My head hurts!
I'm trying to think thru this, now that the day is done and I can concentrate on this.
ICount is right in saying that, if cumulative EV is given by A*N, where A is the per hand EV and N is the number of hands, and if cumulative SD is given by sqrt(N)*B, where B is the per hand SD, then in the limit, as N approaches infinity, the ratio of cumulative SD over cumulative EV goes to zero. At large values of N, A and B are insignificant and can be dropped out of the equation. So we are essentially left with the limit as N->infinity of sqrt(N)/N = 0.
johndoe and
Sonny must also be right when they say that "everyone" always gets to the predicted EV in the limit. After all, if in the limit, cumulative SD/EV is zero, SD can be ignored and we are only left with EV.
I think, but I am not sure now
, that I am correct in saying that at any large N, the expected result will still be described by an EV and a SD, and that therefore half of all players will have realized more than EV and half less (for N < infinity).
And then there is the question of what happens as the number of players (X) approaches infinity. Since the normal distribution curve asymptotically approaches 0, there must still be some area under the curve for very small values of the Z score. Would this not mean that, as X approaches infinity, and as Z also approaches negative infinity, that the product of X times the area under the normal distribution curve from Z = very small to negative infinity is not zero. In other words, that despite EV being very large, and SD being very small in proportion, that there is still some probability that some poor player will have a negative result, if there are an infinite number of players.
I need a drink.
But first...I want to apologize. I was pretty flip earlier today. And you guys were right, I didn't get it (or at least parts of it anyway
). And I was particularly rude to
johndoe, so a double down on apologies to you. Now, suitably humbled, I'm going shopping.
:grin: