[omgwtfnoway]

against my better judgment i'm going to continue this discussion, hopefully helping you to see that short stack poker is lower variance than playing with deeper stacks.

first we're going to try to clear up the definition of variance and standard deviation. when i said that you *have* to think of variance/std deviation in terms of the stakes you're playing, i wasn't just saying it in order to provide perspective for my argument, these are mathematical terms that are expressed in terms similar to the winrate. in fact, in the case of standard deviation, the units are THE SAME as the winrate. so for a cash game player measuring his winrate in BB/100hands, he will also measure his std dev in BB/100hands. Variance is the square of std dev so it will be measured in BB^2/10khands.
var(x)=sum((X-u)^2) where X is the individual result and u represents the mean or expected result.

as i understand it, the definition of variance that you're tacitly using is "how often you get stacked or stack your opponent." examining the difference between the two definitions should go a long way toward clearing up our argument. for what it's worth, i agree that you will put your whole stack in the middle more often the smaller your stack is, but that has less to do with variance than you think it does.

so basically variance is an approximation of how much you can reasonably expect your results to differ from your sklansky bucks.

now we'll take two hypothetical cash game hands. in both hands, hero gets all the money in as a 60-40 favorite. in the first hand the effective stacks are 10bb. in the second hand, 100bb. now we're going to run the board out 5 times and we're going to assume that the most probable outcome (that hero wins 60% of the hands) occurs. this will show the variance that is inherent in making these plays.

10bb case.
hero's equity in this pot is (.6)(20bb)=12bb for a net of 2bb. so u=12bb/hand.
3/5 times hero actually wins 20bb. the other 2/5 he wins 0bb.
var=3*((20-12)^2)+2*((0-12)^2)=192+288=480
std dev=21.9bb/hand

100bb case
hero's equity in this pot is (.6)(200)=120 for a net profit of 20bb. u=120bb/hand.
3/5 times hero wins 200bb. the other 2/5 he wins 0bb.
var=3*((200-120)^2)+2*((0-120)^2)=48000
std dev=219bb/hand

so the variance of the 100bb case is 100 times as large as in the 10bb case.

i know what your arguments are going to be:
1)we're less likely to enjoy a 60% edge on our opponent playing for 10bb than we are for 100bb
2)we're not going to be allin for our stacks as often for 100bb than we are for 10bb

these arguments are valid, but irrelevant to the discussion of variance and i'll show you why.

let's find out how small an edge we'd have to have in the 10bb case for our variance to approach 48000 as it is in the 100bb case.
assume our edge is p
var=5p*((200-120)^2)+5(1-p)((0-120)^2)=48000
320p+720-720p=48000
the solution to this is p<0; you'd have to be drawing less than dead to have the same variance as the 100bb case. obviously this can never happen so it's starting to look like there's inherently less variance the shorter your stack is.

now for the argument that not being allin as frequently in the 100bb case will lower our variance. note that i don't disagree that you will be allin less frequently playing for 100bb than you would for 10bb.
for this point look again at the equation for variance:
var(x)=sum((X-u)^2)
the fundamental difference in variance between the two cases can be surmised by looking at the x-u term. this term has limits placed on it by the game; it is limited by the effective stack of the two players.
let's take the variance equation to a generalized pot. let's assume that in both the 10bb and 100bb cases we're examining the variance over a sample of ten hands.
calculate variance assuming every pot we play is "average size":
var=sum((X-u)^2)
express u in terms of the avg pot size and equity in the pot: u=p*avgpot
var=sum((X-(p*avgpot/2))^2)
now we're going to need to compare the two cases individually. variance will be maximized in each case if both players are allin for their stacks with a random hand each time. however, i'm going to assume that both players are playing optimal jam-fold poker according to the solutions in mathematics of poker. as such over a large amount of hands each player's equity is .5. since we don't have a large number of hands i'm going to use 10hands (an even number, ldo) as a way to negate positional advantage).

now we have to find the average pot size.
at 10bb the sb will shove with 58.3% of hands. once he has shoved, the bb will call with 37.3%. (this information is on page 137 of MOP)
1-.583 of the time sb will fold and bb the pot size will be 1.5bb
(.583)(1-.373) sb will jam and bb will fold resulting in a pot size of 11bb
(.583)(.373) sb will jam and bb will call resulting in a pot size of 20bb.
avg pot=(1-.583)(1.5)+(.583)(1-.373)(11)+(.583)(.373)(20)
avg pot=.6255+4.0210+4.3492=8.99bb ~9bb
since both players have 50% equity in these allin pots, we expect the variance of this game to be generally high for reasons i don't care to go into. (it has to do with maximizing the (X-u)^2 term.)
var=sum((X-(p*avgpot))^2)
when p is .5 and avgpot is 9 the term u=4.5
for simplicity's sake we'll assume that all pots played are "average." half the time hero will win 9 chips and the other half he will win 0.
var=10(.5)((9-4.5)^2)+10(.5)((9-4.5)^2)=202.5

so this is the variance you will experience playing for 10bb if both players are playing optimal jam-fold over ten hands. if your equity in the pot changes in either direction the variance will decrease. try it for yourself if you don't believe me.

now let's look at the 100bb case. we're specifically looking for an equity in the pot (p) and an average pot size (avgpot) that will give us a variance lower than 202.5 (if we can't find values that satisfy this then we will have shown absolutely that you cannot experience lower variance playing for 100bb than for 10bb).
we're trying to find a minimum variance for this deepstack case. variance will be maximized at p=.5 so let's give our hero some sort of healthy advantage in the average pot in any effort to decrease variance. let's assume our hero enjoys the monstrous (and unreasonably high) edge of 50bb/100hands. he wins half a big blind every hand.
this edge is the difference between u and our stake of the avgpot. .5=u-avgpot/2
1+avgpot=2u
remember u=p*avgpot
1+avgpot=2p*avgpot
p=(1+avgpot)/2avgpot
so now, having assumed an absurdly high winrate for our hero, we can find our variance as a function of the average pot size for our sample of five hands.
var=10p((X-(p*avgpot)^2)+10(1-p)((X-(p*avgpot)^2)
where X=avgpot in the first term and 0 in the second term.
this is going to get really messy so i'm going to skip the algebra here and show the simplified variance equation, you can verify that i've done the algebra correctly on your own.
var=(10/4)(avgpot^2-1)
to achieve variance less than 202.5 the avg pot will have to be:
202.5=(10/4)(avgpot^2-1)
avgpot=9.13bb
so if the average pot is just 9.13bb (a fraction of a bb higher than the 10bb case) the variance for our hero will be atleast as large as the maximum variance in the case where we're basically flipping for 10bb at the end of a husng. something interesting to note here is that we've confirmed that changing hero's winrate (from breakeven to .5bb/hand - a huge increase) had very little impact on variance as a function of the average pot.

so what we've found from all this is that as the average pot size (in bb) goes up, so does the variance REGARDLESS OF EITHER PLAYER'S EDGE. at the logical extremes of this you'll see that when stacks approach zero the limit of variance will become very small and as the stacks become deeper variance will increase because as the average pot becomes larger.

in order to confirm that the average pot will become bigger as stacks increase take the case where stacks are very small (as in less than 1bb; both players will be allin for an ante). as stacks go to zero the average pot will also go to zero.
when stacks become larger (say 10-100bb) each player will have enough money to post a full blind. the average pot must be still larger than this so we can safely say that the average pot size increases with increasing effective stack.
this last example is kind of silly but i didn't want to leave anything out.

on to your example:
adjust each stack to be in bb
player A plays w/ 100bb
player B plays w/ 10bb
the first part of my post above shows that player B will experience less variance and a lower standard deviation in bb/100hands than player A.

logical or mathematical enough for you?

not a troll,
wtfnoway