# The Girl Named Florida Problem

Suppose that a family has two children. What is the probability that both are girls? Well, this is straightforward because there are four equally likely possibilities (assuming the chances of a boy and a girl are 50-50).

Let us assume that the two children are concealed from view, one behind a red curtain and one behind a yellow curtain.

Put like this, there are four possibilities:

- Boy behind both curtains.
- Boy behind red curtain and girl behind yellow curtain.
- Girl behind red curtain and boy behind yellow curtain.
- Girl behind both curtains.

So the probability that there is a girl behind both curtains = ¼.

This answers the first question. Given the information that a family has two children, the chance that both are girls is 1 in 4.

Now what if we are told that at least one of the children is a girl. This is like saying that there is at least one girl behind the curtains, possibly two.

This eliminates option 1, i.e. a boy behind both curtains, leaving three equally likely possibilities, only one of which is a girl behind both curtains. So the chance that there is a girl behind both curtains given that you know that there is a girl behind at least one curtain is 1 in 3.

This is equivalent to asking the probability that both children are girls if you know that at least one of the children is a girl. The answer is 1 in 3.

Now what if you are told that at least one of the girls has a chin. This adds little or no new information, insofar as presumably all (or the vast majority of) girls have a chin. So if I tell you that a family has two children, at least one of whom is a girl with a chin, it is giving me effectively no new information. So the probability that both children are girls if at least one is a girl is still 1 in 3.

What if instead I tell you that one of the children is a girl called Florida. This is pretty much equivalent to telling you that the family has a daughter behind the red curtain, insofar as it is not just identifying that there is at least one girl in the family, but identifying who or where she is. When now asked the probability that there is a girl behind the yellow curtain, options 1 and 2 (above) disappear, leaving just option 3 (a girl behind the red curtain and a boy behind the yellow curtain) and option 4 (a girl behind both curtains). So the new probability, given the additional information which identifies or locates one particular girl advance is 1 in 2.

In other words, knowing that there is girl behind the red curtain, or else knowing that her name is Florida, is like meeting her in the street with her parents who introduce her. If you know they have another child at home, the chance it is a girl is 1 in 2. By meeting her, you have identified a feature particular to that individual girl, i.e. that she is standing in front of you and not at home (or behind the red curtain, or named Florida and not simply possessed of a chin).

If, on the other hand, you meet a man in the pub who mentions his two children and you find out that at least one of them is a daughter, but nothing more than that, you are back to knowing that there is a girl behind at least one curtain, but not which, i.e. Options 2, 3 and 4 above. In only one of these equally likely options, i.e. Option 4, is there a girl behind both curtains, so the chance of the other child being a girl is 1 in 3.

So does it matter that the daughter has this unusual name? It does. If you know that the man in the pub has two children and at least one daughter, but nothing more, the chance his other child is a girl is 1 in 3. If you find out that the man in the street has two children, and then he tells you that one of children is called Florida, you are left with (to all intents and purposes) just two options. His other child is either a boy or else a girl not called Florida, which is pretty much equivalent to saying his other child is a boy or a girl. So the probability that his other child is a girl is now effectively 1 in 2.

The different information sets can be compared to tossing a coin twice. The possible outcomes are HH, HT, TH, TT. If you already know there is ‘at least’ one head, that leaves HH, HT, TH. The probability that the remaining coin is a Tail is 2 in 3. If, on the other hand, you identify that the coin in your left hand is a Head, the probability that the coin in your right hand is a Head is now 1 in 2. It is because you have pre-identified a unique characteristic of the coin, in this case its location. Identifying the girl as Florida does the same thing. In terms of two coins it is like marking one of the coins with a blue felt tip pen. You now declare that there are two coins in your hands, and one of them contains a Head with a blue mark on it. Such coins are rare, perhaps as rare as girls called Florida. So you are now asked what the chance is that the other coin is Heads (without a blue felt mark). Well, there are two possibilities. The other coin is either Heads (almost surely with no blue felt mark on it) or Tails. So the chance the other coin is Heads is 1 in 2. Without marking one of the coins, to make it unique, the chance of the other coin being Heads is 1 in 3.

Put another way, there are four possibilities without marking one of the coins:

- Heads in left hand, Tails in right hand.
- Heads in left hand, Tails in right hand.
- Heads in both hands.
- Tails in both hands.

If you declare that at least one of the coins in your hands is Heads, this means the chance the other is Heads is 1 in 3. This is equivalent to declaring that one of the two children is a girl but saying nothing further. The chance the other child is a girl is 1 in 3.

Now if you identify one of the coins in some unique way, for example by declaring that Heads is in your left hand, the chance that Heads is also in your right hand is 1 in 2, not 1 in 3.

Similarly, declaring that one of the coins is a Heads marked with a blue felt tip pen, the chance that the other coin is Heads, albeit not marked with a blue felt tip, is 1 in 2. Marking the coin with the blue felt tip is like pre-identifying a girl (her name is Florida) as opposed to simply declaring that at least one of the children is some generic girl (for example, a girl with a chin).

In other words, there are four possibilities without identifying either child.

- Boy, Boy
- Girl, Girl
- Boy, Girl
- Girl, Boy

If at least one of the children is a girl, Option 1 disappears, and the chance the other child is a girl is 1 in 3.

If you identify one of the children, say a girl whom you name as Florida, it is like marking the Heads with blue felt tip or declaring which hand you are holding the coin in.

Your options now reduce to:

- Boy, Boy
- Boy, Girl named Florida
- Boy, Girl not named Florida
- Girl named Florida, Girl not named Florida.

Options 1 and 3 can be discarded, leaving Options 2 and 4. In this scenario, the chance that the other child is a girl (not named Florida) is 1 in 2. By pre-identifying one of the girls, Option 3 disappears, changing the probability that the other child is a girl from 1 in 3 to 1 in 2.

The new information changes everything.

So what is the probability of the family having two girls if you know that one of the two children is a girl, but no more than that? The answer is 1 in 3.

But what is the probability of the family having two girls if one of the two children is a girl named Florida? Armed with this new information, the answer is, to all intents and purposes, 1 in 2.

Another way to look at this is to consider a set of 4,000 families made up of two children. Choose a single unique identifier of each child, say age (it could equally be height or alphabetical order, anything uniquely identifying one child from the other). 1,000 of these will be two boys – say older boy and younger boy (BB), 1,000 will be two girls – older girl and younger girl (GG), 1,000 will be Boy-Girl – older boy, younger girl (BG), 1,000 will be Girl-Boy – older girl, younger boy (GB). If you identify at least one of the children as boy, there remain 3,000 families (1,000 BB, 1,000 BG, 1,000 GB). 2/3 of these families contain a girl, so the probability the other child is a girl is 2/3.

Now, add into the mix the fact that one girl in a thousand in your set of 4,000 families is named Florida, and there are no families with two daughters named Florida.

In this case, 1,000 of these will be two boys – an older boy and a younger boy (BB), 1 will be a older boy and a younger girl named Florida (BF), 1 will be an older girl named Florida and a younger boy (FB), 1 will be an older girl named Florida and a younger girl not named Florida (FG), 1 will be an older girl not named Florida and younger girl named Florida (GF), 999 will be an older boy and a younger girl not named Florida (BG), 999 will be an older girl not named Florida and a younger boy (GB), 998 will be an older girl not named Florida and a younger girl not named Florida (GG). There will be no families with both girls named Florida.

This can be summarised as (given that B is a boy, G is a girl not named Florida, F is a girl named Florida, and the sequence is older-younger):

1,000 BB; 1 BF; 999 BG; 1 FB; 0 FF; 1 FG; 999 GB; 1 GF; 998 GG).

Given that at least one child is a girl named Florida, 4 possible pairs remain:

BF; FB; FG; GF.

Of these, 2 contain a girl named Florida:

FG and GF.

So, if we know that out of 4,000 families, one is a child named Florida (and 1 in 1,000 girls is named Florida), then what is the chance that the other child is a girl once you are told that one of the children is a girl named Florida. It is 1/2.

Appendix

The solution to the ‘Girl Named Florida’ problem can be demonstrated using a Bayesian approach.

Let P(GG) = probability of two girls if there are two children. Let G be the probability of at least one girl in the family)

Let P(GG I 2 children) be the probability of two girls given there are two children.

Let P(GG I G) be the probability of two girls GIVEN THAT at least one is a girl.

Then, P(GG I 2 children) = 1/4

P (GG I G) = P (H I GG) . P (GG) / P (G) … by Bayes’ Rule

So P (GG I G) = 1 x 1/4 / (3/4) = 1/3

P (GG I 2 children, older child is a girl)

Now there are only two possibilities, GB and GG (Older girl and younger boy or Older girl and younger girl), so the conditional probability of two girls given the older child is a girl, P (GG I Older child is G) = 1/2.

GIRL NAMED FLORIDA PROBLEM

P(GG I 2 children, at least one being a girl named Florida).

B = 1/2

G = 1/2 – x

GF (Girl named Florida) = x

where x is the % of people who are girls named Florida.

Of families with at least one girl named Florida, there are the following possible combinations, with associated probabilities.

B GF = 1/2 x

GF B = 1/2 x

G GF = x (1/2 – x)

GF G = x (1/2 – x)

GF GF = x^2

Probability of two girls if one is a girl named Florida =

G.GF + GF.G + GF.GF / G.GF + GF.G + GF.GF + B.GF + GF.B

= x (1/2 – x) + x (1/2 – x) + x^2 / [x (1/2-x) + x (1/2-x) + x^2 + x]

= 1/2 x – x^2 + 1/2x – x^2 + x^2 / [1/2x – x^2 + 1/2x – x^2 + x^2 + x]

= x – x^2 / x – x^2 + x = x(1-x) / x (2-x) = 1-x / 2-x

Assuming that Florida is not a common name, x approaches zero and the answer approaches 1/2.

So it turns out that the name of the girl is relevant information.

As x approaches 1/2, the answer converges on 1/3. For example, if we know that at least one child is a girl with a chin, x is close to 1/2 and the problem reduces the standard P (GG I G) problem outlined above, i.e.

P (GG I G) = P (H I GG) . P (GG) / P (G) … by Bayes’ Theorem

So P (GG I G) = 1 x 1/4 / (3/4) = 1/3

Further Reading and Links