Thursday, 1 July 2010

Nine is a Harmonious Number

David Bee sent this one to the AP Stats EDG and it caught my eye:
At the end of each Court term the NYTimes has a chart showing how
many agreements there were between each of the 9C2 = 36 possible
pairings of the nine Justices. (There was something someone in the
Forum posted earlier that reminded me of this but I don't recall
it.) [For example, the highest percentage of agreements was between
Justices Scalia and Thomas (92 percent) and the lowest was between
Justices Stevens and Thomas (60 percent). Thus, for the 36 pairings,
60% <= in agreement <= 92% --- math and stat teachers should have it
so good...;^).]

I guess I always thought questions that got to the Supreme Court would be pretty much 50/50 propositions, and was surprised to find out that Justice Stevens, the longest serving of the Justices, who had the lowest percentage of voting with the majority, was still in the majority 73% of the time; and Roberts and Kennedy agreed with the majority a whopping 92%/90% of the time.
So the stats question is: If all nine judges voted randomly on each decision, what percentage of the time would they be in the majority. (a computer simulation is un-acceptable, come let us reason together)..

Those who want the original NY Times article may find it here.


Jeffo said...

I believe the answer is 163/256, about 64 percent.

There is one way that a justice can be a minority of one. The number of ways to be in a minority of two is the number of ways to choose one justice from the remaining eight to join him/her. Similarly, the number of ways to be in a minority of three is 8 choose 2. Proceeding similarly, one adds up a series of binomial terms. There are 2^9 possible outcomes of any case. Also, because a justice could be in the minority either for or against a case, we have to multiply the series by two.

Anyways, in general, for a n-member (n odd) court, we should get the correct probability by summing n-1 choose i as i varies from 0 to (n-1)/2, then dividing by 2^(n-1).

Pat's Blog said...

That's what I got, Jeffo, by much the same way. I just summed (9 choose n) (n/9) from n=5 to 9 and then doubled it for the 512 possible yes/no votes on an issue from nine justices.

Jeffo said...

It appears the probability of siding with the majority approaches 1/2 as the size of the court approaches infinity -- for a court with 1 million plus 1 judges, the probability is approx 0.500399. This seems reasonable, since for large courts most cases will be "close" decisions, and so the probability of siding with the majority will be much closer to 1/2 for more and more of the cases. Unfortunately, I am not able to give a more convincing analytic argument. Also, Mathematica could not evaluate the limit.

Anonymous said...

Let us concern ourselves with a slightly humorous judge, Justice Magoo. Justice Magoo's eccentricities include losing the paper where he writes his vote, and actually voting by pointing, blindfolded, the day after.

So, where are the other 8 votes? Divided. How?

Evenly C(4,4)/256 of the time, unevenly the rest.

Magoo picks the losing side only when the division is uneven, and then half the time.

(1 - C(4,4)/256)/2

Winner? The rest. 1/2 + C(4,4)/512


Pat's Blog said...

I think JD meant to say 1/2 + C(8,4) /512...
I think that because that would make his answer agree with Mine and Jeffo's, and I think he is probably clever, and would get it right, and I want my answer to be right also...
But I love the simplicity of that approach... You can extend it to any (odd) number of justices so easily... I would lie and say I thought of it first, but you wouldn't believe me...

Jeffo said...

And since C(n,n/2)/2^(n+1) approaches zero as n approaches infinity, the probability of siding with the majority does approach 1/2 as court size approaches infinity. Nice!