Ok, you asked for it Beavis. Maybe it's just the phenominal effort it seems to get to the answer in each case, but here ya go....and this is actually gonna be the short version.

The example he gives (and once you understand what he's doing, you can change the cards to fit your situation) is that you raise with JJ and get 2 callers.

He puts those callers on a range [JJ,22] [AK,AJ] [KQ]. (That's not the range I'd put most people on, that's certainly not my calling range and I probably wouldn't put both callers on the same range, but I won't argue this because again, once you understand, you can change to suit your needs)

He gives us 2 flops:
AA3 and 553

The equation he then gives is the % that 1st caller is ahead or P(P1_ahead)
plus the % that P(P1_not ahead)
multiplied by % that the 2nd caller is ahead or P(P2_ahead)

P(P1_ahead)+P(P1_not ahead)P(P2_ahead)=P(beaten)

He figures out that with that range of hands the answers for the 1st 2 parts of that equation are .26 and of course .74 so we have

.26 + .74 x P(P2_ahead) = P(beaten)

Now this is where it gets tiresome.

If we know that P1 doesn't have you beaten (because it really doesn't matter what P2 has if your already beaten by P1) we have to lessen his range to only those hands in that range that don't beat you and then figure out the probability that P2 is holding each of those hands based on what P1 could possibly hold. Please don't ask me to go into the calculations. It's a LOT of effort. For me at least.

Then to figure out for more than 2 opponents, he uses something called and independence approximation that can be found by the equation:

P(beaten) = 1 - (1 - .26)" = 1 - .74"

The quotes reference that this is an approximation and I have no idea how that equation comes about even though he tries to explain it afterward.

I need to remember to come back to this tomorrow - probably at work when I am on a break (which is about 6 hours a day).