CLT Steve, brumar, iggy, et al

[jaydavidson118]

Most any formula that is based on variance is

likely to be an approximation, unless it was variance itself
that you want to compute.

Any opinion on whether variance calculations are based on the CLT?
That question was raised earlier. My impression is it isn't based on
the CLT, although variance is part of the CLT formula itself.

CLT is NOT a formula it refers to CENTRAL LIMIT THEOREM, this is one
of the most used and important theorems in statistical theory, which
allows you to use Normal approxiamations to distributions that are
not Normal when some basic criteria are met.

> --- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
>
> <nightoftheiguana2000@y...> wrote:
> > the formula cited below, credited to Evgeny Sorokin but found by
> > others including Jazbo Burns, does not use variance
> > it is my understanding that it is an exact solution but i have

not

> > attempted to follow the derivation
> > it is possible the derivation is online somewhere
>
> There is no EXACT solution, otherwise it wouldn't be gambling.

Here, "exact solution" means the value computed for risk-of-ruin
is the exact probability of going broke rather than playing

indefinitely.

This method does in fact produce an exact solution. Many other
formulae for risk-of-ruin employ approximations (usually based on
variance).

> Any ror calc. has to use variance parameters in its calculation.

No, that is not true. Using variance to compute RoR gives an
approximation.

> > for full pay deuces wild the R(1) number is 0.999346831403995

and

> > represents the risk of losing a bankroll of one bet,
>

The Normal distribution is completely defined by the first two
moments: mean(ev) and Var or stan dev. now if you don't use Var and
just use ev then you would get the following: any ev greater than
100% ror equals zero, any ev less than 100% ror equals 100%, you need
the var of the distribution for the ror calc to make any sense. What
I am saying is that the Var approximation becomes highly suspect at
the extremes making the calc. value of limited use.

The "0.999346831403995" is a early warning sign that someone is not
following the convention of significant digits, and hence loses
credibility. When you multiply two numbers together you have to
truncate the result to the number of significant digits, otherwise
you give the impression that you don't understand the output of your
calculator or computer.

I would put more credence in a "monte carlo" simulation of ror than
any so called "exact solution" based on faulty approximations.

All this is to say that, these formulaic ror calc. are more rule of
thumb than precise results, and should be understood accordingly.

···

From: "brumar_lv" <brumar_lv@y...>
Date: Fri Jan 7, 2005 1:03 am
Subject: Re: Does High Variance really matter? Utility theory and CLT

--- In vpFREE@yahoogroups.com, Steve Jacobs <jacobs@x> wrote:
--- In vpFREE@yahoogroups.com, Steve Jacobs <jacobs@x> wrote:

On Thursday 06 January 2005 10:16 pm, jaydavidson118 wrote:

[Steve_Jacobs]

The Normal distribution is completely defined by the first two
moments: mean(ev) and Var or stan dev. now if you don't use Var and
just use ev then you would get the following: any ev greater than
100% ror equals zero, any ev less than 100% ror equals 100%, you need
the var of the distribution for the ror calc to make any sense. What
I am saying is that the Var approximation becomes highly suspect at
the extremes making the calc. value of limited use.

I never said "just use EV" or anything remotely like that. RoR can be
computed directly and exactly from the actual probability distribution for
the game. One formula for doing so was given in a recent post. There is
absolutely no need to use variance or any higher moments as an
intermediate step in the computations.

If you use only EV and Var, whatever you get out is very likely to
be an approximation, since you lose information by not including
any influence from higher moments.

The "0.999346831403995" is a early warning sign that someone is not
following the convention of significant digits, and hence loses
credibility.

Nonsense. All of the digits are significant.

Risk of ruin is a probability. It is the probability that you will _fail_ to
play indefinitely. For a given fixed playing strategy, this is completely
and exactly defined.

When you multiply two numbers together you have to
truncate the result to the number of significant digits, otherwise
you give the impression that you don't understand the output of your
calculator or computer.

Nonsense. Whether you say 2 x 2 = 4 or 2.0000000 x 2.000000 = 4.00000
makes no difference. The numbers are all exact, and the act of multiplying
them together doesn't change that.

For any specific combination of VP game and playing strategy, the
exact probability of hitting each payoff can be computed. From this
exact probability distribution we can compute the exact RoR for that
strategy. No approximations are involved in this process. All the digits
are significant.

Perhaps it would help to consider a game that isn't as complex as VP.
Take the proverbial flip of a fair coin. The probability for flipping heads
is exactly 0.500000000000000000..... It isn't "roughly 1/2" it is "exactly
1/2". Now consider a biased coin where the probability of heads is
exactly 51% and you play by always betting on "heads." This game has
an expectation of 2% or an expected return of 1.02. The RoR for this
game is exactly p(lose)/p(win) = 0.49/0.51 which is very nearly
0.970784314. The decimals are all significant.

I would put more credence in a "monte carlo" simulation of ror than
any so called "exact solution" based on faulty approximations.

I see. Then I take it you distrust the VP programs that compute the
exact EV and the max-ER strategy? After all, these programs are
based on "exact solution" and do not involve simulation. They perform
an exact combinatorial analysis of the possible outcomes.

I've written combinatorial analysis programs for VP, blackjack, and
several other games. Personally, I trust CA results much more than
simulation results for one simple reason -- it is much easier to compare
results with independently derived CA results. With CA, two independent
researchers should derive exactly the same results. If two independent
CA programs analyze the same game and come up with different numbers,
then (at least) one of the programs is wrong. Period.

All this is to say that, these formulaic ror calc. are more rule of
thumb than precise results, and should be understood accordingly.

I'm sorry, but you are mistaken. The formula in previous posts that
is often attributed to Sorokin is not an approximation. It is exact, not
a "rule of thumb". You may choose to disbelieve that if you wish.

···

On Friday 07 January 2005 12:38 pm, jaydavidson118 wrote:

[nightoftheiguana2000]

--- In vpFREE@yahoogroups.com, "jaydavidson118" <jaydavidson118@y...>
wrote:

All this is to say that, these formulaic ror calc. are more rule of
thumb than precise results, and should be understood accordingly.

that is mathematically incorrect
the R(1) calculation is exact and its inputs (hand results) are exact

[brucem]

--- In vpFREE@yahoogroups.com, "jaydavidson118" <jaydavidson118@y...>
wrote:

CLT is NOT a formula it refers to CENTRAL LIMIT THEOREM, this is

one of the most used and important theorems in statistical theory,
which allows you to use Normal approxiamations to distributions that
are not Normal when some basic criteria are met.

Of course, you are right. I was careless in my wording. But I
imagine anyone familiar with the CLT knows I was referring to the
formula usually cited in textbooks on the same page where the CLT is
discussed. Unfortunately, my original question still has not been
answered ... are variance calculations based (in any sense) on the
CLT? I don't think it is. But I'll be the first to admit I don't
have the math skills of you Jay, Iggy, and others, which is why I
asked the question. I don't know about others, but I really
appreciate having people with real math skills (expecially statistics)
on this board.

As long as we're discussing the CLT, one of the "basic criteria" you
referred to is it works only for "large sample sizes". Then texts
usually go on to say a "large sample size" is 30 or more! For VP, 30
seems like way too small a sample size. So I assume they mean 30 is
sufficient for some distributions (nearly Normal), but not sufficient
for very non-Normal distributions. Is there a formula to determine
how big a sample is reguired for VP, to apply the CLT concept?
TomSki once posted a message stating it should be at least 50,000
games, but never explained how he arrived at this number.

[Harry_Porter]

brumar_lv wrote:

As long as we're discussing the CLT, one of the "basic criteria" you
referred to is it works only for "large sample sizes". Then texts
usually go on to say a "large sample size" is 30 or more! For VP,
30 seems like way too small a sample size. So I assume they mean 30
is sufficient for some distributions (nearly Normal), but not
sufficient for very non-Normal distributions. Is there a formula to
determine how big a sample is reguired for VP, to apply the CLT
concept?
TomSki once posted a message stating it should be at least 50,000
games, but never explained how he arrived at this number.

50,000 hands clearly is far too small a sample. The royal alone
accounts for a huge amount of game variance and clearly you're looking
at a large likely deviance in return from ER in such a limited sample.

I'm speaking as a layman and in very rough terms, but one might think
of the distribution of vp results as being the cumulative distribution
of individual winning hand outcomes (distribution of high pairs, 2
pairs, etc.)

Observing the large factor the royal plays in determining variance,
one might translate your "30 or more" as being 30 royal cycles, at
minimum. In that case we'd be looking at 1.2 million hands. That
seems roughly consistent with the 1MM+ hands that I've frequently seen
the vp "long term' stated as. (For games such as DB, with a
relatively high variance and a greater proportion of return tied up in
quads, that value reasonably is larger.)

Over such long term periods it seems reasonable to assume the vp
results approach a normal distribution and that variance is a decent
value for expected outcome calculations.

- Harry

[brucem]

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

50,000 hands clearly is far too small a sample. The royal alone
accounts for a huge amount of game variance and clearly you're

looking

at a large likely deviance in return from ER in such a limited

sample.

I'm speaking as a layman and in very rough terms, but one might

think

of the distribution of vp results as being the cumulative

distribution

of individual winning hand outcomes (distribution of high pairs, 2
pairs, etc.)

Observing the large factor the royal plays in determining variance,
one might translate your "30 or more" as being 30 royal cycles, at
minimum. In that case we'd be looking at 1.2 million hands. That
seems roughly consistent with the 1MM+ hands that I've frequently

seen

the vp "long term' stated as. (For games such as DB, with a
relatively high variance and a greater proportion of return tied up

in

quads, that value reasonably is larger.)

Over such long term periods it seems reasonable to assume the vp
results approach a normal distribution and that variance is a decent
value for expected outcome calculations.

- Harry

I waited to respond to your post, hoping someone would submit the
formula for the sample size necessary to assume the Normal curve is
applicable. Actually, I think 50,000 games is sufficient for some
games with a low variance (like 9/6 JorB), but I agree with you, it's
way too few for DDBP and other games with a much higher variance.

Looking at 9/6 JorB, only 1.98% of the overall return is attributed
to the RF, which occurs every 40,391 hands. Extrapolating, that's
about 2.25% in 50,000 games. According to the basic CLT formula, for
50,000 games and a 20 variance, the expection is a return of 96.25%
or better, with 95% confidence. This compares to a 97.29% return if
you play 50,000 games perfectly, never get a RF, but get all other
wins according to expectation. The 96.25% return would occur if you
not only did not get a RF but also get less than the expected quad
and other wins. So I think this suggests 50,000 games is a pretty
good minimum sample size (session size) for applying the CLT to
9/6JorB.

[deuceswild1000]

I waited to respond to your post, hoping someone would submit the
formula for the sample size necessary to assume the Normal curve

is

applicable. Actually, I think 50,000 games is sufficient for some
games with a low variance (like 9/6 JorB), but I agree with you,

it's

way too few for DDBP and other games with a much higher variance.

Looking at 9/6 JorB, only 1.98% of the overall return is

attributed

to the RF, which occurs every 40,391 hands. Extrapolating, that's
about 2.25% in 50,000 games. According to the basic CLT formula,

for

50,000 games and a 20 variance, the expection is a return of

96.25%

or better, with 95% confidence. This compares to a 97.29% return

if

you play 50,000 games perfectly, never get a RF, but get all other
wins according to expectation. The 96.25% return would occur if

you

not only did not get a RF but also get less than the expected quad
and other wins. So I think this suggests 50,000 games is a pretty
good minimum sample size (session size) for applying the CLT to
9/6JorB.

Need my memory refreshed. Would you publish the math for the above
95$ confidence level.

Thanks

DWK

···

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

[brucem]

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

> I waited to respond to your post, hoping someone would submit the
> formula for the sample size necessary to assume the Normal curve
is
> applicable. Actually, I think 50,000 games is sufficient for

some

> games with a low variance (like 9/6 JorB), but I agree with you,
it's
> way too few for DDBP and other games with a much higher variance.
>
> Looking at 9/6 JorB, only 1.98% of the overall return is
attributed
> to the RF, which occurs every 40,391 hands. Extrapolating,

that's

> about 2.25% in 50,000 games. According to the basic CLT formula,
for
> 50,000 games and a 20 variance, the expection is a return of
96.25%
> or better, with 95% confidence. This compares to a 97.29% return
if
> you play 50,000 games perfectly, never get a RF, but get all

other

> wins according to expectation. The 96.25% return would occur if
you
> not only did not get a RF but also get less than the expected

quad

> and other wins. So I think this suggests 50,000 games is a

pretty

> good minimum sample size (session size) for applying the CLT to
> 9/6JorB.

Need my memory refreshed. Would you publish the math for the above
95$ confidence level.

Thanks

DWK

Sure. I took my data from a table I created several years ago, but I
can explain how I computed each cell. In my post I used the table
cell corresponding to a game variance of 20 and session size of
50,000 games. I used 20 in my post because 20 is roughly the
variance of 9/6 JorB.

The CLT formula I used is:
sqrt session variance = sqrt game variance / sqrt games played

We know what the values on the right side are (20 and 50,000). We
need to solve the left side.

The left side represents 1 standard deviation. For VP, this is a
fraction or multiple of a betting unit. This number gets smaller as
the number of games played increases, and larger as the game variance
increases, and vice versa.

But I don't want 1 standard deviation (67% conf. level), I want the
number of standard deviations corresponding to a 95% confidence
level. To do this I have to multiply 1 std. dev. by 1.645 (taken
from a z table), and multiply the right side by 1.645 too.

The 95% conf.level in my table corresponds to 50% of the area of the
Normal curve above the mean and 45% of the area of the Normal curve
below the mean. The table focuses on the worst case scenario
(losing). In other words:

95% conf. level = 1.645 * sqrt session variance.

The equation is then solved:

95% conf. level = (4.4721359 x 1.645) / 223.60679 = .0329

What is .0329? These are betting units (1.645 std dev). The EV of
9/6 JorB is usually written as a percent of a betting unit (99.5439%)
but can also be written in betting units (.995439). Likewise,
the .0329 betting units can be written as a percentage (3.29%).

So I subtracted 3.29% from 99.5439 giving 96.25%. This represents
the session payback percentage (approx) where 95% of 50,000 game
sessions will be higher, and 5% lower.

In the last step I cheated a bit because the 3.29% is based on a 20
variance, not the exact variance of 9/6 JorB (19.51468), but I think
its good enough for government work.

If you see a flaw, please let me know. No one has ever reviewed
these calculations so an error is certainly possible.

···

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

[nightoftheiguana2000]

for what it's worth i think your calculations are correct
but i would have done it this way, which i think is more accurate:
first off, one thing going for us is that ror is the left side of the
curve and the left side is more normal well before the right side
anyway, i'll do 9/6job for 50,000 hands and we'll see how our numbers
compare:
i think it's pretty safe to say the 5% low sample are sessions where
the royal has not been hit:
so, 9/6job without the royal is:
er=99.5%-2%=97.5%
var=19.5-15.8=3.7
sd/hand=sqrt(var/hands)=sqrt(3.7/50000)=0.86%
5%ror=1.645sd
5%ror deviation=0.86% x 1.645=1.4%
97.5%-1.4%=96.1%
so, 5% of hands have worse return than 96.1% assuming hands of
straight flush and less have normal distributions

--- In vpFREE@yahoogroups.com, "deuceswild1000"

<deuceswild1000@y...>

wrote:
>
>
> > I waited to respond to your post, hoping someone would submit

the

> > formula for the sample size necessary to assume the Normal

curve

> is
> > applicable. Actually, I think 50,000 games is sufficient for
some
> > games with a low variance (like 9/6 JorB), but I agree with

you,

> it's
> > way too few for DDBP and other games with a much higher

variance.

> >
> > Looking at 9/6 JorB, only 1.98% of the overall return is
> attributed
> > to the RF, which occurs every 40,391 hands. Extrapolating,
that's
> > about 2.25% in 50,000 games. According to the basic CLT

formula,

> for
> > 50,000 games and a 20 variance, the expection is a return of
> 96.25%
> > or better, with 95% confidence. This compares to a 97.29%

return

> if
> > you play 50,000 games perfectly, never get a RF, but get all
other
> > wins according to expectation. The 96.25% return would occur

if

> you
> > not only did not get a RF but also get less than the expected
quad
> > and other wins. So I think this suggests 50,000 games is a
pretty
> > good minimum sample size (session size) for applying the CLT to
> > 9/6JorB.
>
> Need my memory refreshed. Would you publish the math for the

above

> 95$ confidence level.
>
> Thanks
>
> DWK
>>>>>>>>>>>>>>>>>>>>>>
Sure. I took my data from a table I created several years ago, but

I

can explain how I computed each cell. In my post I used the table
cell corresponding to a game variance of 20 and session size of
50,000 games. I used 20 in my post because 20 is roughly the
variance of 9/6 JorB.

The CLT formula I used is:
sqrt session variance = sqrt game variance / sqrt games played

We know what the values on the right side are (20 and 50,000). We
need to solve the left side.

The left side represents 1 standard deviation. For VP, this is a
fraction or multiple of a betting unit. This number gets smaller

as

the number of games played increases, and larger as the game

variance

increases, and vice versa.

But I don't want 1 standard deviation (67% conf. level), I want the
number of standard deviations corresponding to a 95% confidence
level. To do this I have to multiply 1 std. dev. by 1.645 (taken
from a z table), and multiply the right side by 1.645 too.

The 95% conf.level in my table corresponds to 50% of the area of

the

Normal curve above the mean and 45% of the area of the Normal curve
below the mean. The table focuses on the worst case scenario
(losing). In other words:

95% conf. level = 1.645 * sqrt session variance.

The equation is then solved:

95% conf. level = (4.4721359 x 1.645) / 223.60679 = .0329

What is .0329? These are betting units (1.645 std dev). The EV of
9/6 JorB is usually written as a percent of a betting unit

(99.5439%)

but can also be written in betting units (.995439). Likewise,
the .0329 betting units can be written as a percentage (3.29%).

So I subtracted 3.29% from 99.5439 giving 96.25%. This represents
the session payback percentage (approx) where 95% of 50,000 game
sessions will be higher, and 5% lower.

In the last step I cheated a bit because the 3.29% is based on a 20
variance, not the exact variance of 9/6 JorB (19.51468), but I

think

···

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

> --- In vpFREE@yahoogroups.com, "brumar_lv" <brumar_lv@y...> wrote:
its good enough for government work.

If you see a flaw, please let me know. No one has ever reviewed
these calculations so an error is certainly possible.

[brucem]

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
<nightoftheiguana2000@y...> wrote:

for what it's worth i think your calculations are correct
but i would have done it this way, which i think is more accurate:
first off, one thing going for us is that ror is the left side of

the

curve and the left side is more normal well before the right side
anyway, i'll do 9/6job for 50,000 hands and we'll see how our

numbers

compare:
i think it's pretty safe to say the 5% low sample are sessions where
the royal has not been hit:
so, 9/6job without the royal is:
er=99.5%-2%=97.5%
var=19.5-15.8=3.7
sd/hand=sqrt(var/hands)=sqrt(3.7/50000)=0.86%
5%ror=1.645sd
5%ror deviation=0.86% x 1.645=1.4%
97.5%-1.4%=96.1%
so, 5% of hands have worse return than 96.1% assuming hands of
straight flush and less have normal distributions

Thanks for reviewing my calculations. I'm curious why you believe
your ROR calculation is more accurate. Could you elaborate? I'm not
disagreeing, I simply don't know one way or the other. Evidently the
ROR formula is also based on the Normal distribution. I've never
studied it in detail so I don't understand the 15.8.

Looking at your 96.1% (pretty close to my estimate), this may seem
like a pretty good return for a 50,000 game session and failing to
get a RF. But it's worse than it appears. Playing a 25 cent game
(max bet $1.25) that means total wagering of $62,000 (50,000 x
$1.25), and losing 3.9% of that amount, or $2437.50. That could
easily be an entire bankroll. Fortunately, there is only a 5% chance
it will happen in any given 50,000 game session of 9/6JorB.

[nightoftheiguana2000]

Thanks for reviewing my calculations. I'm curious why you believe
your ROR calculation is more accurate. Could you elaborate? I'm

not

disagreeing, I simply don't know one way or the other. Evidently

the

ROR formula is also based on the Normal distribution. I've never
studied it in detail so I don't understand the 15.8.

in my calculation i did not assume the royal distribution was normal
it clearly is not, yet notice your calculation was fairly close anyway
my only assumption is that the other hands have a normal distribution
the 15.8 is the variance of a royal flush, i adjusted the er and
variance by subtracting the contribution of the royal

for 9/6job the royal cycle is 40391 so:
royal er = 800/40391= 2%
royal variance = (800-.995)^2/40391= 15.8

···

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

[nightoftheiguana2000]

for the heck of it i ran lotspiech's calculator:
http://www.lotspiech.com/GamblersRuin.html
for 50,000 hands of 9/6job with stake of $2440 and retire at $2440,
results are:
less than -$2440: 2.2%
-$2440 to -$1950: 5.0%
-$1950 to -$1460: 10%
-$1460 to -$970: 11%
-$970 to -$480: 13%
-$480 to +$10: 13%
+$10 to +$500: 11%
+$500 to +$990: 9.7%
+$990 to +$1480: 6.9%
+$1480 to +$1970: 3.9%
+$1970 to +$2440: 1.3%
more than +$2440: 9.7%

this would indicate that the 5%ror estimate arrived at below using
variance and clt is overly conservative
(the actual chance of losing more than $2440 is 2.2%, not 5%)
in hindsight this makes sense, the right side (winning side) of the
curve is skewed positive, so should the left side (losing side)
although to a lesser extent

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
<nightoftheiguana2000@y...> wrote:
>
> for what it's worth i think your calculations are correct
> but i would have done it this way, which i think is more accurate:
> first off, one thing going for us is that ror is the left side of
the
> curve and the left side is more normal well before the right side
> anyway, i'll do 9/6job for 50,000 hands and we'll see how our
numbers
> compare:
> i think it's pretty safe to say the 5% low sample are sessions

where

> the royal has not been hit:
> so, 9/6job without the royal is:
> er=99.5%-2%=97.5%
> var=19.5-15.8=3.7
> sd/hand=sqrt(var/hands)=sqrt(3.7/50000)=0.86%
> 5%ror=1.645sd
> 5%ror deviation=0.86% x 1.645=1.4%
> 97.5%-1.4%=96.1%
> so, 5% of hands have worse return than 96.1% assuming hands of
> straight flush and less have normal distributions
>>>>>>>>>>>>>>>>>>>>>>
Thanks for reviewing my calculations. I'm curious why you believe
your ROR calculation is more accurate. Could you elaborate? I'm

not

disagreeing, I simply don't know one way or the other. Evidently

the

ROR formula is also based on the Normal distribution. I've never
studied it in detail so I don't understand the 15.8.

Looking at your 96.1% (pretty close to my estimate), this may seem
like a pretty good return for a 50,000 game session and failing to
get a RF. But it's worse than it appears. Playing a 25 cent game
(max bet $1.25) that means total wagering of $62,000 (50,000 x
$1.25), and losing 3.9% of that amount, or $2437.50. That could
easily be an entire bankroll. Fortunately, there is only a 5%

chance

···

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

it will happen in any given 50,000 game session of 9/6JorB.

[nightoftheiguana2000]

i didn't do my clt estimate correctly, -1.645sd is the 5%ror point but
since i assumed the royal wasn't hit, i need to multiply by the
chances of not hitting a royal in 50,000 hands: (1-1/40391)^50000=0.3
so, 5% x 0.3 = 1.5% which is closer to the lotspiech calculated value
of 2.2%

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
<nightoftheiguana2000@y...> wrote:

for the heck of it i ran lotspiech's calculator:
http://www.lotspiech.com/GamblersRuin.html
for 50,000 hands of 9/6job with stake of $2440 and retire at $2440,
results are:
less than -$2440: 2.2%
-$2440 to -$1950: 5.0%
-$1950 to -$1460: 10%
-$1460 to -$970: 11%
-$970 to -$480: 13%
-$480 to +$10: 13%
+$10 to +$500: 11%
+$500 to +$990: 9.7%
+$990 to +$1480: 6.9%
+$1480 to +$1970: 3.9%
+$1970 to +$2440: 1.3%
more than +$2440: 9.7%

this would indicate that the 5%ror estimate arrived at below using
variance and clt is overly conservative
(the actual chance of losing more than $2440 is 2.2%, not 5%)
in hindsight this makes sense, the right side (winning side) of the
curve is skewed positive, so should the left side (losing side)
although to a lesser extent

>
> --- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
> <nightoftheiguana2000@y...> wrote:
> >
> > for what it's worth i think your calculations are correct
> > but i would have done it this way, which i think is more

accurate:

> > first off, one thing going for us is that ror is the left side

of

> the
> > curve and the left side is more normal well before the right

side

> > anyway, i'll do 9/6job for 50,000 hands and we'll see how our
> numbers
> > compare:
> > i think it's pretty safe to say the 5% low sample are sessions
where
> > the royal has not been hit:
> > so, 9/6job without the royal is:
> > er=99.5%-2%=97.5%
> > var=19.5-15.8=3.7
> > sd/hand=sqrt(var/hands)=sqrt(3.7/50000)=0.86%
> > 5%ror=1.645sd
> > 5%ror deviation=0.86% x 1.645=1.4%
> > 97.5%-1.4%=96.1%
> > so, 5% of hands have worse return than 96.1% assuming hands of
> > straight flush and less have normal distributions
> >>>>>>>>>>>>>>>>>>>>>>
> Thanks for reviewing my calculations. I'm curious why you

believe

> your ROR calculation is more accurate. Could you elaborate? I'm
not
> disagreeing, I simply don't know one way or the other. Evidently
the
> ROR formula is also based on the Normal distribution. I've never
> studied it in detail so I don't understand the 15.8.
>
> Looking at your 96.1% (pretty close to my estimate), this may

seem

> like a pretty good return for a 50,000 game session and failing

to

> get a RF. But it's worse than it appears. Playing a 25 cent

game

···

--- In vpFREE@yahoogroups.com, "brumar_lv" <brumar_lv@y...> wrote:
> (max bet $1.25) that means total wagering of $62,000 (50,000 x
> $1.25), and losing 3.9% of that amount, or $2437.50. That could
> easily be an entire bankroll. Fortunately, there is only a 5%
chance
> it will happen in any given 50,000 game session of 9/6JorB.

[brucem]

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
<nightoftheiguana2000@y...> wrote:

i didn't do my clt estimate correctly, -1.645sd is the 5%ror point

but

since i assumed the royal wasn't hit, i need to multiply by the
chances of not hitting a royal in 50,000 hands: (1-1/40391)

^50000=0.3

so, 5% x 0.3 = 1.5% which is closer to the lotspiech calculated

value

of 2.2%

Do you know how the lotspiech calculator works? I checked his
website but didn't find an explanation.

[nightoftheiguana2000]

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
<nightoftheiguana2000@y...> wrote:
>
> i didn't do my clt estimate correctly, -1.645sd is the 5%ror

point

but
> since i assumed the royal wasn't hit, i need to multiply by the
> chances of not hitting a royal in 50,000 hands: (1-1/40391)
^50000=0.3
> so, 5% x 0.3 = 1.5% which is closer to the lotspiech calculated
value
> of 2.2%
>>>>>>>>>>>>>>>>>>
Do you know how the lotspiech calculator works? I checked his
website but didn't find an explanation.

nope
i imagine it involves matrix manipulation, one hand of job has 10
possible outcomes, two hands have 10x10 outcomes not all unique ...

···

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