A girl named Florida

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Posted by: stak
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Posted on: 2009-03-07 16:32:34

So I came across the "girl named Florida" problem today on somebody's blog. (Part 1, 2, 3). The problem is taken from the book "The Drunkard's Walk" by Leonard Mlodinow and is easily googleable. Here's the problem:

Suppose that a family have two children.
1. What is the probability that both of them are girls?
2. Suppose you know that at least one of the children is a girl. Now what is the probability that both children are girls?
3. Suppose you know that at least one of the children is a girl, and her name is Florida. Now what is the probability that both children are girls?

The answers provided are 1/4, 1/3, and 1/2, respectively (I'm not sure if these answers are actually in the book, but I assume they are based on what everybody is parroting). You can follow the supposed logic in the part 2 link above, or in question 3 here.

The thing is, the solutions given make no sense intuitively. And therefore, I don't think they're right. Knowing the name of one of the girls shouldn't change a thing, because you can assume that the girl is named "x" and the exact same reasoning would apply. Therefore the answers to question 2 and question 3 must be the same.

My answers are 1/4, 1/2, and 1/2. For the second question, there are really four possibilities with equal probability: BG, GB, GG, and GG. And half of them have two girls. Another way of looking it is to simply discount the child you know is a girl. Since the two children are independent events, the second child has a 50% chance of being a girl, and therefore there is a 50% chance that they are both girls.

I'm convinced my solution is right. As a consequence I'm also convinced that everybody else must be stupid, starting with the author of the book (assuming the book says 1/3). Somebody please prove me wrong.

Posted by Dave P at 2009-03-07 18:21:44

Hey I am hopefully not too stupid. The events GG and GG are the same event, yes?

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Posted by stak at 2009-03-08 00:23:14

Not according to my logic, since you don't know which of the two children is the girl you know about. See my other comment below.

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Posted by rohan at 2009-03-07 21:01:52

I haven't thought about 3. yet but on 2. you are definitely wrong. This is a problem that confuses many. Google for example Marilyn vos Savant.

Let X_1 = 1, if child 1 is a girl, 0 if boy
Let X_2 = 1, if child 2 is a girl, 0 if boy

Assume that X_1 and X_2 are independent random variables.
Assume that Prob(X_1 = 1) = Prob(X_2 = 1) = 1/2.

Then we have (note: \cap = intersection, \cup = union)
Prob(both girls | at least one girl)
= Prob( X_1 = 1, X_2 = 1 | X_1 = 1 or X_2 = 1)
= Prob( {X_1 = 1} \cap {X_2 = 1} | {X_1 = 1} \cup {X_2 = 1})
= Prob( {X_1 = 1} \cap {X_2 = 1} \cap ({X_1 = 1} \cup {X_2 = 1})) / Prob({X_1 = 1} \cup {X_2 = 1})
= Prob( {X_1 = 1} \cap {X_2 = 1} ) / Prob({X_1 = 1} \cup {X_2 = 1})
by independence of X_1 and X_2:
= Prob (X_1 = 1) * Prob(X_2 = 1) / Prob({X_1 = 1} \cup {X_2 = 1})
= Prob (X_1 = 1) * Prob(X_2 = 1) / (1 - Prob({X_1 = 0} \cap {X_2 = 0}))
by independence of X_1 and X_2 again:
= Prob (X_1 = 1) * Prob(X_2 = 1) / (1 - Prob({X_1 = 0})*Prob({X_2 = 0}))
= (1/2) * (1/2) / (1 - (1/2)*(1/2))
= 1/4 / (1-1/4)
= 1/4 / (3/4)
= 1/3

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Posted by rohan at 2009-03-07 21:38:15

I didn't believe that question 3 would be different from 2 but according to calculation it's true...
However, it depends on the probability of the parents naming a child Florida.

He assumes that a parent would give two children names with independent probability even if they get the same name... assuming that I get an answer of
(8-p)/(16-4p),
where p = the probability that a girl gets named Florida

When p << 1, this is around 1/2.

I still need to think about it to understand the reason

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Posted by rohan at 2009-03-07 22:03:01

ok this is really weird.
Either my calculations are totally wrong, or the answer to 3 actually becomes 1/3 under the assumption that both children can't be named Florida.

More precisely it becomes (2-p)/(6-2p)

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Posted by rohan at 2009-03-07 22:38:49

ok that last comment was wrong, it is ~1/2 for q3 and 1/3 for q2.

Here's a geometric proof for both of them.

Draw a 1x1 square to represent the probability space.

Draw a vertical line down the middle: points to the right mean the first child is a boy and points to the left mean it is a girl.
Now within the left half of the square, draw another vertical line, whose distance from the edge represents the proportion of girls named Florida. So you now have 3 rectangles, whose areas represent the probability of a child being Florida, a girl not named Florida, and a boy.

Now for the second child, turn your paper 90 degrees ccw and do the same thing to the square.

For question 2, you want the probability of being in the top-left quadrant given that you are in one of the 3 quadrants other than the bottom-right. This is 1/3.

For question 3, it is the probability of being in the topmost-quadrant given that you are in the union of the leftmost and topmost rectangles.
By computing the areas, this is (1-a)/(2-a) where a is the width of the leftmost/topmost rectangles.

Note that if every girl were called Florida, then a = 1/2 so we get 1/3.
If no one is called Florida, then a = 0 and we get 1/2.

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Posted by Eric at 2009-03-07 22:31:58

For all sets of two children, the cases are BB, BG, GB, GG. Note that BG is independent of GB since order matters. For question 2, we know there is at least one girl, so our choices are BG, GB, GG, all equally likely, of which only one has two girls, so the probability is 1/3. The catch here is that in BG sets, the girl can be elder or younger.

Easy. And hopefully can now be intuited.

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Posted by stak at 2009-03-08 00:21:38

So why is it that in the BG sets the girl can be elder or younger, but not in the GG set? All you know is that one of the children is a girl. In the case of two girls, you don't know if the one you know about is the older of the two or the younger.

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Posted by stak at 2009-03-08 00:19:35

Hm. All your arguments seem convincing, particularly rohan's proof. However the end result still seems very wrong to me. Here's how I was thinking about it.. start with a base case and add stuff to it as we go.

Base case: there is a house with two children, call them X and Y. You enter the house and see a child at random. You don't know if the child you see is X or Y; there is a 50% chance of it being X and a 50% chance of it being Y.

Now add gender to the children. There are four equally possible gender assignations, resulting in a total of 8 equally possible scenarios:
1) X is a boy, Y is a boy, and you saw X
2) X is a boy, Y is a boy, and you saw Y
3) X is a boy, Y is a girl, and you saw X
4) X is a boy, Y is a girl, and you saw Y
5) X is a girl, Y is a boy, and you saw X
6) X is a girl, Y is a boy, and you saw Y
7) X is a girl, Y is a girl, and you saw X
8) X is a girl, Y is a girl, and you saw Y
Note that each of these scenarios has a probability of 12.5%. The probability that you saw X is still 50%, and you have no new information as compared to the base case.

Now we add the last component. You realize that the child you're looking at is a girl. This eliminates 4 of the above possiblities, leaving just these 4:
4) X is a boy, Y is a girl, and you saw Y
5) X is a girl, Y is a boy, and you saw X
7) X is a girl, Y is a girl, and you saw X
8) X is a girl, Y is a girl, and you saw Y
Of these 4 equally likely possibilities, two of them are the "two girls" scenario and two are not, giving you a 50% chance that the house has two female children.

Where is this reasoning wrong?

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Posted by rohan at 2009-03-08 01:26:39

yes, I assume this is what you found out in your post below, but you don't enter the house and see a child at random; you are told if there is a girl so you can't miss her by seeing a boy instead.
so the two different GG cases don't have equal probability to the BG and GB

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Posted by stak at 2009-03-08 00:38:13

Ah! The question is ambiguous. I was interpreting it one way and it seems the accepted interpretation is the other.

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Posted by stak at 2009-03-08 01:17:20

Upon further reflection, I stand by my 1/2 answer for question 2, given the formulation of the question as in my post. The problem starts off with a family, and the only restriction is that it has two children. I read the problem as referring to the same family in all three questions. Given that the family is selected from the entire set of two-children families, and that family is later discovered to have one girl child, my answer was correct. If different families are used in the three questions, and the family in question 2 had been selected from the set of all two-children families with at least one girl initially, then the answer would have been 1/3.

However, I no longer think everybody else stupid :)

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Posted by rohan at 2009-03-08 01:33:02

I don't understand, your two formulations sound exactly the same.
Selecting from the entire set of two-child families and then conditioning on having at least one girl is the same as selecting from the set of all two-child families with at least one girl initially.

In the first case you select from GB, BG, GG and BB and then assuming you got one of GB, BG, GG, the probability of GG is 1/3.

In the second you picked from GB, BG, GG and the probability of GG is 1/3.

what's the difference?

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Posted by stak at 2009-03-08 02:33:15

I think there is a difference. If you select from the set of all two-child families with at least one girl initially, then the only possible child combinations are GB, BG, and GG. This means the answer to question 1 is 1/3. We all agree that the answer to question 1 is 1/4, which means that BB must be a possibility at least initially. The other formulation does not have this restriction, because the BB possibility is only removed at question 2. However, you're right that for question 2, the two formulations are the same, so I was wrong there.

The difference, as discovered here, is the method by which the "one girl" condition is ascertained. And given the exact problem statement above, there isn't really any way to determine which method was used. So yeah.. my "further reflection" was flawed, and the question is in fact ambiguous.

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