Tuesday, May 10, 2011

What were the chances?

The Obama administration clearly had some luck on its side today, drawing three Democratic appointees (and in Judge Motz, the intellectual leader of the left on the Fourth Circuit).

This naturally prompts the question: what was the probability of the panel being comprised of three Democratic appointees? In other words, how lucky was the administration?

Lucky, but not remarkably so.

The Fourth Circuit currently has eight Democratic appointees and five Republican appointees (assuming we count Judge Gregory as a Democrat). Thus, the computation (if I am recalling my work from Stats 61 correctly) is as follows:

Probability of draw 1 = 8/13 = .61538

Probability of draw 2 = 7/12 = .58333

Probability of draw 3 = 6/11 = .54545

The product of the three is .195802. In other words, the chances of three Democrats being on the panel was roughly 20 percent.

So the federal government was fortunate, but not in any sort of crazy, outlier way.

UPDATE: This computation assumes that Judge Hamilton, a G.H.W. Bush appointee now on senior status, was excluded from the pool. If we include him fully in the denominator (a conservative assumption), the probability of an all-Democrat panel was roughly 15.4 percent. If Judge Hamilton's chances of being assigned to the panel were less than that of the active judges, then the likelihood of an all-Democrat panel was somewhere between 15.4 percent and 20 percent.