[long and somewhat technical]
> Steve Jacobs wrote:
> >I do an exact calculation.
>>
>> Dan Paymar wrote:
>
> Since the Poisson Distribution formula has been accepted by
> mathematicians for nearly 200 years, I am confident that my method is
> accurate enough provided your bankroll is sufficient to play at least
> one cycle without hitting a royal.
For some reason, neither my post nor Dan's response were among the
other emails from VPFree that I received today. I think Yahoo's group
server is losing stuff.
Dan, the fact that mathematicians have used the Poisson Distribution for
200 years does not imply that it is correct (much less exact) to apply it
to this particular problem. Having a "sufficient bankroll" has nothing to
do with it. Ask yourself this question: If the Poisson Distribution is so
wonderful for problems of this type, why don't you use it to compute
RoR based on bankroll? Answer: because RoR can be computed
exactly by using a formula that has nothing to do with the Poisson
Distribution. One should strive to use the correct tools for the job
at hand.
I'm kind of sorry I brought it up. My only intent was to help you to
produce more accurate results for your program. If my numbers are
correct, then your bankroll of 984 units gives a 60.7% chance of
hitting a royal. Sounds more like a rough guess than something
that I'd call "accurate enough," but perhaps my standards are
unreasonably high.
First, let me jump to the punch line:
neither Dan or Steve is correct in their respective estimates.
Please see my statement above. My claim is that my number
is not an estimate, it is the exact probability of surviving until
hitting a royal.
But it's not really their fault per-se, but rather that the "questions"
fault.
So what actually went wrong? Let's start at the very beginning....
Steve reported two quantities for 9/6 JoB (Max EV strategy):
(a) that starting with a single unit bankroll, a player has a
1/1054.467609892 probability to survive long-enough to hit a royal flush.
(b) He also reported that a bankroll of 731 units is needed in order to
have better than a 50% chance of hitting a royal before going broke.
These two numbers are entirely consistent with each other.
The probability of starting with 1 unit and going broke without hitting
a royal is 0.999051654, so the probability of going broke when starting
with 731 units is 0.999051654^731 = 0.4997887, giving slightly better
than a 50% chance of hitting a royal.
Let's look at each quantities separately:
(a) probability of a player with 1 unit bankroll of a hitting a RF
---------------------------------------------------------------------------
- ----------
The probability of hitting a royal flush in 1 hand is (approx.) : P(RF) =
2.6262e-05 This corresponds to a Royal "cycle" of C = 1/P(RF) = 38077.8
hands. Let's round this to 38079.
I don't know where you got these numbers. The probability for hitting
a royal on one play of 9/6 JoB is 2.4758268e-05, giving a royal cycle
of 40390.5474519. Your numbers here are off by 6%.
Given this Royal cycle, we can compute the likelihood of hitting at least 1
RF in n hands, using the poisson distribution. Keep in mind that hitting
at least 1 RF is means you didn't hit ZERO RF's (don't forget the 1)
The problem with using the posson distribution is the implicit assumption
that you _will_ play n hands. This precludes any possibility that the
player will be forced to halt play due to losing the bankroll. So, using
the poisson distribution here is not correct.
P( at least 1 RF, n hands) = 1 - exp(- n/C) , where exp(a) =
ln(2)^a, where ln(2) is the natural logarithm of 2.
[Let's make sure this formula makes sense: if n = 0, P=0 as it should.
Likewise, of n=infinity, P =1, as it should. Great]
I'll call the probability that a player survives for n hands given a
bankroll of B units , [1- RoR(n,B)]., where RoR is the Risk of Ruin for n
hands an a bankroll B. This can be exactly computed using standard
techniques (see Jazbo's web page) and does not require monte carlo
simulation. In general, these RoR curves look something like a sigmoid,
with an asymptote that goes to 100% RoR (0% survivability) for negative EV
games.
Such a calculation is incompatible with computing the probability of hitting
a royal before going broke. The reason is this: Jazbo's survivability curves
and the value for RoR are both derived, in part, by the _value_ paid for a
royal flush. The higher the royal payoff, greater the survivability and the
lower the RoR for the game. However, the probability of surviving until
hitting a royal is _completely_ independent of the value paid for a royal,
and depends _only_ on the payoffs for non-royal hands. This is also true
when computing the average cost of a royal. Clearly, if the royal payoff
is used in the equations, then the result cannot be exact. My result is
independent of the royal payoff.
A better approximation would use "survivability while excluding the royal"
instead of Jazbo's survivability curve.
To find the probability of a player with a 1 unit bankroll hitting at
least 1 RF we just multiply the two factors to yield:
[1-RoR(n,B)]*[1-exp(-n/C)]
Sorry, but that is at best an approximation. Even if you use a more
appropriate survivability curve, I believe it will still be an approximation.
Now' let's look at the behavior of this result as a function of the number
of hands played, or n.
Interestingly, as n increases the first factor (the survivability)
decreases but the second factor (from the poisson distribution) increases !
But that's not all. When you multiply the two factors, you find some
seemingly bizarre behavior!
For a small number of hands, as you play more, the overall likelihood of
hitting that RF (and not going broke) increases. Then, it starts to level
out... and then finally decreases! Yup, there is maximum value to the
product function. Why? Well, the second factor from the Poisson
distribution as a maximum value of 1 (which occurs for n = inf). The first
factor, the surrivability, has an (inverted) s-like shape and continues to
(asymptotically) decrease to zero at a faster rate. So for a very large
number of hands, the probability must decrease as it must for all negative
EV games (RoR goes to 100%).
It is true that RoR goes to 100% for all negative EV games. However,
probability of going broke before hitting a royal _never_ reaches 100%.
This should be obvious, since there is a non-zero probability of hitting
a royal on the very first trial. Once again, you are trying to use RoR
numbers to compute a quantity that is mathematically incompatible with
RoR.
So.. let's put some numbers on this. A player with a 1 unit bankroll has
about a 16/1000 chance of reaching 10,000 hands. At 10,000 hands, the
likelihood of hitting at least 1 RF is about 0.23, so the overall
probability is then about 0.000375 or about 1/2665 (no where near 1054).
The probability of going 10,000 consecutive plays without hitting a royal is
0.7807, so the probability of hitting ONE OR MORE royals during 10,000
consecutive plays is 0.2193. This is close enough to your 0.23 number.
But, this calculation _never_ gives any consideration to the possibility of
the player going broke. It _assumes_ the player has a sufficient bankroll
to never go broke during this stretch of 10,000 plays.
That same player also has a chance of 0.00121495 of reaching 20,000 hands
(a priori, not after reaching 10K hands!) and a about 0.4086 chance of
hitting at least 1 RF, and and therefore an overall probability of
0.000496 or 1/2014 or so. Yup, this is a lower number than for 10K hands,
as expected.
Well... what is the answer then? I'm not sure. My feeling is the question
is poorly framed, and I don't yet know the best way to answer it with a
"single" number. One way to fix the problem is to rewrite it as "what is
the likelihood of a player with a 1 unit bankroll hitting at least 1 RF
**given that the player survives an average number of hands**?" By
modifying the question in this way, the number of hands becomes computable,
and the solution is easily determined using the formula I gave above.
You're trying to morph the question into something very different. The
concept is quite simple. Start the player with B units and let them play.
If the player goes broke without ever hitting a royal, count the trial as
a failure. If the player hits a royal, end the trial and count it as a
success. Repeat zillions of times to find the fraction of trials that end
with a royal rather than ending broke. That is a simple, clear concept.
It needs no other framing. My method gives an exact solution without
(incorrectly) invoking the Poisson distribution.
For 9/6 JoB, the probability of the player going broke is precisely
p(failure) = 0.9990516541^B for a player starting with B units. The
probability of hitting a royal before going broke is 1 - p(failure).
One could also argue that the problem can't be fixed, since a player never
has a 100% likelihood of hitting at least 1 RF, unless one plays an
infinite number of hands. But, for Job, a negative EV, the survivability
of an infinite number of hands is 0. Oh well.
Negative EV has nothing to do with this specific problem. Any VP game,
whether it is positive EV or negative EV, will have a positive probability
of surviving until you hit a royal, with any starting bankroll of one or more
units. If you dispute this, then you don't understand the original problem.
Perhaps there are other ways to "fix" this problem.... any comments from
the group?
(b) number of units needed in order to have better than a 50% chance of
hitting a royal before going broke
---------------------------------------------------------------------------
- ----------
We can use the same approach I came up with above, with a modification to
handle the 50% number.... or can we (actually no, but I am jumping ahead)
You seem to be admitting here that your approach isn't correct. Perhaps
you should carefully think about just what, exactly, your approach is
computing.
So, how many hands are necessary to have a 50% likelihood of hitting at
least 1 RF? That 's easy... simply solve for n in 1-exp(-n/C) = 0.5
The answer is.... n = - c * ln(0.5) . FOr 9/6 JoB, n= 26394 hands (or
so).
27996 when the correct cycle number is used. But this represents the
number of consecutive hands that can be played to give a 50% chance
of hitting one or more royals.
But do we do with this number? Nothing, it is not relevant (why? it
does not include the survivability factor)!
That is why your formulation is incorrect.
Ok, let's start again, this time, using the correct formula: we need to
find n and B so that
[1-RoR(n,B)]*[1-exp(-n/C)] >=.5
This isn't the correct formula. What _might_ work here is to sum over
all positive values of n, like this:
0.5 <= sum(n:1 <= n <= infinity) {[1 - RoR(n,B)]*[1 - exp(-n/C)]}
Trying to evaluate this for a single value of n is clearly wrong, in
my opinion.
Well, it should be clear to at least some of you that there is more than
one answer. Indeed, there are an infinite number of answers. But, we can
modify the question a just a little to read: what is the *smallest* number
of units needed in order to have better than a 50% chance of hitting a
royal before going broke?
731 units.
I wish Jazbo were here. I can talk to him without being blown off with
comments to the effect that "200 year old math can't be wrong..."
···
On Friday 04 November 2005 06:23 pm, cdfsrule wrote:
