Here’s a riddle.

Question: I donated a kidney anonymously on Wednesday, September 29, 2010. This is a rare act. Perhaps 300 people worldwide did it last year. I also write extensively about kidney donation. I was reading Renal & Urology News May 2011 and I saw a story written by a person explaining why he/she donated a kidney to a stranger on Wednesday, September 29, 2010. But it wasn’t written by me. It was a weird experience reading the story about someone who is very similar to me. (I encourage you to read the article.) What are the odds that two people who enjoy writing also donate a kidney anonymously on the same day?

Short answer: Ex post, p=1.

Long answer: Before the two surgeries occur (called ex ante), the joint probability that my surgery (let’s call it event A) occurs on the same day as the other donor (let’s call it event B) is written as P(*A* ∩ *B*). We want to break this probability into two parts. First is the probability of my surgery happening on a particular day given that the other person donates the same day. This is written as P(A|B). Similarly, the probability of the other donor’s surgery date given mine is P(B|A) and the joint probability is obtained by multiplying the two together, P(*A* ∩ *B*) = P(A|B)*P(B|A).

In this case, I am certain (P>.99) that the date of my surgery was not influenced by the other donor. I was unaware of the existence of the other donor until I saw the story in Renal & Urology News. Thus, we can write P(A|B) = P(A).

Further, I will assume that the other donor’s surgery date was unaffected by my date and so P(B|A) = P(B). Thus, I ignore the possibility that the other donor or his/her surgeons read this blog and selected the donation date to match mine. I will also ignore the possibility of spooky effects like quantum entanglement, ESP, and God’s will forcing the two surgery dates to be identical.

Now we have P(*A* ∩ *B*) = P(A|B)*P(B|A) = P(A)*P(B).

Now, I will assume that the surgery dates for both me and the other donor are random and independent. If this is true, then P(B) = P(A). Substituting gives us P(A)*P(B) = P(A)^2.

Actually, this is not quite true. Elective surgeries are not randomly scheduled. For instance, surgeons like everyone else, want their weekends free and dislike scheduling elective surgeries on Saturday or Sunday. Similarly, surgeons like to visit their patients for two days after surgeries, but want to avoid coming in on weekends. Thus, they don’t schedule elective surgeries on Thursdays or Fridays. Finally, emergency care patients who enter the hospital on weekends are often taken into surgery on Monday. Thus, elective surgeries are nearly always scheduled on Tuesdays and Wednesdays. Eliminating the weeks of New Years, Christmas, and Thanksgiving, the Tuesdays after 3-day weekends, and allowing time off for vacations leaves about 90 possible surgery dates each year.

Now, there are about 300 other nondirected donors, so on average over 3 (300/90) nondirected donors will have surgery on the same day. Note however, that it is unlikely that the doctors at my hospital are on vacation the same dates as the doctors at the other donor’s hospital, or have the same holiday schedule, so this estimate isn’t quite right. Further, not all 300 donors like to write. And not all the writers will be English speakers. Now we have a complicated mess.

Yuck. Let’s start over. Instead, let’s look at the probability that an event will occur after we know the outcome, called ex post. It is always either 100% (it happened) or 0% (it didn’t happen). In this case, we know it happened so P=1.

[Update: I clarified the logic. I also changed the wording to indicate that I don’t know the gender of the other donor. On initial reading of the story, I thought it was written by a man. Now I think it is a woman. But since the writer is anonymous, I can’t be sure. About 60% of anonymous donors are female. (But that doesn’t mean there is a 60% chance that I am female.)]