heidi

Let's say we have these eight kids, four pairs of siblings. One pair of siblings has two girls, one pair has two boys, and two pairs has one of each. Even though two families have a girl-boy pair, and only one has a girl-girl pair, you've still got a 50-50 chance of picking the girl-girl family if you pick out a girl here. Having twice as many boy-girl families as girl-girl families is offset by being twice as likely to pick a girl-girl family. If you pick two of the four girls above, you get the girl-girl family, and if you pick one of the other two, you get a girl-boy family.

It's 50%.

Ann, Courtney
Heidi, Bradley
Gary, Emily
Derek, Fred
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Golden Road posted...

is that GG is twice as likely as GB or BG, since it has two possible girls to choose from,

Which is the same combination.

Just try making a Punnett square.
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We are only dealing with the unknown child. There's a 50/50 chance of it being a boy/girl.
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SunWuKung420 posted...

We are only dealing with the unknown child. There's a 50/50 chance of it being a boy/girl.

Except you do not know whether that is child 1 or child 2.

I remembered once watching a video on the issue. This guy explains it better than me with fancy pictures and animations.
https://youtu.be/go3xtDdsNQM?t=105
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Nichtcrawler X posted...

Golden Road posted...

is that GG is twice as likely as GB or BG, since it has two possible girls to choose from,

Which is the same combination.

Just try making a Punnett square.

But you're ignoring that you're twice as likely to pick that square. Of Ann, Courtney, Emily, and Heidi, you have a 25% chance of picking the first girl who belongs to the GG family, a 25% chance of picking the second girl who belongs to the GG family, a 25% chance of picking the girl who belongs to the GB family, and a 25% chance of picking the girl who belongs to the BG family.
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Nichtcrawler X posted...

SunWuKung420 posted...

We are only dealing with the unknown child. There's a 50/50 chance of it being a boy/girl.

Except you do not know whether that is child 1 or child 2.

That is irrelevant to this Q

Possible family compositions:

BB
BG
GB
GG

With the requirement of "at least one of them is a girl," that leaves us with BG, GB, GG.

As you can see, GG makes up 1/3 of the possible outcomes.

Looks like a lot of yall just got BTFO
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Nichtcrawler X posted...

Except you do not know whether that is child 1 or child 2.

that doesn't matter. we aren't picking a child out of a hat, we're determining the odds for the unknown child. it's 50/50.

knowing the other childs gender is irrelevant to the question.
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joemodda posted...

Possible family compositions:

BB
BG
GB
GG

With the requirement of "at least one of them is a girl," that leaves us with BG, GB, GG.

As you can see, GG makes up 1/3 of the possible outcomes.

Looks like a lot of yall just got BTFO

Except GG is twice as likely as BG or GB. If you randomly pick one of those four girls, you have a 2 in 4 chance of picking a girl from the GG family, not a 1 in 3 chance.
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SunWuKung420 posted...

We are only dealing with the unknown child. There's a 50/50 chance of it being a boy/girl.

This.

Nichtcrawler X posted...

Except you do not know whether that is child 1 or child 2.

Irrelevant. The condition here is simply "they have two girls," not "that the girl has a younger sister" or anything like that. There's one child whose sex is unknown, and there's a 50% chance that child is a girl.
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adjl posted...

There's one child whose sex is unknown

Which one?
From where I'm sitting, it looks like we don't know either child's sex. Just something about their "combined" sexes.
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Lightning Bolt posted...

Which one?

The one that isn't covered by "at least one is a girl."
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adjl posted...

Lightning Bolt posted...

Which one?

The one that isn't covered by "at least one is a girl."

But it was never specified which one that is, and a simple GB vs GG thing would require a fixed G somewhere. Since you don't know which one you don't know, you have to assume that BG is also possible.
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I'd like to ask the following question to those of you who voted for the 33% option:

Do we agree on the following?
-There is a 100% chance the kid we know is a girl is a girl.
-There is a 50% chance the unknown kid is a girl.

If we agree on the above conditions, how can you argue that the odds of both kids being girls are 33% and not 50%? Because one of them is guaranteed to be a girl while the other one has a 50% chance of being a girl. (I am actually on your side, I'd just like to see how you argue for it.)
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DarknessLink7 posted...

Do we agree on the following?
-There is a 100% chance the kid we know is a girl is a girl.
-There is a 50% chance the unknown kid is a girl.

We do not agree. There is no "known child", you don't know the sex of either one specifically.
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Lightning Bolt posted...

DarknessLink7 posted...

Do we agree on the following?
-There is a 100% chance the kid we know is a girl is a girl.
-There is a 50% chance the unknown kid is a girl.

We do not agree. There is no "known child", you don't know the sex of either one specifically.

But we know the sex of one of them. Sure, we don't know which one, but let's call them X. There's a 100% chance X is a girl. And there's a 50% chance the other kid is a girl, right?

What's wrong with this argument?
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That depends. The odds of any particular family with two children, including this one, having two girls is 25%.

The odds of this SPECIFIC family having another girl after knowing that at least one child is a girl is 50%.
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Lightning Bolt posted...

adjl posted...

Lightning Bolt posted...

Which one?

The one that isn't covered by "at least one is a girl."

But it was never specified which one that is, and a simple GB vs GG thing would require a fixed G somewhere. Since you don't know which one you don't know, you have to assume that BG is also possible.

Ah, I'm seeing it now. Okay, it's 33%.
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Everyone picking 33%: why are you ignoring that, while there are twice as many family combinations that have a boy (BG and GB), you are also twice as likely to pick a girl from the GG family as you are from either the BG or GB family?

Ann, Courtney
Heidi, Bradley
Gary, Emily

If you pick Heidi, her brother is Bradley.
If you pick Emily, her brother is Gary.
If you pick Ann, her sister is Courtney.

While that does look like a 33% chance, you're forgetting that you could pick Courtney instead of Ann!

If you pick Heidi, her brother is Bradley.
If you pick Emily, her brother is Gary.
If you pick Ann, her sister is Courtney.
If you pick Courtney, her sister is Ann.

There is a 50% chance the other sibling is also a girl. Believing it's only 33% assumes you're equally likely to have picked all three families, which is not the case.
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Lightning Bolt posted...

But it was never specified which one that is, and a simple GB vs GG thing would require a fixed G somewhere. Since you don't know which one you don't know, you have to assume that BG is also possible.

You don't have to specify anything. If there are only two options then it's flat out 50%. It doesn't matter which one is a girl since it isn't 3 options which is how you would get 33%.
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DarknessLink7 posted...

But we know the sex of one of them. Sure, we don't know which one, but let's call them X. There's a 100% chance X is a girl. And there's a 50% chance the other kid is a girl, right?

What's wrong with this argument?

Because the argument is still assuming that you know which one is a girl. Out of the BG, GB, and GG combinations that were allowed, neither the first nor the second is always a girl.

Golden Road posted...

Everyone picking 33%: why are you ignoring that, while there are twice as many family combinations that have a boy (BG and GB), you are also twice as likely to pick a girl from the GG family as you are from either the BG or GB family?

Who the crap is Courtney? xD
So, by your logic, a family with two random kids has a 33% chance of having two girls?

GG
GG
BG
GB
BB
BB
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Lightning Bolt posted...

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

GG
GG
BG
GB
BB
BB

No. I'll try to explain it in a different way.

Let's assume you have a bag with four discs inside. One disc is black on both sides. One disc is white on both sides. The other two discs are black on one side, and white on the other. At random, you take one disc out of the bag, and look at one side. If it's black, look at the other side of the disc, and record whether it's the single black-black disc, or one of the two black-white discs. If you do that over and over, you'll notice that you get the single black-black disc about the same number of times you get one of the two black-white discs. Even though there are twice as many black-white discs as black-black discs, there are 4 black sides in that bag, and 2 of the 4 black sides belong to the black-black disc.

If you replace the discs with children, the black side is a girl, and the white side is a boy, the same thing happens. 50% of the time you take out a "child" from the bag and see a "girl," you will find another "girl" on the other side.
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Golden Road posted...

Lightning Bolt posted...

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

GG
GG
BG
GB
BB
BB

No. I'll try to explain it in a different way.

Let's assume you have a bag with four discs inside. One disc is black on both sides. One disc is white on both sides. The other two discs are black on one side, and white on the other. At random, you take one disc out of the bag, and look at one side. If it's black, look at the other side of the disc, and record whether it's the single black-black disc, or one of the two black-white discs. If you do that over and over, you'll notice that you get the single black-black disc about the same number of times you get one of the two black-white discs. Even though there are twice as many black-white discs as black-black discs, there are 4 black sides in that bag, and 2 of the 4 black sides belong to the black-black disc.

If you replace the discs with children, the black side is a girl, and the white side is a boy, the same thing happens. 50% of the time you take out a "child" from the bag and see a "girl," you will find another "girl" on the other side.

Wow, that's quite a nice way to explain it. I see your point.
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Lightning Bolt posted...

DarknessLink7 posted...
But we know the sex of one of them. Sure, we don't know which one, but let's call them X. There's a 100% chance X is a girl. And there's a 50% chance the other kid is a girl, right?

What's wrong with this argument?

Because the argument is still assuming that you know which one is a girl. Out of the BG, GB, and GG combinations that were allowed, neither the first nor the second is always a girl.

I think I see what you're trying to say. Though could you explain why it's even important to know which one is the girl? Isn't it enough to be sure one of them is a girl?
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DarknessLink7 posted...

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

50%. Simply because it's 50/50 if you have a boy or a girl.
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100% because who the hell would describe a family of having at least one girl
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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?

DarknessLink7 posted...

I think I see what you're trying to say. Though could you explain why it's even important to know which one is the girl? Isn't it enough to be sure one of them is a girl?

Because which one is which is the difference between the BG and GB possibilities.
Go back to the chart again.

GG
GB
BG
BB

If you say "The one on the left is G", that's different than saying "There exists at least one G".
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Lightning Bolt posted...

DarknessLink7 posted...

I think I see what you're trying to say. Though could you explain why it's even important to know which one is the girl? Isn't it enough to be sure one of them is a girl?

Because which one is which is the difference between the BG and GB possibilities.
Go back to the chart again.

GG
GB
BG
BB

If you say "The one on the left is G", that's different than saying "There exists at least one G".

Yes, that is very different indeed. If it was specified that the "left kid" was a girl, it's obvious the only possibilities are:

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.

I'm pretty tired after a long day of work, so I'm sorry if I'm being stupid. I really appreciate you trying to explain this to me.
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DarknessLink7 posted...

However, if the "left kid" isn't guaranteed to be a girl, we instead have the situation where the "right kid" is the girl.

Neither is guaranteed to be the girl. All you know is that at least one of them is, but you can't tell which is which.
Sort of like if you met them, but they were both too androgynous to read on sight. 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.
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How the hell has this simple question generated such conversation.
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FellWolf posted...

How the hell has this simple question generated such conversation.

Math
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Lightning Bolt posted...

DarknessLink7 posted...

However, if the "left kid" isn't guaranteed to be a girl, we instead have the situation where the "right kid" is the girl.

Neither is guaranteed to be the girl. All you know is that at least one of them is, but you can't tell which is which.
Sort of like if you met them, but they were both too androgynous to read on sight. 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.

I think the reason people (me included) are confused is because the statements "At least one of them is a girl" and "Neither is guaranteed to be the girl" seem to contradict each other. The question which is which doesn't strike everyone as important, even though it is. By using Sample Space and making a list of all the possibilities like you do above we get a nice overview that makes it easy to grasp the situation. But why do we need to use that approach? Why do we need to list all the possible outcomes like that? What's wrong with accepting that one of them is guaranteed to be a girl?

I feel like a 5 year old for asking these things. :P
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T0ffee posted...

FellWolf posted...

How the hell has this simple question generated such conversation.

Math

Stupidity.

You have either 1 roll of 100% (which cancels itself out as there was no chance) followed by 1 roll of 50% or 1 roll of 50% followed by 1 roll of 100% chance (which cancels itself out as there was no chance).
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Kyuubi4269 posted...

T0ffee posted...

FellWolf posted...

How the hell has this simple question generated such conversation.

Math

Stupidity.

You have either 1 roll of 100% (which cancels itself out as there was no chance) followed by 1 roll of 50% or 1 roll of 50% followed by 1 roll of 100% chance (which cancels itself out as there was no chance).

I don't think you should be so quick to call others stupid. You might be surprised by the actual solution to this problem.
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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.