The Florida problem - a new intuition

In #wordpress_migration

1. Introduction

In a prior couple of posts here and here I look into the "evil" probability problem of the girl-named-Florida. This problem compares the following two situations:

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

and

Say you know a family has two children, and further that at least one of them is a girl named Florida. What is the probability that they have two girls?

The former is easy to show is $latex {p(2g|\{L1g\})=1/3}&fg=000000$. The latter is shown to be $latex {p(2g|\{L1g\},F)=1/2}&fg=000000$. Intuition firmly insists that knowing the name shouldn't change the probability, but the math and simulations insist otherwise. Thus, it is our duty, to try to get our intuition around the problem. I was motivated to look at this again when a commenter asked

Can someone tell me what the relevance of the comparative rarity/commonness of the girl's name is? Suppose instead we knew that the girl's name was "Mary". The possibilities would still work out the same:

B GM

GM b

GM GNM

GM GM

GNM GM

After much pondering, I think I have come up with another way to recast the problem that adds to the intuition. I can't say that it makes it completely obvious to me, like the Monty Hall problem is for me now, so I think there still is something missing in my understanding of why the problem is so unintuitive. However, it does seem to push the idea a bit farther forward. In the next section I introduce another problem, with similar properties but is possibly more intuitive. I then describe how it can be used to gain an intuition on the Florida problem, and why the frequency of the name can make a difference.

2. A Card Game

Say I play a game with a very small deck (just so that we can work the numbers well). The deck has 8 cards: Ace, two, three, and four of hearts and the five, six, seven, and eight of spades. Two cards are dealt, and you are given some modest information about the two cards, and asked to determine the probability that the two cards are both hearts. Let's look at three types of information given:

  1. You're only told that there are two cards. Thus, there the probability for two hearts is simply

    $latex \displaystyle p(2H) = \frac{4}{8} \times {3}{7} = \frac{12}{56} \ \ \ \ \ (1)&fg=000000$

    This can be seen pictorially by listing every possible two-card hand and looking at those with two hearts, yielding 12 hands out of 56.

    2. You're told that at least one of the cards is a heart. Now we need to look at the first and second cards, and eliminate the possibility of two spades

    $latex \displaystyle p(2H|\{L1H\}) = \frac{p(H_1,H_2)}{p(H_1,S_2)+p(S_1,H_2)+p(H_1,H_2)} \ \
    \ \ \ (2)&fg=000000$

    where $latex {p(H_1,S_2)}&fg=000000$ is that we drew a heart then a spade. We then have

    $latex \displaystyle \begin{array}{rcl} p(2H|\{L1H\}) &=& \frac{\frac{4}{8}\times\frac{3}{7}}{\frac{4}{8}\times\frac{4}{7}+\frac{4}{8}\times\frac{4}{7}+\frac{4}{8}\times\frac{3}{7}}\\
    &=&\frac{12}{44} \end{array} &fg=000000$

    This can be seen pictorially by listing every possible two-card hand with at least one heart and looking at those with two hearts, yielding 12 hands out of 44.

    2

  2. You're told that at least one of the cards is an ace of hearts. Notice how this changes things. Now, when we outline the possibilities, we get

    $latex \displaystyle p(2H|\{L1H\},A) = \frac{p(H_{1a},H_{2n})+p(H_{1n},H_{2a})}{p(H_{1a},H_{2n})+p(H_{1n},H_{2a})+p(S_{1},H_{2a})+p(H_{1a},S_2)} \ \ \ \ \ (3)&fg=000000$

    Let's look at the numerator first.

    $latex \displaystyle p(H_{1a},H_{2n}) = p(H_{2n}|H_{1a})p(H_{1a}) \ \ \ \ \ (4)&fg=000000$

    Written like this, it is like turning over card one, seeing it's an ace of hearts, and then turning over card 2. $latex {p(H_{1a})=1/8}&fg=000000$, and seeing a heart that is not an ace is really the same as seeing any ol' heart, so it is $latex {p(H_{2n}|H_{1a})=3/7}&fg=000000$. This is not any different than the previous situation. However, the next term in the numerator

    $latex \displaystyle p(H_{1n},H_{2a}) = p(H_{2a}|H_{1n})p(H_{1n}) \ \ \ \ \ (5)&fg=000000$

    is really different, because we are told that there is at least one ace. The probability of drawing a heart that is not an ace on the first card is still $latex {p(H_{1n})=3/8}&fg=000000$. However, drawing an ace of hearts on the second card when we drew a non-ace of hearts on the first is certain, because we have knowledge that at least one is an ace. Thus, $latex {p(H_{2a}|H_{1n})=1}&fg=000000$. This gives, for the numerator,

    $latex \displaystyle p(H_{1a},H_{2n})+p(H_{1n},H_{2a})=\frac{3}{7}\cdot\frac{1}{8} + 1\cdot \frac{3}{8} = \frac{24}{56} \ \ \ \ \
    (6)&fg=000000$

    Notice that if $latex {H_{2a}}&fg=000000$ had been independent of $latex {H_{1n}}&fg=000000$ then we would have gotten the same 12/56 term in the numerator as in the previous situation. Essentially, by giving the information that there is at least one ace, you are really making the value of one card dependent on the other, and thus knowledge of one gives you knowledge of the other and the probability for two hearts goes up. The same thing happens with $latex {p(H_{2a}|S_1)}&fg=000000$, but since we compare to the case where we know there is one heart anyway, this is not a difference.

    Following through with the rest gives us

    $latex \displaystyle \begin{array}{rcl} p(2H|\{L1H\},A) &=& \frac{p(H_{1a},H_{2n})+p(H_{1n},H_{2a})}{p(H_{1a},H_{2n})+p(H_{1n},H_{2a})+p(S_{1},H_{2a})+p(H_{1a},S_2)} \\ &=&\frac{\frac{3}{7}\cdot\frac{1}{8}+ 1 \cdot \frac{3}{8}}{\frac{4}{7}\cdot\frac{1}{8}+1\cdot \frac{4}{8}+\frac{24}{56}}\\ &=&\frac{24}{56} = \frac{6}{14} \end{array} &fg=000000$

    This can be seen pictorially by listing every possible two-card hand with an ace of hearts and looking at those with two hearts, yielding 6 hands out of 14.

    3

3. Back to the Florida problem

The ace-of-hearts problem is exactly like the Florida problem, if you make the deck big enough. The key issue here seems to be that by giving a rare name to one of the girl children, it correlates the two children in a way that the independence assumptions in both the simpler problem and one's intuition break down. If you were to ``draw'' a girl first and not a Florida, then we must have a girl second named Florida. In the same way, the game show host in Monty Hall is forced to give information to the contestant through the rules of the game: 2/3 of the time he is forced to give the contestant the door with the prize.

Another thing to notice is that the frequency of the ``aces'' (or Floridas) in the problem definitely has an effect. You can confirm this by changing the information in the card game to You're told that at least one of the cards has a rank less than three. It is easy to see how this would change the probabilities.

3.1. Intuition

So why is this problem so unintuitive? I think a lot of it is related to the issues that Jeff J states in the comments.

But suppose you learned what you know about this family because you meet the mother walking with her daughter, and asked her how many children she has. When she said “two,” this scenario fits the problem statement just as well as what Brian assumed. You know the family has two children, and that at least one is a girl. But, the probability is 1/2 that she has two daughters, not 1/3 (reference: Bar-Hillel and Falk, or look at Grinstead and Snell's on-line textbook).

There is a certain ``omniscience'' assumed in the card game (not so unrealistically) and in the Florida problem (probably unrealistically) that changes the scope of the problem. Most people's intuitions are shaped by the cases like Jeff J, where we know of a specific child named Florida, and asked about the chances of having another girl which is 1/2, or even know only that there is at least one girl, but a specific one, so you get the 1/3 when thinking about it. The name, therefore, doesn't affect anything...in most realistic situations. However, in this card game it does affect the chances of having two hearts when you are restricted to the hands with at least one ace of hearts. I find that it doesn't seem to violate my intuition as badly in the card game as it does in the Florida , and much more clearly and it doesn't seem to violate our intuitions quite so badly.