There’s a fascinating discussion taking place right now over at Marginal Revolution (and on the WSJ blog) centered around a probability problem I call the “2 daughters” problem posed by Leonard Mlodinow in his book “The Drunkard’s Walk.” I’ve quoted Alex Tabarrok’s summary of the problem here:
Suppose that a family has two children. What is the probability that both are girls? Ok, easy. Probability of a girl is one half, probabilities are independent thus probability of two girls is 1/2*1/2=1/4.
Now what is the probability of having two girls if at least one of the children is a girl? A little bit harder. Temptation is to say that if one is a girl the probability of the other being a girl is 1/2 so the answer is 1/2. That’s wrong because you are not told which of the two children is a girl and that makes a difference. Better approach is to note that without any additional information there are four possibilities of equal likelihood for the sex of two children (B,B), (G,B), (B,G), (G,G). If we know that at least one is a girl we can remove (B,B) so three equally likely possibilities, (G,B), (B,G), (G,G), remain and of these 1 has two girls so the answer is 1/3.
Ok, now here is the stumper. What is the probability of a family having two girls if one of the children is a girl named Florida?
You’ll want to have a look at the discussion over at Marginal Revolution (answers at WSJ), because there’s a lot of interesting and informed opinion being shared. Then I’ll share an even more bizarre example.
Relative Probability in Bizarro-World
I’m going to restate this as a coin-toss, because it makes it easier to create my more-bizarre example. I have taken care to ensure, as best I can tell, that my restatement is isomorphic to the original set of questions. I will present three cases which are isomorphic to parts 1-3 of the “2 daughters” problem, and add a fourth.
Two coins are flipped on a table, at which are four people.
Person A is blind and deaf: he knows there are two coins, but he can’t see either coin or hear comments made by others
Person B is blind: he can’t see the coins, but can hear what others say about them
Person C can see only one of the coins
Person D can see both coins
Now C, who can see one coin, says truthfully of the coin he can see, “one coin is a head.”
QUESTION: What is the probability that both coins are heads?
ANSWER: From which perspective?
Not being privy to this comment from C, the blind and deaf A still only knows that there are two coins on the table. From his perspective, the probability is 1/4 (the isomorph to pt. 1 of the “2 daughters” problem)
B, who is blind, now knows that at least one coin is a head. From B’s POV, the odds both coins are heads is 1/3 (the isomorph of pt. 2 of the “2 daughters” problem). From our point of view as readers, this is also the probability we observe.
C can see the one head, and knows that the other coin must be either a head or a tail. From C’s POV, the odds both coins are heads is 1/2 (the isomorph to pt. 3 of the “2 daughters” problem, though Mlodinow states his solution differently).
D knows the value of both coins. From D’s POV, there is certainty: he knows they are either both heads or one head / one tail. However, we (on the outside of the problem) cannot say at all what the probability is from D’s POV: it’s either 1 or 0, we can’t know which. It’s undefinable.
|Person A||Person B||Person C||Person D|
|A has no information to help rule out any possibilities||B can only rule out the possibility that both coins are tails||C can rule out the possibility that the coin he sees is a tail||D knows the outcome with certainty, but we don’t|
Four people. Two coins. Three probabilities. One anomaly.
Which is the right answer? What is the probability that both coins are heads? The answer: it depends on the quality of the information you possess.
This example fairly clearly illustrates the impact of imperfect information. Let’s tease this apart a little further.
B has only the same information we were given in pt. 2 of the “2 daughters” problem: he knows that one of the coins is a head, but not which one. So he can only rule out the 1-in-4 chance that they’re both tails. There’s still the possibility that they are HT, TH, or TT. Since we readers only have the same information as B, we will have to draw the same conclusion: there is a 1-in-3 probability that both coins are heads.
But wait a second.
Why can’t B can assume that C can see one coin, since he clearly has some information about at least one of the coins. So he will be aware that from where C sits, there is a 50% chance that the other coin is a head.
Why would B rest on his assessment that there is a 1-in-3 chance that there are two heads, when he knows full well that C’s assessment is 50%? B can simply place himself in C’s perspective, and arrive at a 50% probability, even though the only information he has been given is that one coin is a head. Likewise, we as readers can also adopt C’s POV, and arrive at 50%.
The solution is this: if you know that the information you received came from direct observation of one of the individual coins, then you can take this step and arrive at a 1-in-2 chance. If you don’t know how the information was derived, then you are stuck with the original 1-in-3 probability.
I found these problems so mentally tortuous that I ended up writing a computer simulation just to satisfy my intuition. It isn’t that the problems are so very complex; on the contrary, it is their seeming simplicity that makes their counter-intuitiveness so perplexing.
I need an aspirin.