It back ..aargh! One more try at explaining it.

All that matters to distinguish situation 2 (at least one girl) from situation 3 (one is a girl named Florida) is that the two siblings have different names.
In situation 2 either child could be a girl so there a 3 ways it can happen. (gb bg gg)
In situation 3 one child (Florida) is fixed to be a girl so only 2 ways are left (gb gg).



Say the girls are behind a curtain and no names are used and for situation 3, we learn that the one on he left is a girl. Then gb or gg is all that is possible. Same principle as using unique name.

If we take away the identifier in sitaution 3 we are adding an extra way that mixed sexes could occur thus reducing the chances of same sex children so that is why the chances of two girls situation 2 is reduced to 1/3 compared to situation 3 (1/2).

Pokerlogist this is in regards to situation 2 and not situation 3.

My point doesn't have to do with the probability theory of this problem (which is fairly trivial), but why so many people get it intuitively wrong - 85% of MBA students according to one study (wiki article on this). I then came to the conclusion that so many people get the problem wrong because of the way the phrase "at least". In real life when people communicate "at least" to each other, they usually imply that they have looked at a fixed reference and then communicate their lack of knowledge about the rest - or maybe they are just trying to be creative and they are trying to get the other party to guess, but in any case they have a fixed reference in their mind.

The other party trying to guess what they are saying will usually assume that this is the case. But in this problem the reality is that the original party looked at the full data and then said "at least" to the other party (i.e. communicating only partial data of what they know). The other party's assumption is obviously wrong, but the assumption models closer to how people communicate.

For example if I ask my doctor: "what is the sex of my twins" and the doctor says "you will have at least one girl". I will then assume he only looked at one twin (for whatever reason not both): and I will therefore have 50 / 50 chance of having two girls. Why the heck would I assume that he looked at both, scrambled them in his mind, and then randomly said: "you will have at least one girl" to make the probabilty of having two girls 1 / 3 of the time. People just don't communicate like that. That is why so many people get number 2 situation probability wrong. It doesn't have to do with misunderstanding probabilty as with presenting this problem in a way that goes counter to how people communicate.

MBAs try to get "off the hook". There should be a way of wording the problem without any misleading language and re-testing them.

BTW, The use of "twins" as an example makes for another puzzle. Though fraternal twins are more common than identical twins, they account for only about 2/3 of twin pregnancies.

This is wrong. Let's say these 2 troublesome punks are wearing Halloween costumes and are standing right in front of you. Their gender is not apparent.

Their mother is also standing there and informs you that at least one is a girl. This gives us the possibilities of GB, BG and GG. P(GG) = 1/3.

The mother then says her name is Florida. What does this change? Nothing, zero, zilch, nada. You still don't know if the girl is standing on the left or on the right. This gives us the possibilities of GB, BG and GG giving P(GG) = 1/3 still.

Knowing the name of the girl plays no role. The rarity of the name plays no role. This Florida name "twist" is pure BS. The mother would literally have point to the one that is a girl for you to know that one is a girl (duh!). And if she did, only then does this restrict the possibilities to GB and GG giving P(GG) = 1/2.

BTW, as for the names rarity playing no role, once you are informed that the name is Florida, it does not matter if the name is the most common one on the planet or the most rare. The fact of her name being what it is becomes 100% whether it is Florida or Jane.

Gaining the name does give you more information. Without the name, you have three combos:

GG
BG
GB so we agree that this is a 1/3 probability.

But with the name, we now have four possibilities.

FB
BF
FG
GF

Unlike the first combo of girls only, FG and GF are unique solutions, giving us 4 combos, not 3.

Of which, 50% are GG, 1 in 2.

The girl's name makes no difference at all if you make the reasonable assumption that a parent will not name two daughters the same thing. If you instead assume that names are drawn at random from some probability distribution, hearing an uncommon one makes a small difference.

If you are given a girl's name A, and let's denote names other than A with N, then the possibilities for two girl families are:

G-A, G-A
G-A, G-N
G-N, G-A

For exceedingly rare names, G-A G-A is a vanishingly small probability and can be disregarded. Even if we draw names by chance, and we have one girl named Florida, we will almost never have two girls called Florida. This is not true of Jane/Jane. Of course if we assume parents always choose different names for their children it is an entirely moot point.

It seems like it shouldn't matter but it does.
The one named Florida is a girl. What are the chances that the sibling not named Florida is a girl? 1/2, right? Therefore p(gg)=1/2.

I have been thinking about your example a bit...so:

You are correct in this scenario. Because we are letting the mother choose the information, we gain nothing.

Out of all two children trick-or-treaters dressed as unidentifiable ghosts, at least one of whom is a girl, the odds of both being girls is 1/3.

Ironically enough, we are still 1/3. Of all the 'at least one girl' families that come, the mother points and says "That one is a girl" and unmasks her. Two out of three times the other sibling will be a boy (This is similar to the Monty Hall problem. Two out of three times the mother only has one choice if forced to unmask a girl).

The confusion is coming from how the information is acquired. In your example, no new information is gained, because you are giving out random information out. "This one is a girl, her name happens to be Sue".

But, suppose I give a girl a unique status. Out of all girls born first with a sibling, what are the odds of a 2 girl family?

1/2, yes?

Now, lets assume that unique status is the girls name.

Lets say we have 300 two sibling families, all with at least one girl. 100 families have boy/girl, 100 have girl/boy and 100 girl/girl.

There are 400 girls and 200 boys.

We have a big Halloween candy/prize hunt, 400 pink boxes for girls only, 200 blues for the boys.

In one of those pink boxes there is a free ticket to Disney World. On the prize it simplys says 'Florida'.

The kids now all go get in their indentical ghost costumes and come trick-or-treating to your house in sibling pairs.

Every time they come, you ask the mother, did your daughter win the Florida ticket?

Finally one mother says yes, here is the lucky girl. We have nick-named her 'Florida', and she unmasks her.

Now, what are the odds the other sibling is a boy or a girl?


There were 400 girls.

200 had sisters, 200 had brothers.

It is even money that the sibling is a boy or a girl.

Whew.

This is the same question as http://forumserver.twoplustwo.com/25...-poker-530700/

To cut a long story short, the point of disagreement between people is how you model this situation mathematically.

Even if he did that, you'd be wrong assume that the probability of having two girls is 1/3 unless you know that the doctor would never tell you that one of the children was a boy, which seems highly unlikely. If he randomly decided which of the two to tell you about, then you would be, just as in the naive answer, 50/50 to have a boy or a girl as the second one, and the information about one child doesn't tell you anything about the other. It's only if he has a bias that the probability shifts.

The math is simple to work out: 1/4 of the time it's BB and he tells you B, 1/4 of the time it's GG and he tells you G, 1/4 of the time it's mixed and he tells you B, 1/4 of the time it's mixed and he tells you G. So 1/2 the time the other child is a boy, 1/2 the time it's a girl, no matter what he tells you.

The only way you get the 1/3 result is if he tells you nothing at all when there are two boys, and G otherwise.

I've found it enlightening to think about why this logic doesn't affect the original problem (#2) as stated in this post, where the correct answer is, indeed, 1/3...

I can't agree - in that thread, the dispute really boils down to whether you walked up to a random -woman with a child- (that just happens to be a boy), or a random -woman with a boy-, and the problem doesn't specify which is the case. In other words, suppose that woman had shown up with a girl: would we have picked a different woman, one with a boy, to focus on, or changed the problem and said "This woman has a girl, what's the probability that her second child is a girl?" The sample space really matters here, and two people investigating women with children at the park (pervy as that may be) could come to very different probabilistic conclusions about the same woman depending on the populations they drew that woman from - worse, they'd both be "right"!

Here in part 2, it seems much clearer to me that we're drawing a random -family with a girl-, leading to the 1/3 result pretty unambiguously, and in part 3, we're drawing a random -family with a girl named Florida-, approaching the ~1/2 answer asymptotically as the probability of finding a girl named Florida approaches 0. I think this problem is a lot better defined than the other one, probably because the "story" elements that cause so much trouble in these things were left out of it.

All right, I'll bite and take a crack at this. The problem with both 2. and 3. is that the questioner has not made clear whether or not he has looked at both children.

In 2. it is more strongly suggested that he has told you there is a girl after looking at both children.

In 3. it is less strongly suggested.

If he looks at both and can therefore select which child to tell you about --> conditional probability applies and the events are no longer independent. --> 1/3

If he only looks at one, the sex of the 2nd child remains independent --> 1/2.

Just read the tortured arguments made from the original articles to justify a probability of 1/2, regardless. I say tortured, as they make assumptions that aren't neccessarily valid. So then, as a further clarification: the answer to 3. is it depends on what assumptions you make.

LOL
This question is back.
I already posted the answer.

would the mother be a girl in the family?

Yep
the answer is 100%

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