The Necropolitan Sentinel

I see Slate has a nifty calculator for determining the probability of death of at least one Supreme Court Justice with all sorts of simulations, etc., but it's stupid – it does not require a fancy-schmancy simulator to calculate these probabilities. 

And besides, I've got a better mortality table to use.  They're using a general population mortality table, which includes the disabled, mortally ill, etc. The justices, in general, will be in better health than the general population for a variety of reasons, one of the biggest ones being they are still working.  I could factor in known health issues, etc., for the justices, but to give a lower bound for death stats, I'm going to use an annuitant mortality table.

Annuitant mortality tables tend to assume relatively longer-lived people. Also, I'm going to throw in some mortality improvement to see how this does. The spreadsheet I'm using (which I wrote) can be found here, at the Society of Actuaries website.

To make the probabilities minimal, I will put the age to the last birthday of whichever Justice, and look at the probability of surviving 4 years. I'll use the order from the current Wikipedia article.

Roberts (57): 96.3%

Scalia (76): 80.1%

Kennedy (76): 80.1%

Thomas (64): 92.9%

Ginsberg (79): 80.8%

Breyer (74): 83.3%

Alito (62): 94.1%

Sotomayor (58): 97.3%

Kagan (52): 98.5%

Yes, Ginsberg, though older than Scalia and Kennedy by three years, has a higher survival probability using the assumptions of lower female mortality that generally pertains.

If you want the probability of at least one dying, you consider the probability of none of them dying – that is, that all of them survive. If we assume their mortality is independent (not necessarily a bad assumption), we can just multiply all the survival probabilities, which gives us 34.84% — meaning the probability of at least one dying is 65.2%…. which matches Slate's calculation of 64.5% pretty closely.

Also, it didn't require any simulations. Jeez. None of the combinations they ask require it. It's an elementary probability problem.

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