"NO" does NOT depend on any assumption of what the distribution of results looks like. "NO" is a direct result of the definition of "expected return" and "variance". I'll show you that below.
The "approximating normal" part comes in if you want to draw the conclusion that playing "NO" hands means you have about an 84% chance of being ahead. The 84% value comes from the normal distribution.
Regarding "going to be a big winner or a big loser":
If you are playing a positive game with enough bankroll, then
1. The chance you will be ahead will grow.
2. The percent difference between your "expected win" and your actual result will tend to shrink.
That's not inconsistent with what Michael Peck said. The absolute difference will tend to grow the more you play. That's because standard deviation get bigger and bigger. BUT, Expected Return grows faster than standard deviation. So, the percent difference between expected win and actual win will tend to shrink.
Here's how "NO" is derived. It doesn't take more than the simplest of algebra to follow this. First, some definitions:
1. ER = expected return of 1 hand (and I am going to use just the positive fractional amount. For example, the ER for FPDW would be +0.76%*)
2. ER(n) = expected return after n hands
3. Var = variance of 1 hand
4. Var(n) = variance after n hands
5. SD = standard deviation of 1 hand = SQRT(Var)
6. SD(n) = standard deviation after n hands
7. NO = the number of hands where the expected return equals the standard deviation
One of the nice properties of variance is that the variance of independent events, like video poker hands, can be added. Therefore, Var(n) = n * Var
And, since the standard deviation is defined as the SQRT(Var), then this must be true:
SD(n) = SQRT(Var(n)) = SQRT(n*Var), which can also be written as
SD(n) = SQRT(n) * SQRT(Var). And since SQRT(Var) = SD, we get
SD(n) = SQRT(n) * SD. Call this Eq. (1).
What is the expected return after n hands? That's easy, it's n times the expected return of 1 hand:
ER(n) = n * ER. Call this Eq. (2).
Let's look for the number of hands where the expected return, ER(n), is equal to the standard deviation, SD(n). We can find that particular value of n by setting the right hand sides of Eq. (1) and Eq. (2) equal to each other. That looks like this:
n * ER = SQRT(n) * SD. Now we have to solve for n.
First, square both sides:
n^2 * ER^2 = n * SD^2
Divide both sides by ER^2
n^2 = n * (SD^2)/(ER^2)
Divide both sides by n
n = (SD^2)/(ER^2)
We've found the particular value of "n" for which the expected return equals the standard deviation. Brett Harris coined the term "NO" for that value of "n".
NO = (SD^2)/(ER^2)
Note that we've made no assumption about normal distribution or anything else.
--Dunbar
*If you want to use the 100.76% type values for expected return, instead of +0.76%, then you just have to subtract off "1" everywhere you see ER in those equations.
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--- In vpFREE@yahoogroups.com, "deuceswild1000" <deuceswild1000@...> wrote:
--- In vpFREE@yahoogroups.com, "nightoftheiguana2000" Many people think the possible results converge on the average result, but this is incorrect. The longer you play, the less likely it is that you will have an average result. For an even gamble, the eventual outcome is not both sides even but instead one or the other side will bust out (reach their limit), that's why the longterm bankroll requirement for an even gamble is infinite, and also why even slightly positive gambles have gimongous bankroll requirements.
>
You have posted that before, but I did not understand it then and do not understand it now. Especially in light of the NO situation. If you remember back, I said it seems that your above statement contradicts NO. Your answer was to run a risk of ruin program. But, I am looking for the intuitive answer, not a program. I was under the impression that NO depended on a relativelly smooth distribution (approximating normal). Whereas you seem to imply that I am either going to be a big winner or a big loser.
