[Aisthesis]

Interesting result for JJ: I ran this hand in 2 different ways. On one scenario, I have AK and AQ also calling. On the other I have both of those (underdog) hands folding.

Maximum stack for 9 players with the call turns out to be 18.73 times the pot.

Maximum stack for 9 players with AK and AQ folding is only 13.64. Quite a difference there, too.

I really don't think there's much of an alternative to running the individual candidates through pokerstove, as the exact percentages will make quite a difference in the calculation. I already do have the imprecision of not considering the special situation (that you mentioned) with SB and BB--although the JJ calculation indicates that additional callers with inferior hands will (not surprizingly) actually improve your EV and hence make the maximum stack larger.

As to checking the math: Probably the easiest way would be to run an easy hand like JJ against the superior pairs and just see if you get the same results I do (at this point, I'm just copying all the formulae to the next line in my spreadsheet, so if it's right in one instance, it should be right everywhere). In running JJ against the superior pairs, I always picked a pair where exactly one of the cards was the same suit as one of the jacks (evening out the presumed difference with same suits vs. different suits, which probably isn't a big deal in this case anyway).

Here are the formulae I'm using:

B = number of superior hands (just counting through them), so for JJ (counting all the QQ-AA hands left in the deck), there should be 18 of them.

T = number of hands that are a tie (for any pair, there's only one other pair that's tied with it, but for A9s, T=3).

n = number of players to consider (for 9 players, n=9)

f = average percent favorite of the superior hands over the given hand (for JJ, I come up with 81.67% against QQ-AA).

With those variables, my formula for each column is this:

p = [(1325-B-T)/1325]~(n-1) This should be the probability that no one has a superior hand and hence all will fold.

s = 1 - [(1325-T)/1325]~(n-1) This is the probability that there is some hand that's a tie with yours (it's always pretty low).

Stack-size is then figured as:

X = p/[(2*f-1)*(1-p-s)]

I got that by figuring the break-even for stack-size as when

p = (1-p-s)[(f-(f-1)]X

on the one side of the equation are the expected winnings for the full table folding. On the other side are the expected losses when someone calls with a superior hand.

Do my formulae look right to you (again, with the simplification of ignoring the SB and BB specifics)?

If so, I'd be most interested to see if you get the same results on JJ. That's a fairly easy one to do since you only have to consider QQ-AA. If you get the same results as I do on that one (I have it figured all the way through to 3 players, just didn't post since it's clearly going to be good enough already with 9 players), I think everything else will be completely sound under the given assumptions. If not, then it would be a good idea to figure out where I might be off before running a bunch of other hands through the spreadsheet.