Understanding the
nature of the Poisson Distribution, it's clear that it's not
applicable to bankrolls less than what is required to play at least
one cycle without hitting a royal, so the program substitutes the
one-cycle bankroll.
This statement convinces me that you do not understand the
Poisson distribution. When using the Poisson distribution to
estimate the probability of hitting exactly zero royals during
N hands of play, the Poisson distribution gives very good
estimates for any value of N.
Allow me to demonstrate. For 9/6 JoB, the royal cycle is
40390.54745192714, which I will call C. The precise
probability of not hitting a royal is (C - 1) / C, which has
a value of 0.999975241732. The Poisson distribution
gives:
e^(-1/C) = 0.9999752420384
So, for a single trial, these numbers agree to about 9
decimal places.
For N hands of play, the exact probability of not hitting
a royal is:
PNR(N) = [(C - 1) / C]^N
The estimate given by Poisson distribution is:
e^(-N/C) = [e^(-1/C)]^N
These two equations obviously have the same form, and
since we know that e^(-1/C) ~= (C-1)/C to very high
precision, then we can safely say that the Poisson
distribution estimates this exact value very well for ANY
value of N.
The discussion above applies for a specified length
of play, N hands, which are played out consecutively,
so N doesn't represent a bankroll.
Now, what you've done to cobble this into a form that
uses a bankroll, is to say that going for a long time
without hitting a royal is like playing a VP game where
the EV is reduced by an amount equal to the EV
contribution from royals. Fair enough, that means the
effective EV is:
EFF_EV = 0.9954390437 - 0.0198066144 = 0.9756324293
So the effective loss rate is:
Loss = 1 - EFF_EV = 0.0243675707
So the player should expect to lose at a rate of about 2.437%
while playing for a long time with no royal. Losing at this
rate means that on average, each unit of the players bankroll
will allow them to play for the following numbe of trials:
Plays_per_unit = 1 / Loss = 1 / 0.0243675707 = 41.03814911677.
So, if you take an initial bankroll and multiply it by this factor,
then you get the number of trials that the bankroll will allow,
on average, before the bankroll is consumed from losses.
This same factor can be used to express a Royal cycle in
terms of the number of bankroll units need so that the player
can play, on average, long enough to hit a royal. This bank
is:
K = C / 41.038... = 984.2195 units. So, the Poisson distribution
you use to estimate probability of no royal is:
PNR_Dan = e^(-B/K), where B is the player bankroll
This is equivalent to:
RoRBR_Dan(B) = [e^(-1/K)]^B = F^B (with F = 0.9989844824817)
The value of F corresponds to the following probability that a single
unit starting bankroll will survive until hitting a royal:
P(royal) = 1 - F = 0.0010155175183 = 1 / 984.7195329163
Again we see a number close to 984. This number represents a
different kind of "royal cycle". This number says that if you start
a very large number of players with a "sufficient" bankroll, but only
allow them to play 41 hands each (this is the average number of
plays needed for "no royal" outcomes to consume one unit of
bankroll), then on average one player out of 985 will hit one
(or more) royals and the others will not hit any royals.
The concept described by the above paragraph is what your
method computes. It is not a risk of ruin, but something slightly
different.
This value of F corresponds to a 50/50 bankroll of:
B = ln(0.5) / ln(F) = 682.20898
My exact formulation gives:
RoRBR(B) = F^B (with F = 0.99905165413274)
B = ln(0.5) / ln(F) = 730.5546
P(royal) = 1 - F = 1 / 1054.467610
This P(royal) is the precise probability of surviving to hit a royal,
if the player starts with a one unit bankroll. This number says that
if you start a very large number of players with a single bankroll
and let them keep playing until they either go broke or hit a royal,
then on average one player out of 1055 will end up hitting a royal.
Bottom line: both of our methods produce an exponential decay,
but the decay rates are significantly different. So, for a given
bankroll B, if we take your result and divide it by my result, the ration will
shrink ever smaller as B increases toward infinity. What this
means is that your Poisson approximation becomes increasingly
worse for larger bankrolls. It does NOT get increasingly better,
as you claim. In fact, with a bankroll of 9892 units, your estimate
of RoRBR will be off by a factor of two.
Put into simpler terms, if the player wants to play with a 5% risk
of going broke before hitting a royal, then they need to start with
a bankroll of 3158 units. Your formula gives 2949 units, which
underestimates the bankroll by 6.6%. A player using your number
would actually have RoRBR of 6.1% rather than the 5%. This is
a 22% difference in actual RoRBR
If the player wants a 1% RoRBR, they need a bankroll of 4854 units.
Your method gives a bankroll of 4533 units, which gives an actual
RoRBR of 1.36%.
The 984 unit bankroll is a one-cycle bankroll. I
forgot about this when I posted the 984 figure, and I apologize for
that. A 984 unit bankroll yields 36.79% RoRBR. I am modifying the
program to so indicate.
OK, so when you originally said that 984 was a 50/50 bankroll,
that was a mistatement. Thanks for clarifying.
The Poisson Distribution clearly can not be used for a 50/50 bankroll
except on very low payback games where the 50/50 bankroll is greater
than the one cycle bankroll.
I believe you are clearly wrong about that. For this particular problem,
there is no need to place any restriction on bankroll size in order to
get a consistent estimate. However, the number you compute with
your method is not really RoRBR(B). It is "probability of failing to hit
a royal if one plays long enough that losses would, on average,
consume B units of bankroll." To get this average, all players must
play the same number of hands, even if chance causes them to lose
much more than B units.
My simulation suggests a 740 units for a 50/50 bankroll, which is
reasonably close to your number. The interesting point is that the
bigger the bankroll, the closer together your RoRBR values are to my
simulation, suggesting that the Poisson method gives a very good
approximation for large bankrolls. (I have never claimed that it is
"exact," but two significant figures for RoR is adequate for
practical use by most players.)
You keep repeating this same incorrect statement, despite the fact
that I've pointed out that it is wrong. The bigger the bankroll, the
further your number diverges from mine.
In addition, it is deceptive to claim that your simulations imply
anything about your Poisson method. You told me that your
simulations were designed to measure true RoRBR, and that
you give each simulated player B units and allow them to
play until they either go broke or hit a royal. This is the correct
model for RoRBR, but it is NOT a model for what your Poisson
method computes.
I want to replace the use of the Poisson Distribution in OpVP with an
algorithmic method, but before doing that I want to resolve the
question of why my simulation doesn't agree with your calculations. I
have offered to exchange code with you. If you would do that, perhaps
we could resolve the discrepancy.
I've layed out my method completely in posts here. It was described in
sufficient detail that nightoftheiguana2000 was able to reproduce my
results with 14 digit accuracy (once I pointed him in the right direction).
Clearly he understands exactly how my calculations work, and he didn't
need to look at my code to do it. The point is the MATH and not whatever
specific algorithm was used to solve the equation.
In short: "Use the Math, Luke"
My program is simply an iterative solution to a recursive RoR-like
equation. Your program is a simulator. They are like apples and
oragnes. If you want me to look at your simulator to see if I can find
anything that looks suspicious, I'd be happy to do so.
If you want to know whether I've done anything suspicious, all you
have to do is look at the math that I've spelled out in great detail and
try to reproduce my result. You've posted articles on how to solve
RoR-like equations, so you shouldn't have any trouble.
···
On Saturday 12 November 2005 04:11 pm, Dan Paymar wrote: