vpFREE2 Forums

rigged/tampered machines

#21

Basically, it would be good PR and probably bring in more patrons if
the Indians insisted upon outside independent auditing. California
and the Feds do nothing that I can see.

The problem is that with no one watching, even if most casinos are
honest, someone somewhere will cheat, and if caught would hurt all
the tribes. The public knowing that no one is watching can easily
gripe when their luck is poor and say they are cheating.

In California within the Riverside and San Diego county areas there
are many casinos just short distances from each other. This
competition should keep them quite honest. I have played in an area
where there are no other casinos for about a 4 hour drive. This
casino never spreads the cards on a blackjack table. When playing
VP on their 9/6 machines an employee mentioned to me that she had
never seen a royal in 6 months on that particular bank of 12
machines. Do they cheat? Well, being a counter I never saw
continuous high counts (which you would see if aces or 10's were
removed), just the usual spread of different counts, so if they do
not cheat, why not spread the cards and take these doubts away.
As far as their VP goes, perhaps they do because their VP machines
seem to get a lot less play than they should. Then perhaps if you
pay out a few royals, it would keep these machines busy and they
would make more.

Victoria

#22

MHS wrote:

* Bill Velek wrote: �I _CONSTANTLY_ hear that Indian casinos are not
trustworthy. I don't
know if it is simply because of prejudice, or urban legend, or because
of suspicions because they are unregulated (by any state gaming
commission or the Feds), or because there is actual evidence that they
have a propensity to cheat.�
As someone who is highly skeptical about non-regulated casinos, I would have
to say my questions are grounded in none of the above. They arise from the
complete lack of independent outside auditing and supervision.

Not to nitpick, but your suspicions seem to fall into the third clause
that I stated: "because of suspicions because they are unregulated".
Whatever; ... but perhaps a lack of communication on my part. In any
event, I don't play at Indian casinos, and I have nothing against
Indians; my wife is part Indian, and consequently, my eight children are
also part Indian, thus I am surrounded by Indians. :slight_smile: ... and LOVE it. :slight_smile:

Cheers.

Bill Velek

#23

I don't think so. Analyses like these are full of pitfalls. Once
you qualify the blocks to have no 5 cycle droughts and contain a
royal, your blocks are more rich in royals than a completely random
sample. The probablilty of the drought over the gap is overstated by
your formula. The excess richness depends on the block size.

AJ

--- In vpFREE@yahoogroups.com, "dirtyroyal2004a"
<dirtyroyal2004a@y...> wrote:

here's my theory
(not sure it's correct but think it is)
ok, you have two blocks, which contain completely random hands, the
only thing you know, by definition, is that they contain at least

one

royal and no royal droughts of 5 cycles or larger
now you put the two blocks together
the question is, what is the probability of a 5 cycle royal drought
from the rightmost royal of block 1 to the leftmost royal of block

2?

isn't it (1-Pr)^(5/Pr)?
or the more general:
probability of a royaless block of length L or greater:
(1-Pr)^L
there is only one royaless block across the seam, and it starts

with

the rightmost royal of block one, which i know exists by

definition,

···

therefore P=1, and it ends with the leftmost royal of block two,
ditto, the only question is what is its length
i think (1-Pr)^L is correct
comments?

#24

i don't think you understood the example
*by definition* these blocks are completely random, *except* that,
*by definition* they must contain at least one royal, in this way
they are not blocks of 5 royaless cycles, they are less
and one royal in 5 cycles is not what i'd call rich in royals
it's probably too complicated to explain, maybe Harry can explain
better than i can, i assume he understands the logic
the reason for the block construction is pretty straightfoward, the
only issue is how to cover the seams created when blocks are joined,
i think my logic is correct but i haven't thought it out enough to be
totally sure

I don't think so. Analyses like these are full of pitfalls. Once
you qualify the blocks to have no 5 cycle droughts and contain a
royal, your blocks are more rich in royals than a completely random
sample. The probablilty of the drought over the gap is overstated

by

your formula. The excess richness depends on the block size.

AJ

--- In vpFREE@yahoogroups.com, "dirtyroyal2004a"
<dirtyroyal2004a@y...> wrote:
> here's my theory
> (not sure it's correct but think it is)
> ok, you have two blocks, which contain completely random hands,

the

> only thing you know, by definition, is that they contain at least
one
> royal and no royal droughts of 5 cycles or larger
> now you put the two blocks together
> the question is, what is the probability of a 5 cycle royal

drought

> from the rightmost royal of block 1 to the leftmost royal of

block

···

--- In vpFREE@yahoogroups.com, "AJ" <mile_5280@y...> wrote:

2?
> isn't it (1-Pr)^(5/Pr)?
> or the more general:
> probability of a royaless block of length L or greater:
> (1-Pr)^L
> there is only one royaless block across the seam, and it starts
with
> the rightmost royal of block one, which i know exists by
definition,
> therefore P=1, and it ends with the leftmost royal of block two,
> ditto, the only question is what is its length
> i think (1-Pr)^L is correct
> comments?

#25

"Harry Porter" <harry.porter@v...> wrote:

However, if it should turn out that the probability of hitting such

a

dry streak over 20 cycles is something like 3%-5%, then that hardly
puts them in rarified company. (Well, if the hand in question is

the

royal there is the question of just how many players will stick that
drought out through the full 5 cycles rather than run home with

their

tail between their legs.)

by the way, this has ruined many otherwise competant gamblers, the
classic is the gambler who has a couple of lucky or even just average
years, and instead of building up their bankroll they buy the trophy
house and trophy car and trophy spouse and then that 5 royal drought
hits and wipes them out

the longer you play, the greater your chances of eventually hitting
one of those rare 5 royal droughts, if you are lucky it will occur
later in your career *and* you will be prepared for it by always
maintaining a sufficient bankroll

#26

dirtyroyal2004a wrote:

here's my theory
(not sure it's correct but think it is)
ok, you have two blocks, which contain completely random hands, the
only thing you know, by definition, is that they contain at least
one royal and no royal droughts of 5 cycles or larger
now you put the two blocks together
the question is, what is the probability of a 5 cycle royal drought
from the rightmost royal of block 1 to the leftmost royal of block
2? isn't it (1-Pr)^(5/Pr)?
or the more general:
probability of a royaless block of length L or greater:
(1-Pr)^L
there is only one royaless block across the seam, and it starts with
the rightmost royal of block one, which i know exists by definition,
therefore P=1, and it ends with the leftmost royal of block two,
ditto, the only question is what is its length
i think (1-Pr)^L is correct
comments?

Sorry for the delay in reply. I wanted to have my head on straight
before attempting a response.

Well, I think your equation is absolutely dead on, for what it
describes. But, in fact, all you've stated is the probability that
any given "L" hands will be royalless. You've simply dressed it up a bit.

···

------

In your initial "5 royal" example, you don't define the length of the
"blocks"; you simply note that each one contains at least one royal.
As you construct this, you're really looking at a very general case:
once you hit a royal in play, what's the probability that the next N
cycles will be royalless.

That indeed is going to be (1-Pr)^(N * 1/Pr). When you multiply the
number of cycles, N, by the cycle length, 1/Pr, you get "L" - the
total royalless block exprssed in hands. (1-Pr)^L.

And, since it really doesn't matter whether this string of hands led
off with a royal or not, this is the very general probability that any
given string of L hands will be royalless. No surprise here.

------

In assuming that there's at least one royal, this example isn't
sufficiently non-specific to draw conclusions generally applicable to
the defined problem.

I'm becoming satisfied that there is no straightforward solution to
the question of how probable a given drought is over a fixed span of
hands. I appreciate your help in working this through.

- Harry

#27

The mere fact that no solution has been presented does not imply
that none exists.

···

On Friday 23 April 2004 07:33 am, Harry Porter wrote:

I'm becoming satisfied that there is no straightforward solution to
the question of how probable a given drought is over a fixed span of
hands. I appreciate your help in working this through.

#28

Harry Porter wrote:

> I'm becoming satisfied that there is no straightforward solution to
> the question of how probable a given drought is over a fixed span of
> hands. I appreciate your help in working this through.

Steve Jacobs replied:

The mere fact that no solution has been presented does not imply
that none exists.

And I expect that when one does surface, it will be relatively complex
... not straightforward :wink:

- H.

#29

You can't define a block as both "completely random" and containing a
royal and no 5 cycle drought. Once you put qualifications on a
random block it will differ statistically from the unqualified random
block. To analyze the statistics of your blocks you must define a
method for generating the blocks.

Suppose we choose a block size of 400K hands with a royal cycle of
40K hands. We expect an average of 10 royals per block. We run
simulations and generate 1M blocks of 400K hands. We expect 10M
royals in the entire sample. About 45 of the blocks will have no
royals. We exclude these blocks to meet your requirement. We still
expect that the entire sample (1M*400K hands) contains 10M royals.
The royal density of the remaining 999,955 blocks is
(1/40000)*(1000000/999955). When you exclude the 5 cycle dought
blocks the remaining blocks get a further boost in royal density.

I agree with Harry that these simple analyses are unlikely to get the
answers you seek.

AJ

--- In vpFREE@yahoogroups.com, "dirtyroyal2004a"
<dirtyroyal2004a@y...> wrote:

i don't think you understood the example
*by definition* these blocks are completely random, *except* that,
*by definition* they must contain at least one royal, in this way
they are not blocks of 5 royaless cycles, they are less
and one royal in 5 cycles is not what i'd call rich in royals
it's probably too complicated to explain, maybe Harry can explain
better than i can, i assume he understands the logic
the reason for the block construction is pretty straightfoward, the
only issue is how to cover the seams created when blocks are

joined,

i think my logic is correct but i haven't thought it out enough to

be

totally sure

> I don't think so. Analyses like these are full of pitfalls.

Once

> you qualify the blocks to have no 5 cycle droughts and contain a
> royal, your blocks are more rich in royals than a completely

random

> sample. The probablilty of the drought over the gap is

overstated

by
> your formula. The excess richness depends on the block size.
>
> AJ
>
>
> --- In vpFREE@yahoogroups.com, "dirtyroyal2004a"
> <dirtyroyal2004a@y...> wrote:
> > here's my theory
> > (not sure it's correct but think it is)
> > ok, you have two blocks, which contain completely random hands,
the
> > only thing you know, by definition, is that they contain at

least

> one
> > royal and no royal droughts of 5 cycles or larger
> > now you put the two blocks together
> > the question is, what is the probability of a 5 cycle royal
drought
> > from the rightmost royal of block 1 to the leftmost royal of
block
> 2?
> > isn't it (1-Pr)^(5/Pr)?
> > or the more general:
> > probability of a royaless block of length L or greater:
> > (1-Pr)^L
> > there is only one royaless block across the seam, and it starts
> with
> > the rightmost royal of block one, which i know exists by
> definition,
> > therefore P=1, and it ends with the leftmost royal of block

two,

···

--- In vpFREE@yahoogroups.com, "AJ" <mile_5280@y...> wrote:
> > ditto, the only question is what is its length
> > i think (1-Pr)^L is correct
> > comments?

#30

"Harry Porter" <harry.porter@v...> wrote:

In assuming that there's at least one royal, this example isn't
sufficiently non-specific to draw conclusions generally applicable

to

the defined problem.

that's not an assumption
your original specification called for a 5 cycle block that was not
royaless, hence not a 5 cycle royal drought, *by definition* this
block of 5 cycles must contain at least one royal, otherwise it would
be a 5 cycle drought, which your original specification precluded
occuring

just to review:
your original specification:
assuming 40k royal cycle, how likely is a player to suffer a 5 cycle
drought or greater during 20 cycles of play
i solved for the case of *not* getting a 5 cycle drought:
5 cycle block with one or more royal: 99%
4 consecutive 5 cycle blocks with one or more royal each: 97%
and no 5 cycle droughts where each block is joined to the next: 95%
hence a player is 5% likely to suffer a 5 cycle drought or greater
during 20 continuous cycles of play

#31

the formula would be:
1-(1-(1-1/Cr)^CrXa)^(2Xb/Xa-1)
for Cr royal cycle of 40,000 and Xa royaless cycle 5 and Xb 20 cycles:
1-(1-(1-1/40000)^40000x5)^(2x20/5-1)=0.046=4.6%

#32

I happened to stumble across a math site that caught my interest, and
it had a link to a freely downloadable book on Generating Functions.

The math site is www.cut-the-knot.org (in case anyone cares) and
it is a wonderful site for math freaks. The link that started me down
this path is www.cut-the-knot.org/ctk/GeneratingFunctions.shtml
which is itself worth looking at, and points to
www.math.upenn.edu/~wilf/DownldGF.html where the book
can be downloaded.

Problem 20 on page 28 of the books say:

20. Let f(n,m,k) be the number of strings of n 0's and 1's that contain
exactly m 1's, no k of which are consecutive.

(a) Find a recurrence formula for f. It should have f(n,m,k) on
the left side, and exactly three terms on the right.

(b) Find, in simple close form, the generating functions

Fk(x,y) = sum{n,m >= 0} f(n,m,k)x^n*y^m (k=1,2,...)

c( Find an explicit formula for f(n,m,k) from the generating function
(this should involve only a single summation, of an expression
that involves a few factorials).

The good news: I think this would exactly solve, possibly in closed
form, the problem posed by Harry for finding the probability of going
20 cycles without a 5 cycle drought.

The bad news: although the book provides solutions for many problems,
this particular solution isn't included.

Generating Function appear to offer an extremely power method for
solving complex questions such as this. I'm reading the book and
trying to absorb it, so I hope to be able to offer an exact solution
some time soon.

#33

AJ wrote:

You can't define a block as both "completely random" and containing a
royal and no 5 cycle drought. Once you put qualifications on a
random block it will differ statistically from the unqualified random
block. To analyze the statistics of your blocks you must define a
method for generating the blocks.

Suppose we choose a block size of 400K hands with a royal cycle of
40K hands. We expect an average of 10 royals per block. We run
simulations and generate 1M blocks of 400K hands. We expect 10M
royals in the entire sample. About 45 of the blocks will have no
royals. We exclude these blocks to meet your requirement. We still
expect that the entire sample (1M*400K hands) contains 10M royals. The royal density of the remaining 999,955 blocks is
(1/40000)*(1000000/999955). When you exclude the 5 cycle dought
blocks the remaining blocks get a further boost in royal density.

I agree with Harry that these simple analyses are unlikely to get the
answers you seek.

Thanks. Something didn't quiet 'smell right' to me either, but I didn't have the time to digest this thread and think it through. Your illustration is, in my opinion, perfectly clear (whether the 45 block figure is correct or not ... about which I have no idea). I generally have a lot of interest in these threads because they help me sharpen my math reasoning. I appreciate everyone taking the time to exchange their views, and for doing so on this forum rather than via private email. Great stuff.

Cheers.

Bill Velek