Poker Poker Poker!: [vpFREE] Re: Here we go again?

It's a bit simpler than that. For the moment just ignore the covariance-- it just
complicates the main issue.

Indeed if the covariance was zero (that is we are ignoring it), the variance would decrease
like 1/N [and the standard deviation 1/sqrt(N) ] where N is the number of hands played
in total. Total means total and means we can understand the fundamentals of the concept
not worrying about the difference between deals and draw..

So, if V is the variance for 1 hand, we would have the variance for N hands = V/N . Take
V = 49. and N = 100. Variance for 100 hands would be 0.49 = 49/100, and standard
deviation would be 0.7 = 7/10 = sqrt(49/100) . This so called "1 over square root of N"
behavior for the standard deviation is the hallmark of all uncorrelated random events. It's
so famous many people confuse it for the central limit theorem.

Now, we know Video Poker Deals are uncorrelated random events. But what about the
draws?

Well, The draws are correlated-- which means the covariance is not 0. Now, there's this
funny law out there that says if the random event is correlated its variance goes down
slower than 1/N. The more the events (the draws) are correlated-- the slower the
variance goes down. The less correlated they are, the faster the variance decrease-- but
never faster than 1/N. Nice and simple (well we ignored a couple of fine points)

So, due the fact the draws are correlated, we know that the variance of multi play goes
down slower than 1/N -- but it goes down. How much slower (on average)? For that you
need the covariance. But you don't need the covariance to understand that the variance
goes down with increased number of hands.

Play a game: Grab some dice (you can do this with 1 die, but its a pain). Roll I die over
and over again recording the results. Compute the mean and variance of those results.

Now, take a few dice and roll them as a group the same number of time you did for the
single die. Record the average results for each "roll" (the sum of all the dice divided by the
number of dice). Now compute the mean and variance of this set of number. Viola-- you
should have proven the 1/N behavior (or close to it) , where N is the number of dice.

Notice that I said average the results. That's because I wanted the answer to come out
correctly-- in the same units as the 1-die question so I could compare things. If I added
the results from the individual die-- perhaps I would find something different for the
overall variance (but I wouldn't be comparing the same thing). This situation is akin to the
issue raised by others about normalizing the vp results for a constant bet amount -- or
not. I leave working out this case to the interested reader (if there are any)

--- In vpFREE@yahoogroups.com, "cmayhem2001" <chandler_re@...> wrote:

>
> You need to understand the concept of covariance. Because so many
> hands share the same deal and only the draw is separate, it
> increases total variance vs a game where every deal and draw is a
> separate event. A single play FPJB game has a variance of a bit
> over 19, as I recall. The same game in hundred play has a variance
> of something like 214. See the Jazbo or Wizard of Odds links to N
> Play discussions.
>
> Chandler
>