Irony Alert: I got an alert & hide in the Hillary Clinton group for talking about the alert stalking

[Jim Lane]

91. You don't need a Ph.D. to do basic arithmetic.

You're correct that, if there were only four people on a jury, the chance of picking four Clinton supporters would be .15^4. On my calculator that's .00050625, which, as I said, is about 1 in 1,975 (odds against of 1,974 to 1, not 50,000 to 1).

Are you getting 50,000 by just moving the decimal enough places to accommodate the nonzero digits? That's not how it works. To see that your approach is wrong, suppose that there were only 10% Clinton supporters. Then the probability would be .1^4, or .0001. If you're using the move-the-decimal method, then the odds against the four-out-of-four Clinton jury would be 1,000 to 1. That would mean that, if we up the proportion of Clinton supporters from 10% to 15%, the chance of the all-Clinton jury drops by a factor of 50. That can't be right.

Remember that .00050625 isn't odds or a percentage; it's the probability, on the normal zero-through-one scale. If you want percentages, then .00050625 is .050625%. The fraction .05 equals one twentieth, so .050625% is a bit more than one-twentieth of one percent, i.e. a bit more than one-twentieth of one-one hundredth, i.e. a bit more than one in two thousand.

Of course, all this is for the nonexistent four-person jury. Having seven slots instead of four greatly increases the chance. I claim no Ph.D. but I'll set forth my method, and I'd be very interested in having it vetted by your expert friend.

Using H to represent a Hillary Clinton supporter and N to represent a non-Clinton supporter, we could get a Clinton-majority jury if the seven jury slots were filled this way: HNHNHNH. For each pick the software does, the chance of H is .15 and the chance of N is .85, so the chance of this particular alignment is .15 x .85 x .15 x .85 x .15 x .85 x .15, or .15^4 x .85^3, or .00031090078125. The key point about a seven-member jury, though, is that there are so many different ways to get a 4-3 split. Listing the jurors as #1 through #7, we might find HNHNHNH, as above, but we might also find HHHHNNN or NHNHNHH etc. Each of these has the same probability of occurring, namely .00031090078125, so we have to add all those probabilities.

How many are there? Out of slots 1 through 7, how many different ways are there of making four of the slots be H? This is the number of possible combinations of 7 objects taken 4 at a time. This is sometimes written as ₇C₄ and sometimes as a 7 over a 4 with this vertical pair encased in big parentheses that we probably can't do on DU. It's sometimes voiced as "seven choose four". The formula for _nC_r is n!/r!(n-r)!, where the exclam means factorial (multiply that number by every integer below it down through 1). For ₇C₄ this means 7!/4!3! which equals 35. If you take the trouble to list all the different strings of H's and N's that have precisely four H's and 3 N's, you'll find 35 of them. Each has the probability of .00031090078125. The probability that any particular jury will follow one of these 4-3 patters is 35 x .00031090078125, which is just under .011 (I weary of writing out these long numbers).

That's the probability of precisely a 4-3 split. We could go through the same exercise for juries of 5, 6, or 7 Clinton supporters, which collectively add a bit more than a tenth of a percent to the chance. That's how we get from just under 1.1% for the 4-3 split to just over 1.2% that any particular jury will be a Clinton-majority jury.