VLS Roulette Forum
Resources => Reference Area => Topic started by: Tangram on November 30, 2009, 02:57:59 PM

The short answer is  "when it makes a profit". :D
Ok, but there is a simple calculation you can do on your results which can help tell you if your profits are due to luck or something more, and can indicate that more testing is needed, whether a method is "showing promise" or should be binned, and also gives a standard way of comparing the merits of methods which may be betting on different outcomes (for example, a single number system and a dozens system). Note that this is covered briefly in KonFuSed's excellent probability tutorial, but it's so important that it deserves more emphasis and a fuller explanation.
Introducing the "zscore"
The calculation generates a number called the "zscore", which tells you how far away your actual results are from what you would expect to get (on average) if you were simply betting randomly for the same number of decisions. The higher the score is, the lower the chance that your results are due to random variability, or "luck".
In the statistical literature, there are 3 zscore values which are commonly quoted. For our purposes they can be expressed as the following:
zscore of 1.0  there is a 16% chance that your results are due to luck.
zscore of 2.0  there is a 2.3% chance that your results are due to luck.
zscore of 3.0  there is a 0.13% chance that your results are due to luck.
Scores higher than 3.0 mean that it's highly unlikely that luck is involved at all. If you have a true advantage the score will (and must) continue to rise as you get more results.
To calculate the zscore for your system you need 3 pieces of data:
 The number of placed bets. It's important that these are actual bets  you don't count any "virtual" bets.
 The number of wins (or hits).
 The probability of a win on a single outcome. e.g. if you're betting 5 numbers, the probability is 5/37, if dozens it's 12/37, two columns is 24/37 etc.
Note that this measurement is only valid for bet selections  it doesn't apply to systems using a progression.
This isn't really a problem because you can still get the zscore by just counting the number of wins without taking account of the actual profit (In fact you could argue that finding the zscore is more important for those systems which do use a progression, precisely because a progression can easily give the illusion that a method has merit, whereas in all probability you have no advantage at all, and are surviving on borrowed time...). For example, if you were using a 4 step grand martingale of 1,3,6,12 on R/B then after a series of wins and losses like this:
LLLWLLWLWLLLWLLLWLWLLWWW
You would have made +16u, but the number of wins is 9 and the number of placed bets is 24.
Similarly if you were betting on 2 numbers you might get a series like this:
LLLLLLLLLLLLLLLLLLLLLLLWLLLLLLLLLLLLLLLLLWLLL
If using a progression you may well be in profit at this point by an amount depending on the progression, but for the purpose of calculating the zscore all you need to know is that the number of wins is 2, and the number of placed bets is 45.
As such, the zscore can be regarded as a "figure of merit" for bet selections. Mathematically, there is no way to win consistently using a progression alone. The bet selection is what's important. If you can win flat betting, then you can boost profits with progressions, but you can't use a progression to paper over the cracks of a poor bet selection. This happens to also be my opinion, but not everyone will agree. Regardless though, even the most hardened progression players are looking to shorten those losing runs, so it makes sense to pay at least some attention to bet selection.
Note that the score can come out negative. What does this mean? Well, it depends, because if the score is between 3.0 and +3.0 then you are within the accepted interval governed by chance. In other words, if your score is consistently between these two numbers, you can't assume that your strategy is giving you any advantage. The vast majority of systems and methods will give scores which fluctuate between these values  that's just randomness at work. It's only when your score moves outside these values (and continues to move away from one or the other of them) that you know you're on to something.
In general, a negative score means you're doing worse than average (a score of zero means your strategy is right about where it should be, according to chance). But note that a score of 5 means you're doing much worse than you should be, and this is just as significant as a score of +5. If you you're getting scores which are increasingly negative, logic suggests you should (if possible) bet the opposite strategy. :thumbsup:
[box title=Z] 4 > 3  Interval of Chance (Luck)  +3 < +4 [/box]
It's important to realize that your score needs to keep increasing as you make more bets, to be sure that you're operating "outside" randomness. Fluctuations will still occur even if you have a real edge, so it doesn't matter if the score climbs to 3.0, then drops back to 2.5 over the next session, as long as the overall trend is upwards.
Here is the formula for z:
z= (w  np)/sqrt(np(1  p)) Note: "sqrt" is the squareroot function
Don't panic, I'll go through some examples.
There are 3 variables in the formula:
 w  the number of wins in the sequence
 n  the number of placed bets (wins + losses)
 p  the probability of winning a single bet
example 1
Let's take the first sequence of W/L above (betting R/B) 
LLLWLLWLWLLLWLLLWLWLLWWW
There are 9 wins, so w = 9
The total number of placed bets is 24 (9 wins, 15 losses), so n = 24
The probability of a winning a single outcome is 18/37, so p = 18/37 = 0.4865
Plug these numbers into the formula:
z = (9  24*0.4865)/sqrt(24*0.4865*(10.4865)) = 1.093
Notice that even though the score is fairly poor, a profit of 16u was made in this sequence. The danger with progressions is they can lead you into a false sense of security, making you think you have an advantage (you're making money right?) , but the "advantage" is often illusory.
example 2
Ok, we'll take the 2nd sequence above 
LLLLLLLLLLLLLLLLLLLLLLLWLLLLLLLLLLLLLLLLLWLLL
Here there are 2 wins  w = 2
There are 45 bets, so n = 45
If the sequence comes from betting 2 numbers, then p = 2/37 = 0.0541
So z = (2  45*0.0541)/sqrt(45*0.0541*(10.0541)) = 0.286
So for this sequence, the bet selection was running a little below average.
example 3
Here is a sequence of W/L playing 2 dozens 
LWWWWWLWWLLWWWWWWLLWWWWWWWWWWWWWWWLWLLWLLWWWLWWW
There are 36 wins, so w = 36
There are 48 bets in total, n = 48
The probability is 24/37, p = 0.6486
And z = (36  48*0.6486)/sqrt(48*0.6486*(10.6486)) = 1.472
Not bad, more testing needed...
Ok, that's enough examples. One question though, what about if you're not betting on the same group (size) of numbers? suppose your method calls for a number of bets on a dozen, then maybe a couple on streets, or perhaps you start off on 24 numbers and gradually work you way down to a single number (as in the socalled "parachute" strategies), is there a way of calculating the zscore for that kind of system?
The answer is yes, but it can get a little tedious working it out, so a spreadsheet is highly recommended. Actually, it's not a bad idea to use one anyway, then you can update the zscore quickly every bet or few bets as you do the testing. By doing this you might be able to spot more easily any weaknesses in the strategy as they occur.
Suppose you make a series of 20 bets:
5 on double streets (p = 6/37 = 0.162),
5 on singles streets (p = 3/37 = 0.081),
3 on dozens (p = 12/37 = 0.324),
7 on an EC (p = 18/37 = 0.486)
To use the zscore formula, you need an average value for p, so you have to multiply each separate p by the number of bets, add up the total, and divide by the total number of bets made (20).
So,
5 x 0.162 = 0.81
5 x 0.081 = 0.405
3 x 0.324 = 0.972
7 x 0.486 = 3.402
Add them up  which gives 5.589, and 5.589/20 = 0.279. This is the average p which you use in the formula for z.
You can find more information on statistical testing here: http://www.fourmilab.ch/rpkp/experiments/statistics.html (http://www.fourmilab.ch/rpkp/experiments/statistics.html)
Note:
There are actually 2 interpretations of the zscore. In both cases it measures outcomes, but in one case the outcome represents a win/loss result and in the other it represents what the wheel actually produced. There's no conflict  for example, if you were betting Red, and the wheel generated a sequence of 10 Reds in a row, then the zscore would be the same for both your wins vs losses (10 consecutive wins), and for the outcomes generated for the wheel (10 reds in a row). This is simply a consequence of the probabilities being the same for an individual outcome  ie; the probability of a win betting on Red is the same probability of a Red occurring.
Alternatively, you might be using a bet selection which bets on Red for 5 spins, then switches to Black for 5 spins, alternating this way. So if you start a session and the first 10 outcomes generated by the wheel are:
RRRRRBBBBB
Then your zscore would be more than 3.0, but the zscore with respect to Red would be zero, because there are as many Blacks as Reds in the sequence.
It might be more helpful to think about what the zscore actually measures; it measures the difference between actual results, and what the results should be, on average, according to the normal distribution. Whether they are your results (from a particular bet selection), or results "from the wheel" is irrelevant. In that sense the zscore is a relative measure.
Another way to think about it (which might be a little less confusing) is to interpret the zscore in only one way  that of what your results would be if you were betting on the outcomes that the wheel actually produced. So an increasing zscore would mean that the outcomes generated by the wheel coincide with what you are choosing to bet on.

Here's a little App which measures your zscore:
(http://img691.imageshack.us/img691/4289/screenshotzmeter.png)
Select a number from the list on the left which corresponds to the amount of numbers you're betting on, then click Win or Loss as required. The "meter" at the bottom gives a visual indication of the flow; a red bar means the current score is lower than the last, and green means it has increased. Checking "Save Results" will cause a csv file (zcore.csv) of the scores to be saved in your current directory.

Topic stickied.
Thanks Tangram. :thumbsup:

Hello all,
I wonder who's confusing things  you or me?
SD and zscore are used for different things. I have learned that...
The SD is used to show how probable one testresult (MAYBE one from a set of testresults) is, compared to PROBABILITY
The zscore is used to show how probable one testresult FROM A SET of testresults, is compared to the OTHER RESULTS IN THE SAME SET
To get the zscore you have to have these variables:
SD: The Standard Deviation for ONE testresult  calculated as Tangram shows above, but he calls it "z"
PM: The MEAN RESULT for ALL tests in the SET of testresults
RS: The SPECIFIC RESULT for the ONE testresult, of the SET of testresults, you want the zscore for
To get the zscore you subtract PM from RS and divide the difference by the SD
This, from Wiki, is what I interpret as above:
[attachimg=#1]
Not correct?
Best regards,
Homeito Bemek
BTW  some was copied from GG...

Thanks Marv. :)
Homeito,
I wonder who's confusing things  you or me?
You are. ;D
Seriously though, there's no conflict between what you posted and my version. I wanted to keep things as clear and simple as possible. The formula you posted and the one I used are essentially the same.
The SD is used to show how probable one testresult (MAYBE one from a set of testresults) is, compared to PROBABILITY
This could be confusing, because the SD (standard deviation) isn't really this at all. It's a measure of the dispersion of a distribution. You use the SD to compute the zscore, but by itself it doesn't tell you how probable a set of results is relative to the normal distribution.

Hi Tangram,
Thanks for your reply.
So I'm wrong and that's OK.
Out of curiosity: How is the SD calculated (what's the formula) if we have
* Singlezero wheel
* 200 bets
* All bets are on 3 numbers (3/37)
* 25 bets hit (175 lost)
Thanks in advance,
Homeito Bemek

Out of curiosity: How is the SD calculated (what's the formula) if we have
You mean how is the zscore calculated? :)
The formula is given in the first post of this thread:
z= (w  np)/sqrt(np(1  p))
The bottom bit (denominator) of the fraction is the SD (sqrt(np(1  p))
Here, n = 200, p = 3/37 = 0.0811 and w = 25
so z = (25  200*0.0811)/sqrt(200*0.0811(1  0.0811)) = 2.274
So I'm wrong and that's OK.
I wouldn't say wrong, it's common to speak of "the number of standard deviations from the mean", which is what the zscore is, but people often just say (or write) "the SD". Maybe I'm being pedantic but statistics can be confusing enough without careless use of the terminology, so I wanted to make a clear distinction between them.

So I'm wrong and that's OK.
It could indeed get confusing.
When for instance bias/vb players refer to the SD of a pocket/sector, they are in fact referring to its zscore in a particular data sample. It is the correct technical term, and it's what we use to evaluate methods and establish relevance.
As for the actual "standard deviation", it simply refers to the measurement of the spread/deviation of data from the mean. This is its formula:
(http://geographyfieldwork.com/standa5.gif)
So the SD simply lets you know how much did the values deviate from the mean within a data sample, and therefore possibly give you an idea on how much you'd expect them to deviate from the mean in the future.
PS. I just noticed you replied already Tangram. :)

Marv,
To make matters worse, it says in the wiki entry for "standard deviation":
Standard deviation may serve as a measure of uncertainty. In physical science, for example, the reported standard deviation of a group of repeated measurements should give the precision of those measurements. When deciding whether measurements agree with a theoretical prediction the standard deviation of those measurements is of crucial importance: if the mean of the measurements is too far away from the prediction (with the distance measured in standard deviations), then the theory being tested probably needs to be revised.

Interesting. Now I know how one can establish a "margin of error" of a particular measurement tool/method.
Seeing as SD is the "measurement of the spread/deviation of data values from the mean", I also think it's what we would implement in (for example) measuring the volatility of a particular market in a chosen period, and stuff like that.

Tangram and Marven,
In my first post, I wrote that the SD is calculated like what Tangram called "z" in the first post.
In my second I asked you to show how the SD was calculated.
That was waaay wrong and done because I didn't notice the first part of your formula (sloppy reading  I'm sorry).
We agree that the "sqrt(np(1p))" part is the calculation of the SD.
Thanks for the info!
Homeito Bemek

Here's a spreadsheet I made for zscore calculation for the following games:
No zero roulette
Single zero roulette
Double zero roulette
Baccarat
@Victor: I couldn't attach the spreadsheet here, it says:
zscore calculator.xlsx.
You cannot upload that type of file. The only allowed extensions are pdf,doc,gif,jpg,odf,png,txt,zip,rar,xls,docx,xslx

"If Black has hit 30 times in a row there is both a possibility and a probability for it to hit again.
But maybe it is not very likely?"
Aren't these mutually exclusive statements? (Hell, I don't know what that means; it just sounded good!)
Let's try again: Don't these two statements contradict each other? Both cannot be true, can they? Because if they CAN, I'm going back to my original system which won thousands until I learned it wouldn't work!
TwoCat
Hmmmmmmm...an afterthought....
Perhaps his word "maybe" is his wiggle room. Huh?

@Victor: I couldn't attach the spreadsheet here, it says:
Hi marv,
Please try again, as you know files are better hosted at our server in the long run since those file hosters delete and we don't.
Thanks mate, it was a misspell of the xlsx extension.

It's working. Thanks Victor. :thumbsup:

"How to know when you have a good strategy." >>> Good question. This is how I gage it and YES, it might sound a bit odd. During play (assuming I'm ahead) I actually pay off a bill in my head or a partial bill depending on when I need to leave or I'll set a goal of how much the bill is.
EXAMPLE: Lets say I owe $1,200 for 'whatever'. I get to that amount (net) and leave, BILL PAID! Thats how I know I have a good strategy. NOT the neat, cool chips in front of me but how can I BETTER MYSELF once I have LEFT the casino...doing what with the net amount? Ken

Hi all this is so great, thank you Tangram @ all.
LS

[box title=Z] 4 > 3  Interval of Chance (Luck)  +3 < +4 [/box]
I just had to post this chart again :)
LS

How do you know when you have a good strategy? When the thought crosses your mind....I think I'll post this on VLS. Then you decide the majority of the members there are not worthy of receiving it.
Spike was right. You say you'll give away your winning method. You say and swear that....until you find it!
There is one person on this forum who has done a foolish thing, although I know he is not a fool. He has given away something he should have locked away for his grandchildren. I am totally certain there are many who decide not to join his club.
See what all your hate and negativism has done to VLS?
This forum has made me sick to my G.U.T for a year now. Somebody ban me, for God's sake, and put me out of my misery!
Sam

Here's a spreadsheet I made for zscore calculation for the following games:
No zero roulette
Single zero roulette
Double zero roulette
Baccarat
@Victor: I couldn't attach the spreadsheet here, it says:
Hi Marven, your spreadsheet is nice but it equals the zscore to the SD.
For instance: 1000 trials, strait bet, 48 hits, result= +4,09
Is the same result for the zscore and SD.
The formula I use for the binomial SD is: Square root of the sample times the prob to win plus the prob to lose.
This formula gives us the value of 1 sd.
square root of 1000x(1/37)x(36/37)=5,12
The mean is 27

The formula I use for the binomial SD is: Square root of the sample times the prob to win plus the prob to lose.
This formula gives us the value of 1 sd.
square root of 1000x(1/37)x(36/37)=5,12
The mean is 27

Thank you very much Albertojonas
A mistake in the formula but, the result was correct.

(http://vlsroulette.com/index.php?action=dlattach;topic=13193.0;attach=7482;image)

(http://i499.photobucket.com/albums/rr351/skakus/Stinkerbell.jpg)

.. and I told you, my feet are perfectly sanitized. Have a closer look:
(http://www.foreverlookingood.com/wpcontent/uploads/2010/11/prettytoenails.jpg)

.. and I told you, my feet are perfectly sanitized. Have a closer look:
(http://www.foreverlookingood.com/wpcontent/uploads/2010/11/prettytoenails.jpg)
You have the feet of a teenaged girl.

My wife's foot...
(http://i499.photobucket.com/albums/rr351/skakus/roulettefoot.jpg)

is that really her foot?
I bet these are your feet:
(http://2.bp.blogspot.com/_1Q3CzPT8QTk/SXPPNQBnucI/AAAAAAAAFcQ/WC70SdkiHs/s400/penis.jpg)

LOL!
Yes they're my feet. :lol:

My wife's foot...
(http://i499.photobucket.com/albums/rr351/skakus/roulettefoot.jpg)
Was she born with the disfiguring birthmark?

Rofl Steve.