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Bob Dancer's CasinoGaming Column - 5 MAY 2009

Sampling the Evidence

http://www.casinogaming.com/columnists/dancer/2009/0505.html

<a href="http://www.casinogaming.com/columnists/dancer/2009/0505.html">
http://www.casinogaming.com/columnists/dancer/2009/0505.html</a>

···

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Bob wrote:
"over the next million hands or so, you're results will be very close to 99.54%"

Depends what the meaning of "very close" is, but assuming a near normal distribution, one standard deviation alone is more than $5,000 on a quarter job machine at one million hands. The more you play, the less likely are your odds of hitting an average result. That's what the math says, counter to the standard word on the street.

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--- In vpFREE@yahoogroups.com, "vpFREE Administrator" <vpfreeadmin@...> wrote:

Sampling the Evidence

http://www.casinogaming.com/columnists/dancer/2009/0505.html

<a href="http://www.casinogaming.com/columnists/dancer/2009/0505.html">
http://www.casinogaming.com/columnists/dancer/2009/0505.html</a>

************************************************

This link is posted for informational purposes and doesn't
constitute an endorsement or approval of the linked article's
content by vpFREE. Any discussion of the article must be done
in accordance with vpFREE's rules and policies.

************************************************

I've done a lot of simulations with WinPoker. Seems very common to end up with between 98.80 and 100.2 after a million hands of JoB. Seems like a million hands is not really a very meaningful sample. Maybe 3 million is.

-BB

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

Bob wrote:
"over the next million hands or so, you're results will be very close to 99.54%"

Depends what the meaning of "very close" is, but assuming a near normal distribution, one standard deviation alone is more than $5,000 on a quarter job machine at one million hands.

N0 is the better metric that you search for:
http://members.cox.net/vpfree/Bank_NO.htm
http://www.youtube.com/watch?v=IMxWLuOFyZM

···

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

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000" <nightoftheiguana2000@> wrote:
>
> Bob wrote:
> "over the next million hands or so, you're results will be very close to 99.54%"
>
> Depends what the meaning of "very close" is, but assuming a near normal distribution, one standard deviation alone is more than $5,000 on a quarter job machine at one million hands.

I've done a lot of simulations with WinPoker. Seems very common to end up with between 98.80 and 100.2 after a million hands of JoB. Seems like a million hands is not really a very meaningful sample. Maybe 3 million is.

-BB

Oh, I remember that. See? This is why they need to leave the rest of vpfree on the internet even with this new vpree2 thingy going now. There's a lot of good info on here.

Thanks for posting that!

-BB

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

N0 is the better metric that you search for:
http://members.cox.net/vpfree/Bank_NO.htm

Does your NO calculations agree with this. It seem contradictory. Please explain in your own words at a level of non stat majors.

···

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

Bob wrote:
"over the next million hands or so, you're results will be very close to 99.54%"

Depends what the meaning of "very close" is, but assuming a near normal distribution, one standard deviation alone is more than $5,000 on a quarter job machine at one million hands. The more you play, the less likely are your odds of hitting an average result. That's what the math says, counter to the standard word on the street.

Agree with what? The N0 for 9/6 jacks with no cashback is variance/advantage^2 = 19.5/(-.0046^2) = 921,550 hands. But N0 is NOT the point where "your results will be very close to the expected or average results". Instead, at N0 you can expect to be ahead in a positive game or behind in a negative game about 84% of the time. 9/6 jacks without cashback or progressive is a negative game (-.46%), so you can expect to be a loser about 84% of the time after N0 hands played.

Here's a results calculator, run it for yourself and see:
http://www.lotspiech.com/poker/GamblersRuin.html

···

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

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000" <nightoftheiguana2000@> wrote:
>
> Bob wrote:
> "over the next million hands or so, you're results will be very close to 99.54%"
>
> Depends what the meaning of "very close" is, but assuming a near normal distribution, one standard deviation alone is more than $5,000 on a quarter job machine at one million hands. The more you play, the less likely are your odds of hitting an average result. That's what the math says, counter to the standard word on the street.

Does your NO calculations agree with this. It seem contradictory. Please explain in your own words at a level of non stat majors.

Another point:

the point where "your results will be very close to the expected or average results".

For most definitions of "very close", this point does not exist. In fact, according to the math, the more you play, the greater is your standard deviation in dollars from the expected or average result.

$Standard Deviation = sqrt(variance x number_of_hands_played) x bet_per_hand

FPDW 5 coin quarters:

at 1 million hands, $standard deviation is about $6,000
at 5 million hands, $standard deviation is about $14,000
at 10 million hands, $standard deviation is about $20,000
at 20 million hands, $standard deviation is about $29,000
...

But, the ratio of standard deviation to coin-in goes down, does it not?

I suppose I do not understand what you are getting at.

And, I am probably getting in at the end of the thread... my apologies.

..... bl

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

Another point:

>the point where "your results will be very close to the expected or average results".

For most definitions of "very close", this point does not exist. In fact, according to the math, the more you play, the greater is your standard deviation in dollars from the expected or average result.

$Standard Deviation = sqrt(variance x number_of_hands_played) x bet_per_hand

FPDW 5 coin quarters:

at 1 million hands, $standard deviation is about $6,000 = 0.48% of coin-in
at 5 million hands, $standard deviation is about $14,000 = 0.22%
at 10 million hands, $standard deviation is about $20,000 = 0.16%
at 20 million hands, $standard deviation is about $29,000 = 0.12%
...

Well it seems you say that either you are ahead or behind, but never in the middle. Is that correct. If so, then doesn't NO depend on a distribution from way ahead to way behind with a clumping near the middle? In otherwords a distribution resembling a normal distribution and where the 2 sigma point set up the 84% point, and thus depending on a predictable distribution.

No further response from you is asked, as no one else seems to challenge that, so as usual I must be wrong.

···

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

--- In vpFREE@yahoogroups.com, "deuceswild1000" <deuceswild1000@> wrote:
>
> --- In vpFREE@yahoogroups.com, "nightoftheiguana2000" <nightoftheiguana2000@> wrote:
> >
> > Bob wrote:
> > "over the next million hands or so, you're results will be very close to 99.54%"
> >
> > Depends what the meaning of "very close" is, but assuming a near normal distribution, one standard deviation alone is more than $5,000 on a quarter job machine at one million hands. The more you play, the less likely are your odds of hitting an average result. That's what the math says, counter to the standard word on the street.
>
>
> Does your NO calculations agree with this. It seem contradictory. Please explain in your own words at a level of non stat majors.

Agree with what? The N0 for 9/6 jacks with no cashback is variance/advantage^2 = 19.5/(-.0046^2) = 921,550 hands. But N0 is NOT the point where "your results will be very close to the expected or average results". Instead, at N0 you can expect to be ahead in a positive game or behind in a negative game about 84% of the time. 9/6 jacks without cashback or progressive is a negative game (-.46%), so you can expect to be a loser about 84% of the time after N0 hands played.

Here's a results calculator, run it for yourself and see:
http://www.lotspiech.com/poker/GamblersRuin.html

Addendum: (since we cannot edit on here)

Your simulator has a fixed bankroll and a fixed goal.

I was under the impression that inorder for NO to apply that the bankroll was infinite and no quit point, thus giving a unimodal distribution centered near the expected value.

···

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

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000" <nightoftheiguana2000@> wrote:
>
> --- In vpFREE@yahoogroups.com, "deuceswild1000" <deuceswild1000@> wrote:
> >
> > --- In vpFREE@yahoogroups.com, "nightoftheiguana2000" <nightoftheiguana2000@> wrote:
> > >
> > > Bob wrote:
> > > "over the next million hands or so, you're results will be very close to 99.54%"
> > >
> > > Depends what the meaning of "very close" is, but assuming a near normal distribution, one standard deviation alone is more than $5,000 on a quarter job machine at one million hands. The more you play, the less likely are your odds of hitting an average result. That's what the math says, counter to the standard word on the street.
> >
> >
> > Does your NO calculations agree with this. It seem contradictory. Please explain in your own words at a level of non stat majors.
>
>
> Agree with what? The N0 for 9/6 jacks with no cashback is variance/advantage^2 = 19.5/(-.0046^2) = 921,550 hands. But N0 is NOT the point where "your results will be very close to the expected or average results". Instead, at N0 you can expect to be ahead in a positive game or behind in a negative game about 84% of the time. 9/6 jacks without cashback or progressive is a negative game (-.46%), so you can expect to be a loser about 84% of the time after N0 hands played.
>
> Here's a results calculator, run it for yourself and see:
> http://www.lotspiech.com/poker/GamblersRuin.html
>

Well it seems you say that either you are ahead or behind, but never in the middle. Is that correct. If so, then doesn't NO depend on a distribution from way ahead to way behind with a clumping near the middle? In otherwords a distribution resembling a normal distribution and where the 2 sigma point set up the 84% point, and thus depending on a predictable distribution.

No further response from you is asked, as no one else seems to challenge that, so as usual I must be wrong.

No. I wrote: "The more you play, the less likely are your odds of hitting an average result". If you want to define the "middle" as plus or minus one standard deviation, the standard deviation increases with the number of hands played, it doesn't decrease. And the probability of hitting exactly an average result decreases with the number of hands played, it doesn't increase.

You can see it visually in Jazbo's curves:
http://www.jazbo.com/videopoker/curves.html

···

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

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000" <nightoftheiguana2000@> wrote:
>
> --- In vpFREE@yahoogroups.com, "deuceswild1000" <deuceswild1000@> wrote:
> >
> > --- In vpFREE@yahoogroups.com, "nightoftheiguana2000" <nightoftheiguana2000@> wrote:
> > >
> > > Bob wrote:
> > > "over the next million hands or so, you're results will be very close to 99.54%"
> > >
> > > Depends what the meaning of "very close" is, but assuming a near normal distribution, one standard deviation alone is more than $5,000 on a quarter job machine at one million hands. The more you play, the less likely are your odds of hitting an average result. That's what the math says, counter to the standard word on the street.
> >
> >
> > Does your NO calculations agree with this. It seem contradictory. Please explain in your own words at a level of non stat majors.
>
>
> Agree with what? The N0 for 9/6 jacks with no cashback is variance/advantage^2 = 19.5/(-.0046^2) = 921,550 hands. But N0 is NOT the point where "your results will be very close to the expected or average results". Instead, at N0 you can expect to be ahead in a positive game or behind in a negative game about 84% of the time. 9/6 jacks without cashback or progressive is a negative game (-.46%), so you can expect to be a loser about 84% of the time after N0 hands played.
>
> Here's a results calculator, run it for yourself and see:
> http://www.lotspiech.com/poker/GamblersRuin.html
>

Well it seems you say that either you are ahead or behind, but never in the middle. Is that correct.

Yeah, it does, but is that ratio really useful to a gambler? What do you tell Charlie the gambler: "Hey sure, you've already lost a lot, but if you keep playing at least the ratio of your standard deviation to coin-in will decrease"? Basically, if you've lost 3% of $100,000 coin-in ($3,000), if you keep playing by $1,000,000 coin-in you may only be down by 1% ($10,000). This is good news?

···

--- In vpFREE@yahoogroups.com, "bornloser1537" <bornloser1537@...> wrote:

But, the ratio of standard deviation to coin-in goes down, does it not?

Basically, if you've lost 3% of $100,000 coin-in ($3,000), if you keep playing by $1,000,000 coin-in you may only be down by 1% ($10,000). This is good news?

In some sense, in my opinion at least, it is good news. I think that all results have to be normalized, so that you are always matching apples to apples.

My "rate of loss" seems to be going done the more I play, meaning my loss per unit time is decreasing, even though my rate of coin-in is staying the same.

In your example, I have increased my coin-in by a factor 10 and (I play 10 times longer), yet my loss only went up a factor 3.

What am I doing wrong?

I know that I am a recreational player. But, that is exactly why this sounds like good news to me.

..... bl

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

For all of us who have WinPoker, it might be awakening to do simulations of 3 or 4 million hands. I've done that a lot, with many different games. When I do "only" a million hands, the results are typically higher or lower by about 3/4ths of a percent of what it's "supposed" to be. When I do simulations of 3 million hands, it's generally a lot closer to that number, but still can be "off" a little bit.

Anyway, do you know how much 3 million hands is in reality? That's a lot of playing. Do the math. That's playing full time like a job, maybe for about 21 months. And expect to have occasional periods of 200,000 hands where no royal is hit. That will test anyone's faith in mathematics.

-BB

···

--- In vpFREE@yahoogroups.com, "bornloser1537" <bornloser1537@...> wrote:

In some sense, in my opinion at least, it is good news. I think that all results have to be normalized, so that you are always matching apples to apples.

My "rate of loss" seems to be going done the more I play, meaning my loss per unit time is decreasing, even though my rate of coin-in is staying the same.

In your example, I have increased my coin-in by a factor 10 and (I play 10 times longer), yet my loss only went up a factor 3.

What am I doing wrong?

I know that I am a recreational player. But, that is exactly why this sounds like good news to me.

..... bl

Yeah! Like 625 eight-hour days, at a comfortable rate of 600 hands/hour.

Or, 125 forty hour weeks.

More than 2 working years. What a job to have!

LOL.

..... bl

···

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

Anyway, do you know how much 3 million hands is in reality? That's a lot of playing. Do the math.

-BB

Faith in mathematics:"
Faith has nothing to do with it.

It is important to understand that a calculation of "percent payback" yields a prediction which is approached as the number of trials grows large without bound. That last is a fancy way of saying "in the long run".

In the short run, the actual consequence of a session will only approximate the long term prediction. It may be higher. It may be lower.

A group of short-term experiments will yield a distribution around the long-term prediction. The width of that distribution is determined by the statistical variance of the long-term prediction.

Two different video poker games may have somewhat similar long-term paybacks, but will exhibit quite different variances. This is sometimes called "volatility". When playing on a limited bankroll, other things being more-or-less equal, it is better to play a low-volatility game. For instance. Full Pay Deuces Wild has a higher payback than does Full Pay Jacks-or-Better, but the deuces wild game has a much higher volatility. This means your bankroll will experience wilder swings.

Having said all that, what exactly is "long term"? The only meaningful answer is that, the longer you play, the closer your results will approach the probability computation of "expected payback".

It is certainly true that, in a contest between you and the casino, the casino is hands-down playing in the long term. You are not.

One final note: It is also important to understand that, as the number of trials increases, the payback can be expected to approach the theoretical prediction, but only in relative terms. The absolute difference between the theoretical prediction, and reality, may diverge.

This is illustrated by simple coin tossing. The theoretical prediction is that in the long term, heads will equal tails. If you actually try this (or program a computer to simulate it) you will find, as the number of trials increases, that the percentage difference between what you get, and "50-50", decreases, but the absolute difference increases! This surprises some people who are statistically naïve.

But think about it. If at some point in the proceedings heads leads tails by, say, 10, then it is improbable (less than even odds) that the tails will ever catch up. This can be shown by arguing that a new sequence can be thought of as starting with heads already 10 ahead. The theoretical expectation is 50-50 FROM THAT POINT in the proceedings.

Hope this helps.

- - Norma Posy

···

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

--- In vpFREE@yahoogroups.com, "bornloser1537" <bornloser1537@> wrote:
>

>
> In some sense, in my opinion at least, it is good news. I think that all results have to be normalized, so that you are always matching apples to apples.
>
> My "rate of loss" seems to be going done the more I play, meaning my loss per unit time is decreasing, even though my rate of coin-in is staying the same.
>
> In your example, I have increased my coin-in by a factor 10 and (I play 10 times longer), yet my loss only went up a factor 3.
>
> What am I doing wrong?
>
> I know that I am a recreational player. But, that is exactly why this sounds like good news to me.
>
> ..... bl
>

For all of us who have WinPoker, it might be awakening to do simulations of 3 or 4 million hands. I've done that a lot, with many different games. When I do "only" a million hands, the results are typically higher or lower by about 3/4ths of a percent of what it's "supposed" to be. When I do simulations of 3 million hands, it's generally a lot closer to that number, but still can be "off" a little bit.

Anyway, do you know how much 3 million hands is in reality? That's a lot of playing. Do the math. That's playing full time like a job, maybe for about 21 months. And expect to have occasional periods of 200,000 hands where no royal is hit. That will test anyone's faith in mathematics.

-BB

What I see are a distributions that would predict that you should expect to end up somewhere under those curves each time the given number of hands. Further if I played enough of those given number of hands, I would expect to end up near the peak more often than out at any of the tales.

Your explanations just do not resonate with me. That is why I said that you did not need to responed, as I cannot understand your explanations. I can only conclude that I am wrong again.

FWIW, If I run the simulator that you show I end up more or less a big winner or a big loser, but that simulator fixes a bankroll and a quit point, whereas what Jazbo published assumes that a given number of hand will be playhed, regardless of the amount of coin in required. Therefore, I do not see how one explains the other, and thus how your conclusions and your promotion of NO are compatible.

Once before I got into asking a question on here and it was suggested that I forget asking it until one could talk face to face. I see that as the only means whererby we would be able to understand each other as this "finger talking" is not getting it done.

Thanks, and just drop it, unless you can think of a better way to help.

···

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

--- In vpFREE@yahoogroups.com, "deuceswild1000" <deuceswild1000@> wrote:
>
> --- In vpFREE@yahoogroups.com, "nightoftheiguana2000" <nightoftheiguana2000@> wrote:
> >
> > --- In vpFREE@yahoogroups.com, "deuceswild1000" <deuceswild1000@> wrote:
> > >
> > > --- In vpFREE@yahoogroups.com, "nightoftheiguana2000" <nightoftheiguana2000@> wrote:
> > > >
> > > > Bob wrote:
> > > > "over the next million hands or so, you're results will be very close to 99.54%"
> > > >
> > > > Depends what the meaning of "very close" is, but assuming a near normal distribution, one standard deviation alone is more than $5,000 on a quarter job machine at one million hands. The more you play, the less likely are your odds of hitting an average result. That's what the math says, counter to the standard word on the street.
> > >
> > >
> > > Does your NO calculations agree with this. It seem contradictory. Please explain in your own words at a level of non stat majors.
> >
> >
> > Agree with what? The N0 for 9/6 jacks with no cashback is variance/advantage^2 = 19.5/(-.0046^2) = 921,550 hands. But N0 is NOT the point where "your results will be very close to the expected or average results". Instead, at N0 you can expect to be ahead in a positive game or behind in a negative game about 84% of the time. 9/6 jacks without cashback or progressive is a negative game (-.46%), so you can expect to be a loser about 84% of the time after N0 hands played.
> >
> > Here's a results calculator, run it for yourself and see:
> > http://www.lotspiech.com/poker/GamblersRuin.html
> >
>
> Well it seems you say that either you are ahead or behind, but never in the middle. Is that correct.

No. I wrote: "The more you play, the less likely are your odds of hitting an average result". If you want to define the "middle" as plus or minus one standard deviation, the standard deviation increases with the number of hands played, it doesn't decrease. And the probability of hitting exactly an average result decreases with the number of hands played, it doesn't increase.

You can see it visually in Jazbo's curves:
http://www.jazbo.com/videopoker/curves.html

Yes. This is what I was trying to say in my other posting.

As the number of trials increases, notice (in the computations given below) how the Standard Deviation percentage contracts, while the absolute value of the dollar spread increases.

For my simple example of coin tossing: For a million coin tosses, the ratio of heads to tails will be very close to 50/50, but an absolute difference might be (say) something like one hundred or so, more one than the other. For several million coin tosses, the ratio will be indistinguishable from 50/50, but the absolute difference might now be several hundred.

This surprises some people, who haven't thought about it.

- - Norma Posy

···

--- In vpFREE@yahoogroups.com, "bornloser1537" <bornloser1537@...> wrote:

--- In vpFREE@yahoogroups.com, "nightoftheiguana2000" <nightoftheiguana2000@> wrote:
>
> Another point:
>
> >the point where "your results will be very close to the expected or average results".
>
> For most definitions of "very close", this point does not exist. In fact, according to the math, the more you play, the greater is your standard deviation in dollars from the expected or average result.
>
> $Standard Deviation = sqrt(variance x number_of_hands_played) x bet_per_hand
>
> FPDW 5 coin quarters:
>
> at 1 million hands, $standard deviation is about $6,000 = 0.48% of coin-in
> at 5 million hands, $standard deviation is about $14,000 = 0.22%
> at 10 million hands, $standard deviation is about $20,000 = 0.16%
> at 20 million hands, $standard deviation is about $29,000 = 0.12%
> ...
>

But, the ratio of standard deviation to coin-in goes down, does it not?

I suppose I do not understand what you are getting at.

And, I am probably getting in at the end of the thread... my apologies.

..... bl