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Wheel Poker

cdfsrule wrote:

Harry,

I was responding to something your wrote in message 87890.
Here it is again:

>When playing a negative expectation game such as the examples above
>(before factoring other play incentives), if the return is
> reasonably close the higher variance game will yield the greater
> percentage of winning sessions.

I was taking some issue with your statement (above), which I
believe is false in general for VP games. Sure, one can certainly
find plenty of hypothetical situations, like the zero-
variance game you sited, for which your statement is true, but
those situations are just not, well, relevant to any casino VP
games I know.

It is pretty easy to show how your statement isn't always true for
a real VP game. But first
let me make sure I have what you were claiming correct...

Let me see if I can place my earlier statements into context:

Williams essentially said that someone who gives up ER for the sake
of higher variance is off their rocker. I asserted that was not
necessarily so, depending upon what they valued most in play. I then
went on to note that in switching to a higher variance game with near
similar (but smaller) ER, you could expect a stronger probability of
a winning session (in terms of an hour or two of play).

Now, my only intent was to suggest that rational behavior could be at
play. Period. As an example, I've cited the choice of 9/6 DDB over
9/6 JB play. I believe my suggested relationship holds true in that
case.

However, I overstepped the bounds in choosing to make a generalized
statement. That was an inadvertant slip. I agree that higher
variance doesn't make for a higher expectation of a session profit in
all cases. And it's unfortunate in having done so that I surfaced a
distraction from the key point I wished to make.

- Harry

Likewise, I would strongly encourage those who do bother to compute
all

the EVs and VAR's for the every growing variation of games out there

then go on to

compute the CDF's for various sessions lengths (and RoR's). Its a

straight forward process

that requires just a table of the final hand probabilities and

payouts.

I would love to compute CDFs for the games I play. How do I do that?

···

--- In vpFREE@yahoogroups.com, "cdfsrule" <vpfree_digests@...> wrote:

I'll outline the conceptual steps involved in the computation then discuss some
implementation strategies. To keep things simple, I'll start ignoring ruin.

I recall writing about this before-- but whatever...

Conceptually you start with the PDF for the "game" or the PDF for 1-hand of the game.
You then compute the PDF for N-hands of the game and then add up (integrate it) to get
the CDF. To be fair, this PDF is really just a table consisting of all the possible outcomes
from the game and their probabilities of occurring.

So-- where do you get the PDF for the game?
It's precisely the payout table with a 0 added in for Nada (don't forget the zero) and the
final hand probabilities. Where to get the payout table is obvious. The final hand
probabilities (for each payout, including zero) come from your favorite VP simulator or you
find them on the web, etc.

This PDF looks like to columns of numbers,
0, 0.60
2, 0.13
and so on
The first column here is the payout, the second is the probability. I prefer to use fractional
units rather than percentage though here. The second column, the probability, must sum
to exactly 1. if it doesn't your cdf with be wrong.

The procedure of turning a PDF into a CDF is relatively straight forward. First, you sort the
table in increasing order of the payout (smallest first). If it is sorted, skip this step.
Second, you replace rows in the same that have the same payout (this can happen) with a
single row that has a probability equal to the sum of the rows you are removing. Third,
you create a new column in the table that is the running sum of all the probabilities. This
third column has units of probability. Its first values should be the value for nada and its
last value should be 1 (or 100) or something is wrong already. The first and the third
(new) column comprise the CDF-- computed here as the integral (sum) of the PDF. If you
plot the CDF, you probably want to have a horizontal payout axis (x) and a vertical
probability axis (y)

So that's how you get the CDF for 1-hand of any VP game. To get the CDF for 2 hands,
you need the PDF for 2 hands and then apply the same procedure on it as described
above. For 3 hands you need the pdf for 3 hands, and so on.

But where to you get the PDF for 2 hands? This is where the fun begins. Conceptually,
you need to figure out all the possible outcomes (and their probabilities) after two hands
of play (ignore ruin for now). Let's see, you could have lost the first hand, and lost the
second; lost the first, and got a high pair, lost the first, got 2 pair,... etc, all the way to
getting 2 RF's... (yes it could happen, so you need to include it). You all need to get the
probabilities of all this things occurring. To do that you take the probability of the first
event and multiply it by the probability of the second event occurring. You already have a
table of these probabilities-- its the PDF for the game (don't forget nada!)-- all you have
to do is create the new, much bigger table with everything multiplied by everything else.
Then clean it up (sort it and combine duplicate rows)-- and you have a nice 2-hand PDF--
one step away from the CDF (which just needs the running sum column)

I suspect you can see that finding the CDF (or PDF) for n- hands, where n is large, would
be a pain following this procedure. So don't try to do it this for more than for hands
maybe 10 if you are using a spreadsheet. Instead you are going use standard techniques
to compute these kinds of things much more quickly.

BTW, what you did is compute the convolution of the 1-hand PDF with itself. It turns out,
the n-hand PDF can always be computed by convolving the 1-hand PDFs with itself n-
times... but its a pain, as you know, to compute convolutions by hand.

It turns out, that, if we can ignore ruin, there is mathematical trick that allows us to quickly
compute the n-hand PDF. Explain why this works is beyond the scope of this post, but
trust me, it works fine:

PDF (n-hands) = Inverse Fourier Transform{ [Fourier Transform ( PDF 1-hand)] ^ n}

(ok, don't trust me; here's the first proof that popped up for me on google http://www-
structmed.cimr.cam.ac.uk/Course/Convolution/convolution.html#conv_theorem)

[now it gets really ugly...]

To be complete, I'll discuss one possible implementation, using excel, though, I don't
really expect this comments to be complete enough for you to easily follow them. Excel,
BTW, has a fourier transform function. That's the good news. Other than that, its a pain
to use, and It's not always load be default-- so you may have to search around you
installation to find it. Excel is not the right tool here, though it can be coaxed to work and
you can use it to compute the n-hand PDFs, etc. But you will have to be very careful. You
are going to have to do the FFT's (fast fourier transforms) on a very BIG table of all the
possible outcomes (& probabilities) for the n-hands. This table should include a long list
of possible total payouts, starting with the biggest loss possible (make it VERY big, as in
you actually lost each hand) running up to at least a couple of royal flushes. The list of
payouts a must be equally spaced with no missing values (as in 1,2,3,4 NOT 1,4,5), so
most of the probability entries in it the table will be zero!. [Note: the length of your table
should either be a power of two, like (2, 4, 8,...1024,2048,4096, etc) or a product of
primes -- or maybe it doesn't matter for excel. I don't have it in front of me, you should
check on this to get the best performance]. You then take the FFT on this big table
(acutally just on the probability column), take the n'th power of the result (you have to do
this on a complex number!) and then transform back. In my old version of excell, I had to
start with complex numbers so that means one more column in your table (full of zeros).
Maybe some excel guru out there can help you with this.... I'm also sure I've made a lot of
typos and errors here sorry...

···

--- In vpFREE@yahoogroups.com, "deuceswild1000" <deuceswild1000@...> wrote:

I would love to compute CDFs for the games I play. How do I do that?

Harry,

Let me try to help you out here. It seems to me that game choice (at least in my
experience) seems better correlated with a higher probability of "significant" wins rather
than a higher probability of having an overall winning sessions (winning at all). For VP, a
higher variance (generally) means that there is a higher probability of larger wins over a
game with the same (or similar) EV and lower variance.

This explains or is correlated well with the seemingly higher popularity of high variance
games. Games that offer a the same or higher probability of a winning session like
Pick'em just don't seem to be as popular as say DDB. This suggests that If high variance
games are popular that are not so because of the higher probability of a winning session-
- and I just don't see how probabiliy of a winning session helps explain the EV vs variance
trade-off

Now, Williams question had to do with taking a Lower EV for a Higher Variance. I kinda
agree with Williams here. Generally (in gambling, in the stock market, whatever), one
should expect the trade to go the other way, you get a Higher EV in exchange for a
higher Variance. People are of course free to do what they want, but that (Variance/EV for
comparable things) is generally how the universe of "investment" opportunities is judged.
In other words, the risk/reward ratio seems like a reasonable metric to minimize and to
use to compare opportunities. Why anyone would want to maximize that ratio seems to
be at the heart Williams' comments....
perhaps because they are interested only in a big hit rather than a higher probability of a
wining session? In other words, they value a big win disproportionately more than a small
win or loss. For these people, you have to include that "valuation" in the reward part of
the metric rather than the EV. So it seems that they are playing for something else other
than EV. I can't speak to whether that's rational or not for them, since, I'm not them. I
could, however, compute what that valuation must be in order for their choice to be
rational.

···

--- In vpFREE@yahoogroups.com, "Harry Porter" <harry.porter@...> wrote:

Let me see if I can place my earlier statements into context:

Williams essentially said that someone who gives up ER for the sake
of higher variance is off their rocker. I asserted that was not
necessarily so, depending upon what they valued most in play. I then
went on to note that in switching to a higher variance game with near
similar (but smaller) ER, you could expect a stronger probability of
a winning session (in terms of an hour or two of play).

Now, my only intent was to suggest that rational behavior could be at
play. Period. As an example, I've cited the choice of 9/6 DDB over
9/6 JB play. I believe my suggested relationship holds true in that
case.

- Harry

"William Canevari" <wcanevari@...> wrote:

Why would anyone with an IQ above 85 want to play a VP game with a
greater variance but an identical or lower payout, when the original
game is also available on the same machine for the same

denomination?

Each of us has an internal "risk thermostat." We get uncomfortable
whenever the risks in our lives are too high OR too low. Consciously
or un-, we manage our risks in order to keep them close to our
individual set-points. The beauty of casino gambling is that it
allows us to be very precise about taking on the level of risk we
seek.

No one with any IQ seeks to eliminate risk or variance. (If you don't
like variance, you'd prefer working for a salary to gambling.)
Rather, we all seek to meet our personal Daily Risk Requirement. So
it's perfectly reasonable for some people at some times to choose a
greater variance over a lesser variance game.

Stuart (RandomStu)
http://stuart-randomthoughts.blogspot.com/