[TNixon]

Ok, I came up with one more way of stating the problem that isn't saying something exactly the same way that I already have.

In fact, this is how I'd want to frame the question for anybody coming over from the probability forum to settle a bet.

The following question is asked:

"I have a $1000 bankroll, that I don't mind replacing if I go broke, so I want to play it very agressively, but I'd still prefer *not* to go broke if I can avoid it, while keeping in mind that I want to maximize value by playing agressively.

I've narrowed my choices down to two:

Choice A: Playing full buyins at 100NL.
Choice B: Buying in short at 1000NL, for $100 or 10 big blinds.

Which of these 2 choices is going to be lower variance? (and please don't try to talk me out of playing above my bankroll. I've already stated that I want to play agressively. Just tell me which of these two choices is lower variance)"

What is your answer, and why?

My answer is Choice A is obviously lower variance, with a smaller chance of going broke, as long as you don't play a supper-aggresive style where you're playing "optimal" push-or-fold.

Your answer, if you're going to be consistent with everything you've already said, should be that choice B is lower variance. You even "proved" mathematically that choice B would be lower variance, despite the fact that you would be getting the chips all-in more often as the 10BB stack, because the average pot size in choice A would have to be less than around 9BB for choice A to be the lower-variance choice.

Is that your final answer?

If so, do we have a bet? I'm even willing to give you 2 to 1 odds. Your $50 to my $100, or your $100 to my $200.