July 16, 2007
Wall Street Journal law blog blooper on court size
The author of the Wall Street Journal law blog brought up an old hobby horse -- splitting the Ninth Circuit. Ashby Jones, reporting on some calculations (of dubious applicability) in a recent Los Angeles Times op-ed piece by law professor Brian Fitzpatrick, wrote that "as a court grows larger, it is increasingly likely to issue extreme decisions."
Larger courts are more likely to issue extreme rulings than smaller ones? This cannot be true in general. The effect that Professor Fitzpatrick identifies in his numerical example (which I won't describe here) is a subtle consequence of the finite size of the population (the judges) from which the three-judge panels that decide the appeals are drawn. Let N be the size of the full court, and let s be the number of "extreme" judges. It should suffice to consider the probability of selecting three extreme judges at random. This probability is
s(s–1)(s–2) / N(N–1)(N–2). (1)
As N grows larger (and s grows proportionately, as Fitzpatrick posits), the subtractions matter less and less. In the limit, the chance of an extreme panel is just (s/N)3. Contrary the the claim in the Journal's blog, this quantity is less -- not more -- than (1). The proof is left as an exercise to the reader.
July 16, 2007 | Permalink
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I'm afraid that your mathematical point is wrong. Using your equation, the probability of pulling three extreme judges under s=6 and N=28 (Fitzpatrick's numbers) is (6*5*4)/(28*27*26)=0.61%. But (6/28)^3 gives a probability of 0.98%, clearly higher than the prior number. An analytical proof would be rather tedious, but think of analyzing the probability on the second and third draws. What is bigger, 5/27 or 6/28? 4/26 or 6/28?
Anyway, why should it suffice to consider the probability of selecting three extreme judges, when the relevant probability is selecting two or three? Fitzpatrick is using the correct method for analyzing the probability of two or three extreme judges, the hypergeometric distribution. If you have N=28 judges and s=6 extreme judges, then the probability of pulling two or three extreme judges is 10.68%. If you have N=14 judges and s=3 extreme judges, then the probability falls to 9.34%.
As N grows, the distribution tends toward the binomial. The binomial distribution gives an 11.81% chance of pulling two or three extreme judges. Again, more evidence of a higher chance of an extreme panel with a larger circuit.
I think one flaw in what Fitzpatrick is saying is the assumption that the extreme judges would be divided equally between the old and new circuits. What if all the extreme judges go to one circuit? We can obviously assign a 0% probability to the other one. In the extreme circuit, you have N=14 judges and s=6 extreme judges. Now, the probability of pulling an extreme panel is 38.46%. On average between the two you a 19.23% chance of an extreme panel, far worse than before the split.
I know very little about the merits of the Ninth Circuit, but I don't think your mathematical critique holds up.
Posted by: Eric Chason | Jul 25, 2007 7:24:23 PM
My thanks to Professor Chason for the correction. In the proof I originally had in mind, I divided both sides of an inequality by (s-N) but neglected to reverse the direction of the inequality, as is required when dividing by a negative quantity.
It is easy to prove that the probability of drawing three extreme judges in a very large court with a proportion s/N of extremists is greater than the probability of drawing them in a smaller court with the same proportion of extremists. (I assume that a similar result will apply to a panel of two out of three; if so, it is not necessary to fuss with this probability. Also, it is not clear that a panel should be considered extreme when the two extremists on it are of opposite polarity.)
For a very large N, the binomial probability of a panel of three judges containing all extremist judges is (s/N)^3. The hypergeometric probability (applicable to small N) is (s/N)[(s-1)/(N-1)][(s-2)/(N-2)]. The question I posed is whether
(s/N)^3 > (s/N)[(s-1)/(N-1)][(s-2)/(N-2)] (1),
as the WSJ blog seemed to claim. Indeed, it is. The proof is short:
By definition, s < N
as < aN, where a = 1 or 2
-as > -aN
sN - as > sN - aN
s(N-a) > N(s-a)
s/N > (s-a)/(N-a)
That is, s/N > (s-1)/(N-1) and s/N > s/N > (s-2)/(N-2)
Hence, (s/N)(s/N)(s/N) > (s/N)[(s-1)/(N-1)][(s-1)/(N-2)]
Professor Chason also is correct about Professor Fitzpatrick’s numerical example comparing the situation with N=14 to that with N=28. I did not discuss this because I was only denying the intimation in the WSJ blog that larger courts always are more likely to have extreme panels than are smaller ones. Talking about two small sizes (28 and 14), however, is far more appropriate to modeling the effect of splitting the Ninth Circuit. In other words, my original mathematical claim, even if it were true, has no practical significance.
I also agree with Professor Chason's observation that the assumption that the number of extreme judges is proportional to the court size is restrictive. There are other potentially troublesome assumptions in Professor Fitzpatrick's model and its parameters. (I have described them in an essay submitted to a few law reviews. For other discussion and for Professor Fitzpatrick's defense of some of his assumptions, see the SCOTUS blog.)
Posted by: DHK | Aug 2, 2007 11:10:01 AM