vpFREE2 Forums

Long Run Index, N0

#41

Well, most certainly that will depend on how closely you want to be to
a real normal distribution. it's never "exactly" normal, it only gets
arbitrarily close as N increases (and not necessarily uniformly).

The "30" number you quoted might be enough when studying a coin flip,
but as you noted it's grossly inadequate for VP.

Notice that the "easy" proofs of the CLT use convolutions, and reduce
them to multiplications via a fourier transform - there's absolutely
no theory of probabilities or statistics involved there.

Interestingly, the million-hand simulations we're looking at are
approximately 25 royal cycles, which is not far from 30. I will not
attempt to venture much further down that path, but I'll note that the
point where we enter the "long term" for practical VP purposes (i.e.
within a few percent or so of a normal distribution for most outcomes)
might lie somewhere between 10 and 100 cycles, assuming of course that
you play a single denomination.

JBQ

···

On 1/26/06, brumar_lv <brumar_lv@yahoo.com> wrote:

So how big is big enough?

#42

If your interest is +/-2sd, 9 royal cycles is probably enough since 9
+/-6 is 97% of a poisson distribution which is pretty close to 95% of
a normal distribution.

If you're a gambler, I would propose that N0 better suits your interests.

···

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

So how big is big enough?

#43

The number for 2 standard deviations seem overly optimistic and I

imagine it would be

interesting to see the actuall numbers from the PDF.

If we're still talking about the example I posted of a million hands
of 9/6 Jacks, here are a few numbers. I'm not trying for phony
precision here but Mr. Iguana was being a little too approximate.

The expected return in bets is about -4561, and the standard
deviation is sqrt(1E6 * 19.5147) = 4417.5. For a normal distribution
a symmetrical 95% confidence limit is +- 1.96 sd (that's the more
precise version of the usual 2 sigma approximation). So, the 2.5 %ile
return would be -4561 - 1.96*4417.5 = -13219. The 97.5 %ile would be
4097.

For the "true" distribution I get 2.5 %ile = -12906 and 97.5 %ile =
4399. Also the median return is -4670, slightly worse than the mean.

These differences don't seem very significant to me, but they're
consistent with the slightly skewed appearance of the histogram.

[Aside: While the formula is very simple (see below) it is not an

trival thing to compute the

"exact" PDF for so many hands. The problem arrises not so much

from the limited

precision of computers, but rather their limited memory.

Eventually, the tails of the PDF

The computer I have with the most memory will do an FFT on a 32
million element vector without choking. I just try to pad out to far
enough in the tails of the distribution that aliasing isn't going to
be a problem. There's going to be some error, but hopefully it's
small and way out in the tails. For this example I padded out to a
maximum net return of about 60 royals worth. If I did the calculation
right the probability of getting 60 or more royals in a million hands
is less than 1 in a billion.

Mike

···

--- In vpFREE@yahoogroups.com, "cdfsrule" <groups.yahoo@v...> wrote:

#44

For what it's worth, 9+/-6 (or 16+/-8, 25+/-10, 36+/-12 ...) works on
cycles of just about anything, for example, for 9 quad cycles
(423x9=3807 hands) you will get 9+/-6 (3 to 15) quads about 95% of the
time, if you get less than 3 quads you can state with about 2.5%
sampling error that you are either misplaying or getting cheated.

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

> So how big is big enough?

If your interest is +/-2sd, 9 royal cycles is probably enough since 9
+/-6 is 97% of a poisson distribution which is pretty close to 95% of
a normal distribution.

If you're a gambler, I would propose that N0 better suits your

interests.

···

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

#45

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

For what it's worth, 9+/-6 (or 16+/-8, 25+/-10, 36+/-12 ...) works on
cycles of just about anything, for example, for 9 quad cycles
(423x9=3807 hands) you will get 9+/-6 (3 to 15) quads about 95% of the
time, if you get less than 3 quads you can state with about 2.5%
sampling error that you are either misplaying or getting cheated.

Just to make sure I understand, with 9+/-6 ... does the 9+ represent 9
quad cycles and the -6 represent what you add or subtract to the 9
(9+6=15 and 9-6=3) to get the 95% confidence level for quads received
in 3807 hands? Also, does the 16+/-8 represent a different confidence
level, such as 3SD? If not, what does it mean?

Also, as I mentioned earlier, I haven't been following the NO
discussion closely (as you can probably tell). Please refer me to the
web site(s) (or whatever) that explain it.

#46

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
<nightoftheiguana2000@y...> wrote:
>
> For what it's worth, 9+/-6 (or 16+/-8, 25+/-10, 36+/-12 ...) works on
> cycles of just about anything, for example, for 9 quad cycles
> (423x9=3807 hands) you will get 9+/-6 (3 to 15) quads about 95% of the
> time, if you get less than 3 quads you can state with about 2.5%
> sampling error that you are either misplaying or getting cheated.
>>>>>>>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>>>>>>>>
Just to make sure I understand, with 9+/-6 ... does the 9+ represent 9
quad cycles and the -6 represent what you add or subtract to the 9
(9+6=15 and 9-6=3) to get the 95% confidence level for quads received
in 3807 hands?

Yes, 9 plus or minus 6.

Also, does the 16+/-8 represent a different confidence
level, such as 3SD? If not, what does it mean?

2SD.
Cycles plus or minus two times the square root of cycles is 2SD

Also, as I mentioned earlier, I haven't been following the NO
discussion closely (as you can probably tell). Please refer me to the
web site(s) (or whatever) that explain it.

http://members.cox.net/vpfree/Bank_NO.htm
http://www.bjmath.com/bjmath/refer/N0.htm

···

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

#47

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

>
> --- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
> <nightoftheiguana2000@y...> wrote:
> >
> > For what it's worth, 9+/-6 (or 16+/-8, 25+/-10, 36+/-12 ...)

works on

> > cycles of just about anything, for example, for 9 quad cycles
> > (423x9=3807 hands) you will get 9+/-6 (3 to 15) quads about

95% of the

> > time, if you get less than 3 quads you can state with about

2.5%

> > sampling error that you are either misplaying or getting

cheated.

> >>>>>>>>>>>>>>>>>>>>>>>>
> >>>>>>>>>>>>>>>>>>>>>>>>
> Just to make sure I understand, with 9+/-6 ... does the 9+

represent 9

> quad cycles and the -6 represent what you add or subtract to the

9

> (9+6=15 and 9-6=3) to get the 95% confidence level for quads

received

> in 3807 hands?

Yes, 9 plus or minus 6.

> Also, does the 16+/-8 represent a different confidence
> level, such as 3SD? If not, what does it mean?

2SD.
Cycles plus or minus two times the square root of cycles is 2SD

I need further help here. I do not understand what you mean. I
understand (I think) th 9 plus/minus 6 and thought I did for the
other numbers, but now I don't (the other numbers). Please try it
again in different or more words.

DWK

···

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

#48

I think what NOTI is trying to say is this: In VP parlance, the Poisson distribution only
depends on the number of cycles. It does not depend on the size of the cycle, so the
Poisson distirbution doesn't care if you are computing something for Quads or RF's , etc.
IMHO, its an exellent point that NOTI made.

Prob (m) = Poisson(m, X) where m = number of occurrences of something and X is number
of cycles.

Now suppose you play for X cycles of something and you want to know your so-called
SYMETRIC 95% confidence interval . That is, 95% if the time you would have gotten
between (X-A) and (X+A) of whatever that something was, an you want to know A

Well, since we know that the Poisson distributoin only depends on the number of cycles X,
we can go ahead find the value of A for various values of X that gives 95% . We do not
need to know how many handw we played, whether the something is Quads or RF's. Nice.
But, this is a trial an error procedure that involces computing sums like

fP(3)+P(4)+P(5)+P(6)+P(7)+P(8)+.... P(15) = NOTI's 9+/-6 %

You have to do this numerically (there is no formula for the sum).
Here are the individual numbers you need for X=9
M P(M)
0 0.00012341
1 0.00111069
2 0.0049981
3 0.0149943
4 0.0337372
5 0.0607269
6 0.0910903
7 0.117116
8 0.131756
9 0.131756
10 0.11858
11 0.0970201
12 0.072765
13 0.0503758
14 0.0323844
15 0.0194307

The answer comes out to 97.1732% or so. (Not NOTI's 95%, but close.. perhaps he meant
97%?), but then again, maybe I made a math error. You can then compute the other sums
and convice yourself that 9+/-6 is the best answer. (But it isn't becuase 9+/-5 = 93.7306
%, oh well... did I make another error? Could be. My point here is just to show how the
computation is done.)

while we are at it:

18+/- 8 = 96.7685 %
25+/- 10 = 96.514%
36+/-12 = 96.3407%
BTW (36+11/-11 = 94.5596% , which is closer to 95%?)

···

> --- In vpFREE@yahoogroups.com, "nightoftheiguana2000"
> <nightoftheiguana2000@y...> wrote:
> >
> > For what it's worth, 9+/-6 (or 16+/-8, 25+/-10, 36+/-12 ...) works on
> > cycles of just about anything, for example, for 9 quad cycles
> > (423x9=3807 hands) you will get 9+/-6 (3 to 15) quads about 95% of the
> > time, if you get less than 3 quads you can state with about 2.5%
> > sampling error that you are either misplaying or getting cheated.
> >>>>>>>>>>>>>>>>>>>>>>>>

#49

oops... one more thing. As noti already stated, he didn't pick his +/- number out of thin
air. You see, for a poisson distirbution, the number of cycles (mean) = the variance
ALWAYS.

So, given X cycyles, you have a variance of X. Thus a standard deviation of sqrt(X). NOTI
chose X+/- 2*SQRT(X) as the size of the symetric interval, where the sqrt(x) was an
integer. It turns out this trick gives numbers close to 95% percent, and gets close as X
gets bigger....

100+/-20 = 0.959879

But does it ever hit 95% exactly? Does it go too far? (And why would anyone care?)

[Computing this much above X=100 requires some tricks. Most computers can't handle
the large factorirals involved. Instead, though one can use the gammaln(n) function,
which for integer n, is effectively the same as the ln(factorial(n)). Nice. But problems will
soon show up with the gammaln function as n get too big for it. Oh what to do, what to
do? LOL]

I think what NOTI is trying to say is this: In VP parlance, the Poisson distribution only
depends on the number of cycles. It does not depend on the size of the cycle, so the
Poisson distirbution doesn't care if you are computing something for Quads or RF's , etc.
IMHO, its an exellent point that NOTI made.

Prob (m) = Poisson(m, X) where m = number of occurrences of something and X is

number

of cycles.

Now suppose you play for X cycles of something and you want to know your so-called
SYMETRIC 95% confidence interval . That is, 95% if the time you would have gotten
between (X-A) and (X+A) of whatever that something was, an you want to know A

Well, since we know that the Poisson distributoin only depends on the number of cycles

X,

we can go ahead find the value of A for various values of X that gives 95% . We do not
need to know how many handw we played, whether the something is Quads or RF's.

Nice.

···

--- In vpFREE@yahoogroups.com, "cdfsrule" <groups.yahoo@v...> wrote:

#50

oops... one more thing. As noti already stated, he didn't pick

his +/- number out of thin

air. You see, for a poisson distirbution, the number of cycles

(mean) = the variance

ALWAYS.

So, given X cycyles, you have a variance of X. Thus a standard

deviation of sqrt(X). NOTI

chose X+/- 2*SQRT(X) as the size of the symetric interval, where

the sqrt(x) was an

integer. It turns out this trick gives numbers close to 95%

percent, and gets close as X

gets bigger....

100+/-20 = 0.959879

But does it ever hit 95% exactly? Does it go too far? (And why

would anyone care?)

[Computing this much above X=100 requires some tricks. Most

computers can't handle

the large factorirals involved. Instead, though one can use the

gammaln(n) function,

which for integer n, is effectively the same as the ln(factorial

(n)). Nice. But problems will

soon show up with the gammaln function as n get too big for it. Oh

what to do, what to

do? LOL]

--- In vpFREE@yahoogroups.com, "cdfsrule" <groups.yahoo@v...>

wrote:

>
> I think what NOTI is trying to say is this: In VP parlance, the

Poisson distribution only

> depends on the number of cycles. It does not depend on the size

of the cycle, so the

> Poisson distirbution doesn't care if you are computing something

for Quads or RF's , etc.

> IMHO, its an exellent point that NOTI made.
>
> Prob (m) = Poisson(m, X) where m = number of occurrences of

something and X is

number
> of cycles.
>
> Now suppose you play for X cycles of something and you want to

know your so-called

> SYMETRIC 95% confidence interval . That is, 95% if the time you

would have gotten

> between (X-A) and (X+A) of whatever that something was, an you

want to know A

>
> Well, since we know that the Poisson distributoin only depends

on the number of cycles

X,
> we can go ahead find the value of A for various values of X

that gives 95% . We do not

> need to know how many handw we played, whether the something is

Quads or RF's.

Nice.

  cdfrules,

Followed your explanation completely. I had figured out the other
size numbers correctly, Thought I had, but the 2SD was throwing
me, but now I understand.

Thanks

···

--- In vpFREE@yahoogroups.com, "cdfsrule" <groups.yahoo@v...> wrote:

#51

So, given X cycyles, you have a variance of X. Thus a standard

deviation of sqrt(X). NOTI

chose X+/- 2*SQRT(X) as the size of the symetric interval, where the

sqrt(x) was an

integer. It turns out this trick gives numbers close to 95%

percent, and gets close as X

gets bigger....

100+/-20 = 0.959879

But does it ever hit 95% exactly? Does it go too far? (And why

would anyone care?)

According to the CLT, it approaches the same as a normal distribution,
in other words: 95.44997361...%

···

--- In vpFREE@yahoogroups.com, "cdfsrule" <groups.yahoo@v...> wrote:

#52

Just for the record, I never would have raised my doubts about the CLT
applying to VP if I had known this topic was hashed over in great
detail last September. Thanks for bringing these posts to my
attention.

···

--- In vpFREE@yahoogroups.com, "jeffcole2003oct" <jeff-cole@...> wrote:

In message #48297, cdfsrule said that the CLT does not apply to VP.
However, this was debated by others (nightoftheiguana2000 and JBQ
for sure), and I'm not sure there was ever a definitive conclusion
drawn.