Bizarre Probability Problem (1 Viewer)

[superchuck500]

That's a lawyerly perspective. However, mathematically he would be wrong if he said that one of his children is a girl, if he actually has two girls, because mathematically he has two girls. Since this is a math problem, mathematical language trumps semantics. Besides it would make the question wrong, since it would result in both probabilities being equal to 50%, so the obvious intent was to identify the portion of the whole

 

[BobE]

 

[DaveXA]

You know you're doing something right when BobE drops a gif in a Bizarre Probability Problem thread.

 

[SystemShock]

It is just the way the question was worded, and the answer the person who came up with the riddle was looking for. The question really was "What is the probability that his other child is also a girl given that the first child is a girl?" but the actual question is hidden. And yes, it is a riddle, because the actual question is hidden in the paragraph.

But when people read the question, "What is the probability that his other child is also a girl?", uncommon chromosome combinations aside, it could only be either boy or girl.

So, all things equal, whether the 2nd child is a boy or a girl is 50/50.
Given that the first child is a girl and he only has 2 , then it is 1/3.

 

[lapaz]

No, he didn't. He didn't say that he only had one girl. He said that he had two children, and that one of them is a girl. There are two possibilities for the other child. The other child could be a girl, or the other child could be a boy. However, having only two possibilities does not translate into a 50% probability.

I'll explain Part 1. The correct answer is 1/3, or 33%. With two children, there are four possible combinations
Boy/Boy
Boy/Girl
Girl/Boy
Girl/Girl

Since he said that one of his children is a girl, that means there are three possible combinations (B/G, G/B, G/G). Of those three, in only one case is the other a girl.





I didn't say it was a trick question. I said it was a bizarre probability problem. The bizarre part is not in any kind of trickery in the wording. The bizarre part is that telling us the name of his daughter actually changes the probability that the other child is also female.

There are actually only 3 possible combinations. BB, BG and GG, in other words either 2 boys, 2 girls or 1 of each. BG=GB. Each is an independent variable. The question wasn't "what is the probability of having 2 girls, if you know 1 is a girl". Then you would use the concept of given 1 gender, what is the probability of the other being a girl.

Telling us the name wasn't what changed the answer. What changed the answer is that he identified another characteristic of the one female child. Now it is no longer mathematically wrong to say you have 1 girl with a particular characteristic, because you really only have 1 girl that also fits this other characteristic.

 

[Madmarsha]

My brain hurts.

 

[DaveXA]

My brain hurts.

Drink more.

 

[Dre]

He could just be a smart-arse that plays word games - my dad does that crap with me all the time

So your other child must be a boy? No, I didn't say that. I just said one of them is a girl. The other one is too, but was I wrong?

“I used to do drugs. I still do, but I used to, too”
Mitch Hedberg

 

[guidomerkinsrules]

My brain hurts.

Want my helmet?

 

[FullMonte]

There are actually only 3 possible combinations. BB, BG and GG, in other words either 2 boys, 2 girls or 1 of each. BG=GB. Each is an independent variable. The question wasn't "what is the probability of having 2 girls, if you know 1 is a girl". Then you would use the concept of given 1 gender, what is the probability of the other being a girl.

Except that BG doesn't equal GB. There are two children, let's call them child 1 (born first) and child 2 (born last). They can each be one of two sexes. That means that there are two children, with two possible sexes. A simple array would show that there are four possible combinations. But, even if you lump the B/G and G/B groups into a single group, the math still works out.

If you took 100 families with two children, and asked you to give them the sexes of their two children, and you put it in a chart form, you would end up with (assuming equal distribution): 25 who said B/B, 25 who said G/B, 25 who said B/G, and 25 who said G/G. Now, if you want to lump the G/B and B/G together into one group, you would end up with 25 pairs that are G/G, and 50 pairs that are BG/GB. So, the probability of having two girls when one of them is a girl would still be 1/3.

Changing the conditions so that you know that name of the girl changes the odds to 1/2.

Let's look at the above group of 100 families. And, let's assume that in a given group of girls, there is a certain percentage that are named Debra. That means that for any number of girls, 1 out of X are named debra.

So, when we look at the groups that have a girl, we end up with

25 B/G with 25G/x being named Debra
25 G/B with 25G/x being named Debra
25 G/G with 50G/x being named Debra

So out of the three possible combinations that have a girl, we have 50/X girls named Debra when there is one boy and one girl; and 50/X girls named Debra when there are 2 girls. So, it's a 1/2 chance at that point.

 

[lapaz]

Except that BG doesn't equal GB. There are two children, let's call them child 1 (born first) and child 2 (born last). They can each be one of two sexes. That means that there are two children, with two possible sexes. A simple array would show that there are four possible combinations. But, even if you lump the B/G and G/B groups into a single group, the math still works out.

If you took 100 families with two children, and asked you to give them the sexes of their two children, and you put it in a chart form, you would end up with (assuming equal distribution): 25 who said B/B, 25 who said G/B, 25 who said B/G, and 25 who said G/G. Now, if you want to lump the G/B and B/G together into one group, you would end up with 25 pairs that are G/G, and 50 pairs that are BG/GB. So, the probability of having two girls when one of them is a girl would still be 1/3.

Changing the conditions so that you know that name of the girl changes the odds to 1/2.

Let's look at the above group of 100 families. And, let's assume that in a given group of girls, there is a certain percentage that are named Debra. That means that for any number of girls, 1 out of X are named debra.

So, when we look at the groups that have a girl, we end up with

25 B/G with 25G/x being named Debra
25 G/B with 25G/x being named Debra
25 G/G with 50G/x being named Debra

So out of the three possible combinations that have a girl, we have 50/X girls named Debra when there is one boy and one girl; and 50/X girls named Debra when there are 2 girls. So, it's a 1/2 chance at that point.

That’s a strange analysis and I’m fairly sure it is wrong. The only thing that matters is whether you know there is only 1 girl or whether the question reveals a clue of how many girls there are. Chuck made a good case that the question doesn’t give that away due to word play. That is the only basis for anything other than a 50% chance the second child is a girl. You’re adding qualifiers/characteristic of birth order, which I stated does change things. If the question was what are the odds of having 2 girls given the first born was a girl, then I would use the matrix of 3 possibilities (bb, bg, gg), and the answer would be 1/3rd. If you don’t know they have 2 girls, then the odds of the next child being a girl is 50%, no matter what name the other child has.

 

[Krodwhodat]

Where is @zeetes he has a big brain right?

 

[FullMonte]

That’s a strange analysis and I’m fairly sure it is wrong. The only thing that matters is whether you know there is only 1 girl or whether the question reveals a clue of how many girls there are. Chuck made a good case that the question doesn’t give that away due to word play. That is the only basis for anything other than a 50% chance the second child is a girl. You’re adding qualifiers/characteristic of birth order, which I stated does change things. If the question was what are the odds of having 2 girls given the first born was a girl, then I would use the matrix of 3 possibilities (bb, bg, gg), and the answer would be 1/3rd. If you don’t know they have 2 girls, then the odds of the next child being a girl is 50%, no matter what name the other child has.

I'm not using birth order as a qualifier, I'm simply using it as a way to show that one particular child (in my example the oldest), has two possible sexes; and for each of those two possibilities, there are two possibilities for the sibling's sex. That means there are four possibilities. You are correct that if one particular child is female, then the probability that their sibling is female is 1/2. But the question doesn't ask that. The question says that there are two children, and one of them is female. That means that there are two possibilities for which is female.

Here's another way of looking at it. Let's take two seats at a Saints game. Seat 1 and Seat 2. Two people sit down. What are the possible combinations of sexes in those two seats?
Seat 1 Seat 2
Male Male
Male Female
Female Male
Female Female

So, if someone in one of the seats is Female, what is the probability that the other seat is occupied by a female? There are three combinations where one of the two is female, but only one combination where both are female. So, there is a 1 in 3 probability that the second seat is taken by a female.

If the question was what are the odds of having 2 girls given the first born was a girl, then I would use the matrix of 3 possibilities (bb, bg, gg), and the answer would be 1/3rd.

I don't think that's right either. Because if the first born was a girl, then your BB possibility doesn't exist. If the first born is a girl, the probability of a second girl is 1/2 because there are only two possibilities: G/B, G/G.

 

[tomwaits]

This is the Let's Make a Deal thing all over again. 2 of his children are mules and the other one is a car.

Something like that.

 

[lapaz]

I'm not using birth order as a qualifier, I'm simply using it as a way to show that one particular child (in my example the oldest), has two possible sexes; and for each of those two possibilities, there are two possibilities for the sibling's sex. That means there are four possibilities. You are correct that if one particular child is female, then the probability that their sibling is female is 1/2. But the question doesn't ask that. The question says that there are two children, and one of them is female. That means that there are two possibilities for which is female.

Here's another way of looking at it. Let's take two seats at a Saints game. Seat 1 and Seat 2. Two people sit down. What are the possible combinations of sexes in those two seats?
Seat 1 Seat 2
Male Male
Male Female
Female Male
Female Female

So, if someone in one of the seats is Female, what is the probability that the other seat is occupied by a female? There are three combinations where one of the two is female, but only one combination where both are female. So, there is a 1 in 3 probability that the second seat is taken by a female.



I don't think that's right either. Because if the first born was a girl, then your BB possibility doesn't exist. If the first born is a girl, the probability of a second girl is 1/2 because there are only two possibilities: G/B, G/G.

You’re right about the latter if given the first is a girl. The possibilities would only be gb or gg, so it would be 50% chance the next one is a girl. I had said that in a prior post, and hastily added that last piece as I was parked at a fast food restaurant. I still say the answer to the original question only depends on what you know based on the question. If the question is what are the odds of gg, without knowing anything else, then it is 1/2 x 1/2. The name doesn’t matter except to add clues about the number of children and their gender.