Ok, I think I've got all my ducks in a row now, and I'm *very* disappointed nobody was willing to take me up on the bet, because it would have been such easy money. [img]/images/graemlins/frown.gif[/img]
There are two comparisons under consideration here. We want to compare the variance over 10 hands of a 100BB $100 stack vs a 10BB $10 stack (Case A), and then a 100BB $100 stack vs a 10BB $100 stack (Case B).
I'm going to borrow from the math conveniently provided in this thread:
http://forumserver.twoplustwo.com/sh...page=&vc=1
Which proved that the variance of a short stack is lower than that of a big stack.
Only I'm going to take a slightly different tack, without making any assumptions about our player's edge, and see how it comes out.
The assumptions for the 10BB stack are that the player and the opponent are playing optimal push-or-fold poker, which gives the variance formula:
p is our player's "edge", the percentage of hands he will win
A is the average pot size, which for this situation is 8.99 big blinds, but we're going to leave it as a variable for a minute.
V = 10p((A-pA)^2) + 10(1-p)((0-pA)^2)
Which reduces to
V = 10p(1-p)A^2
Now for the 100bb case, we assume that every pot is average, which leads to the variance formula:
V = 10p((A-pA)^2) + 10(1-p)((0-pA)^2)
Which should look familiar. The variance calculation over 10 hands will be exactly this formula anytime you assume that every pot is going to be equal to the "average" pot, which is an assumption that was made in both the 10BB jam-or-fold and the 100BB situations.
The reduction is again
V = 10p(1-p)A^2
The conclusion we can draw from this is that variance depends on both the player's edge and on the average pot size. Which seems fairly obvious, but it's nice to see that the math backs up things that are intuitively obvious.
Time to plug some numbers in.
In the 10BB case, the average pot is going to be 8.99 big blinds. But since we want to make direct comparisons, and because we eventually want to be able to compare our results against HUSNGs (where measuring by big blinds doesn't even make sense), so we're going to calculate variance in dollars.
In case A, the 10BB player has $10, and in case B, the 10BB player has $100.
The maximum variance will occur when p=0.5.
In case A (bb = 1):
V = 10(.5)(.5)(8.99)^2 = 202.05
In case B (bb = 10):
V = 10(.5)(.5)(89.90)^2 = 20205
Obviously, the 10BB stack in case 2 has a much higher variance than the 10BB stack in case 1, because he's playing $100 at a time rather than $10 at a time.
In case 1, since the formula for the 100BB stack is exactly the same, varying by edge and average pot, it should be fairly obvious that given the same edge, the 100BB player will have a higher variance unlesss his average pot is smaller than the 10BB player.
But it's safe to assume that the 100BB player is going to have some sort of edge over optimal jam-or-fold, since with 100BB, we're actually playing poker, not shoving all the time.
If we give the 100BB player a very reasonable edge of 10BB/hundred hands (again, using the formulas from the other post, but using a more reasonable number):
p = (A + .2)/2A;
Subsituting this into our variance formula for p:
V = (10A^2 - 0.4)/4
Solving for A:
A = sqrt( .4V + .04 );
So for case 1, (V = 202.05), the average pot would have to be just over $8.99 for the variance to match.
Which, again, is intuitively obvious. Since we already know that variance only depends on the average pot size and the edge, if a player with $100 is playing pots with an average of $8.99, and has the same edge, his variance is going to be exactly the same as a player with $10 who plays average pots of $8.99. In this case, we've given the 100BB player a very slight edge, so he gets the same variance as the $10 player with a *slightly* larger pot size.
At this point, to have any way of determining whose variance is truly higher, we need to know the average pot size. We know that if the 100BB player generally has an average pot size greater than 8.99 big blinds, then his variance will be higher than the 10BB stack playing optimal push-or fold. But how does 8.99BB compare to *real* average pot sizes?
A quick sampling of some of the currently running games at full-tilt (11 $2/$4 tables and 22 $1/$2 tables) shows that the average pot is 7.87 big blinds at 2/4 and 7.47 big blinds at 1/2, for an overall average of 7.6BBs.
Of course, there's a very wide range here, from 2.22BB to 18BB, but with an average of 7.6, it could reasonably be said that cash-table play is, on average, lower variance than that of a player playing 10BB optimal jam-or-fold.
Which is actually somewhat of a surprising result, because I was totally willing to accept the argument that 100BB stack poker would be higher variance than 10BB stack poker, even when the blinds were the same, but that does not appear to be the case in the real world.
On to case B, where the 10BB player has $100 instead of $10:
In this case, v = 20205, giving us an average pot of $89.90.
So to have the same variance of a player playing $100 at a 5/10 table, our 100BB player with $100 at a 0.5/1 table would have to be playing average pots of $90,
even if (and this is very important) we don't give him *any* edge at all over the 10BB player. If we increase the edge at all, the average pot is going to have to be bigger and bigger to give the same variance as the 10BB player playing jam-or-fold. Without going through the trouble of using "real" edge numbers (such as 10BB/hundred hands) here is a demonstration of how the average pot will have to grow.
p = 0.5, A = 90
p = 0.55 A = 90.45
p = 0.6 A = 91.855
p = 0.65 A = 94.346
p = 0.7 A = 98.198
So, if the player's edge happened to be as high as .7 (which is really, really, unreasonably high), he would basically have to be pushing every single hand before his variance would be as high as somebody playing optimal jam-or-fold with 10BB.
So, in case B, *even if his edge is only 50%, giving him the maximum variance possible*, the 100BB player's variance is clearly going to be lower than the 10BB player's variance, because he'd basically have to be playing optimal jam-or-fold to hit an average pot of $89.90.
And this is why you can't "simplify" the problem when trying to compare the variance between case A and case B by assuming that every 10BB stack is equivalent to every other 10BB stack, when comparing to a $100 100BB stack.
Attack at will, but this is not longer a matter of 'I say' vs "you say' anymore. The math is there, and if it is not correct, then point out where I've calculated something incorrectly. But if it is correct, then the people in this thread and the previous one who have tried to "simplify" the problem in a way that actually changes the result are the ones who have been lacking in understanding.
Conclusion: when throwing some real-world data into the mix, it seems that a player playing with 100 big blinds is *always* going to have a lower variance than a player playing with 10 big blinds, unless the 10BB player is playing lower stakes. (For example, a 100BB player at 5/10 would obviously have a higher variance than a 10BB player at 0.5/1)
A surprising result, but, as far as I can tell, a correct one.
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The reason I've been fighting for so hard on this point, and from so many different angles, is this:
We want to be able to compare the variance between a $100 sit-n-go and a session at a cash table where you buy-in with $100. (something you simply cannot do if your variance calculations are in units of big blinds, by the way, which is why it's silly to calculate them that way for the intended purpose)
At the beginning of the cash table, you start with 100BB. At the beginning of the sit-n-go, you start with 50-100BB, and can advance to a point where your stack, even if it hasn't changed at all, is now worth 10BB, or even 5BB.
If every 10BB stack is equal when trying to compare variance (which is a statement that is so obviously false that I'm *still* flabbergasted that anybody could argue), then at the end of a sit-n-go, when you have 10BB instead of 75, you would be in a lower variance situation than you are in a cash table. And, in fact, since you started out with 75BB (which is a "smaller" stack than 100BB), you would be in a lower variance situation throughout the entire sit-n-go. If this were the case, then sit-n-gos would *obviously* be lower variance than a cash game. To the point that no further discussion is even necessary.
However, if, as I have just shown, there are significant differences between a $10 10BB stack and a $100 10BB stack (which, again, seems so mind-numbingly obvious that I'm amazed this discussion made it past "you can't compare variance values directly when the value of the BB is different"), then while it is true that your overall variance depends only on your winrate, that winrate will depend on the variance of the play of individual hands during the sit-n-go. And since we've already shown that a $100 10BB stack is higher variance than a $100 100BB stack (and that by comparison, a $100 75BB stack is also likely to be higher variance than a $100 100BB stack), you start the sit-n-go in a higher variance situation, and end it what is likely to be a *much* higher variance situation than a cash session that lasts as many hands as your sit-n-go does.
Of course, that doesn't necessarily mean that cash is higher variance than sit-n-gos, because there are other factors that affect the variance during a sit-n-go, for example, the fact that players can rebuy, and the fact that you're never playing for a double-stack or deeper in a sit-n-go, while you certainly can on a cash table.
But it does mean that there's actually a discussion to be had, and other factors to consider. If every 10BB stack compared to a 100BB stack is equivalent, then there's no discussion at all.
I'm going to ask BruceZ (a mod in the probability forum, who helped me get the final duck I needed in the right spot) to review this post and comment on it, but please let me know when it's safe to say "neener neener, I told you so".