You mean the one which opened a little window showing a list of your bets? yeah, I can't find it either. I actually use the mobile version on BV now, it's lighter and has 6 separate windows (use arrow keys to navigate). The history window there doesn't show the individual bets, but it does show a lot of stats -

Thanks, Teo.

I wrote a similar program a while ago to generate these kinds of stats. There's a classic problem in probability called the "Coupon Collector's Problem":

Given N coupons, how many draws do you have to make before you've drawn each coupon at least once?

Obviously, in a roulette context, the coupons are spins, but you could apply it to streets, double-streets etc, just replace 37 with 12, 6 or whatever.

The program can find the average number of trials in cases where not all coupons are collected, and also the probability that at least X coupons will be collected within Y trials. e.g. The "law of the third" says that roughly 24 unique numbers will hit in 37 trials, but what is the probability that at least 24 numbers will hit in 37 trials?

The program input is:

Quotecoupons 24 37 37

output:

QuoteAverage number of trials needed to collect
24 coupons from a set of 37 coupons = 38

Maximum number of trials needed was 69
Minimum number of trials needed was 24

Probability of collecting at least
24 coupons in 37 trials = 51.489 %

Or, what is the probability that in 20 spins, at least 8 streets will have shown up?

Input:

Quotecoupons 8 12 20

Output:

QuoteAverage number of trials needed to collect
8 coupons from a set of 12 coupons = 13

Maximum number of trials needed was 35
Minimum number of trials needed was 8

Probability of collecting at least
8 coupons in 20 trials = 98.77 %

This program was purely for my own use, so it only runs on a Linux command line, but if anyone's interested, here's the code (it would be quite easy to re-write in the language of your choice):

'-----------------------------------------------------------
' Given n coupons, how many coupons do you expect you need
' to draw with replacement before having drawn each coupon
' at least once?
' http://en.wikipedia.org/wiki/Coupon_collector%27s_problem
' The program finds the average number of trials in cases
' where not all coupons are collected, also the maximum
' and minimum number of trials, and the probability that
' at least m coupons will be collected within v trials.
'-----------------------------------------------------------
'xfce4-terminal -H --geometry 90x30 -e @
OPTION BASE 1

'can take a long time for large N.
CONST RUNS = 100000

SPLIT ARGUMENT$ BY " " TO arg$ SIZE dim
IF dim < 4 THEN
   PRINT "Usage: coupons <m> <n> <v>"
   PRINT "m: Number of coupons to be collected (m <= n)"
   PRINT "n: Number of coupons in the set."
   PRINT "v: Number of trials within which the probability"
   PRINT "is P that all coupons are collected."
   END
ENDIF

M = VAL(arg$[2])
N = VAL(arg$[3])
V = VAL(arg$[4])

GLOBAL coupons TYPE NUMBER ARRAY N
DECLARE t[RUNS]

trials = 0
sum = 0
class = 0
max_trials = 0
min_trials = 99999

FOR i = 1 TO RUNS
   REPEAT
      x = RANDOM(N) + 1
      INCR trials
      INCR coupons[x]
      drawn = 0
     
      FOR j = 1 TO N
         IF coupons[j] > 0 THEN INCR drawn
      NEXT j
     
      IF drawn = M OR drawn = N THEN
         t[i] = trials
         INCR sum, trials
         
         IF trials <= V THEN INCR class
         
         FOR k = 1 TO N
            coupons[k] = 0
         NEXT k
         
         ' get max & min
         IF trials > max_trials THEN
            max_trials = trials
         END IF
         IF trials < min_trials THEN
            min_trials = trials
         END IF
         
         trials = 0
      END IF
   UNTIL drawn = M OR drawn = N
NEXT i

p# = class/RUNS

PRINT
PRINT "Average number of trials needed to collect "
PRINT M, " coupons from a set of ", N, " coupons = ", sum/RUNS + 1
PRINT
PRINT "Maximum number of trials needed was ", max_trials
PRINT "Minimum number of trials needed was ", min_trials
PRINT
IF V NE 0 THEN
   PRINT "Probability of collecting at least "
   PRINT M, " coupons in ", V, " trials = ", 100 * p#, " %\n"
END IF


And I've attached the binary, in case there are any Linux users here.
[attachurl=1]

Teo, I wasn't trying to hijack your thread or steal your thunder, so feel free to delete if you like! (I just thought it seemed like an on-topic contribution).

Teo,

You do know that thread starters can remove other member's posts, right? 

Quote from: Albalaha on December 13, 2014, 07:57:59 AM
Is it a research? Even the dumbest members on forums know "law of third". You can not use it for any gain, whatsoever. It is as useless info as Martingale. This reminds me of an analysis written by someone in rf.cc that in 37 spins, there will be atleast two numbers repeating from one dozen. None could tell, what to do with this revelation.
and I am not trying to close the door but open eyes. I would love to be proved wrong.

You could say the same for any probability or statistic, including those which you supposedly use in your own holy grail.  And a Martingale is not "info" but a progression. If you're going to design a progression, it should ideally be based on probabilities and statistics.

I have to echo XXVV - why so negative?

A raw statistic is just the beginning, that's all. As sqzbox points out, you have to use your creativilty in order to apply it, and yes, there are endless possibilities.

By the way, the "law of the third" also applies to other bets, so on average:

In 18 spins, 12 splits will have hit.
in 12 spins 8 streets will have hit.
In 6 spins, 4 double streets will have hit.

Al,

Boy, you have some nerve. The point of my post was not simply to be a joke, but to show that anyone can upload a great looking graph and boast that they have beaten the game over millions of spins. It proves nothing. And when that someone has a track record of selling systems which don't work, and also has an email in their signature, suggesting that others can "earn", I think one is justified, to say the least, in being a little suspicious.

Also, I'd like to remind you of rule 7:

Quote#7  Since this is a space for posting and analyzing bets for making conclusions: No bloviating, no hinting-only.

And FYI, I did not change my name to Slacker. I re-registered at this forum after a break, and "Bayes" was no longer available. It was only later that it was suggested that I change it back to Bayes.

Al, you just don't get it. We only have your word for it. Where is the mathematical proof or the rules of the system so that it can be independently verified?

You say "I have done it, now your turn". and in your signature EARNsumit@gmail.com.

But you say you no longer sell systems.

Sorry, it doesn't add up.

Quote from: Albalaha on December 11, 2014, 04:55:15 PM
Once you told me that there is a virtual limit of it and now denying it. If you believe that randomness can deliver any kind of variance and till any stretch, you are mistaken.

I'm not sure what you mean by virtual limit. There are limits but they cannot effectively be exploited because they are unpredictable. Keep in mind that I'm talking about a mathematical proof here, not empirical results. It may seem academic to you, but there is a big difference - the casinos rely on it.

Turbogenius posted more than one million-spin system on Gambler's Glen years ago, and I've produced a couple myself, but they are unplayable and generate so little profit that they're not worth the trouble.

Yes, believe it or not the win over millions of spins was a coincidence. Post the rules of your system; I'll write a simulation and prove it to you. 

Hi Gr8player,

Thanks for your input. I'm actually in broad agreement with you, although I have to add that consistent winning at any negative expectation game lies on a knife edge, and the devil is in the detail.

@ Al,

The title of this thread is "but why?".  The fact that you've "beaten" several million spins/hands doesn't constitute a proof in the mathematical sense, unless you can answer your own question - why?

A mathematical proof, once proved, cannot be "unproved". So a theorem such as Pythagoras' theorem, which states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides, is true for every right-angled triangle, for eternity. The proof of negative expectation is similarly true for every system, for eternity. So what grail seekers are looking for is a right-angled triangle in which the square on the hypotenuse is not equal to the sum of the squares on the other two sides - a fruitless and illogical task.

On the other hand, it is not fruitless and illogical to search for triangles in which the square on the hypotenuse is not equal to the sum of the squares on the other two sides and which are not right-angled (analogous to advantage play).

I'm not saying this to be "negative" or a naysayer, but only because I think many forum members really do believe that if only they search hard enough and long enough, they will find some chink in the armor of roulette or baccarat, some asymmetry or "limit" of randomness which they can exploit.

It might seem paradoxical, but I truly believe that the reason I've been able to stay ahead of the game as long as I have is that I know - and I mean really know - that outcomes are completely unpredictable. Maybe it's necessary to search for a  long time looking for the system in order to realize this, so perhaps the effort is worthwhile after all.

Computer simulation is another way of getting to the mathematical proof that no system can win, without the actual maths (which may well be too complex). So although your systems may have "done well" over a few million spins, they will not continue that way. And besides, I've seen a few million-spin winners in my time, and they are all pretty much unplayable, even with a bot.

Without a logical reason why your system has won, the millions of spins don't count for much, I'm afraid. Do you have one?

Quote from: Albalaha on December 10, 2014, 04:05:00 PM
Will someone dare to include the reasoning or logic to play this or that?

I doubt it, because there is no valid reason or logic to play any system. I think most people realize this, even if they're totally ignorant of the mathematical details, because if that weren't the case, the casinos would not exist.

All systems, no matter how complex or ingenious, when you work through the maths, come out negative. It cannot be otherwise.

If you assume such-and-such an event won't happen, then you can back up your system with mathematical "facts", but the problem is, you can never make such an assumption.

The law of large numbers is a mathematical theorem which says that as you take more trials, the average value approaches the expectation. So betting one number, the "theoretical" expectation is 1/37 or 0.027, and the LLN says that after say 100 spins, the average might only be 0.015, but after 1000 spins, it will be closer to 0.027, and after 10,000 spins, it will be closer still.

The "law of averages" isn't a theorem, but more of a fallacy, and you could say it's akin to the gambler's fallacy. i.e., a gambler might think that "red has hit 10 times in a row, therefore, by the law of averages, black is now more likely to hit". Basically, it's saying that what is true in "the long term" is also true in the short term, which isn't the case.

Thanks esoito, it looks like a good read. There are very few books on Bayesian Stats which aren't bristling with advanced maths, so this is quite a find. Of course, you have to know some Python, but I think most would find it a more more palatable learning curve than the stuff currently out there.

If anyone want a quick overview of the topic from an elementary mathematical perspective, go here.

Sputnik,

Basically, yes. The formula (Laplace's Rule of Succession) is validation for those who like to play trends. 

But it's not easy to create a good trending system because you can get caught by the "whip-saw" effect, which is why I think it's better to work with rolling frames which target different histories. Events which are dominant in the short term can fade away in the middle or long-term, and conversely, events can be dominant long-term but not particularly so in the short term, so this technique aims to keep you locked on to all "hot" events, whether they are occurring in the short, medium, or long-term.

You might remember I posted this formula on betforum.cc, just before it disappeared. I had a couple of emails from members saying they were getting good results using it, which prompted me to investigate it more thoroughly.

Hi BEAT-THE-WHEEL,

Quote from: BEAT-THE-WHEEL on November 22, 2014, 03:14:23 AM
But I think u mean, as the standard "should" be, is, at last, they will "equilibrium, minus the pesky green", and our method,whatever it is,  should take advantages of these 'known' fact?

No, that's completely wrong.  I was just making the point that the formula will, in the long run, reflect the long term results as given by the standard formula for the probability of red/black, which is 18/37.

The formula says to bet on the event which has the highest probability, which is usually the event which is the "hottest".

The procedure is this:

1. Identify your events. They must be mutually exclusive and exhaustive, meaning that only one of the events can occur and one must occur. So all the roulette standard roulette bets, streets, dozens, numbers, but also compound EC bets, for example, in baccarat there are 8 different permutations in 3 hands:

BBB
BBP
BPB
BPP
PBB
PBP
PPB
PPP

In any 3 hands, only one of these can occur and one must occur.

2. Get some data and use the formula to calculate the probability of each event. The event with the highest probability is your bet. That's it.

In practice, it's better to define some "event windows" for example, short term, medium term, longer term over which to calculate the probabilities. e.g. suppose you were betting single numbers, you could look at the last 25, 50, and 75 spins on a rolling basis. On each spin, calculate the probabilities for all numbers within these spin frames. Usually you will get some of the same numbers in each frame which have the highest probabilities. As you get more spins, numbers will drop out of the frames and be replace by new "hot" numbers.

Needless to say, there is a lot of tracking and sorting to be done, so this method isn't really suitable for B & M play; you'll need a spreadsheet, software tracker or bot. I'll post a detailed example later.

[math]p(A_i|n_i) = \frac{n_i+1}{n+m}[/math]

where

[math]p(A_i|n_i)[/math]

means "the probability that the next event will be in category i, given that n such events have occurred.

[math]n_i [/math]

is the number of times each category was observed.

[math]m[/math]

is  the number of categories, and

[math]n [/math]

is the total number of observations.

When nothing has been observed,
n_i = 0, and n = 0, so to use an example from roulette, there are 37 "categories" (numbers that could hit) so m = 37 and the probability of say the #13 hitting is

[math]P(13|0) = \frac{0+1}{0+37} = \frac{1}{37}[/math]

which is what you would expect. And obviously it would be the same for any other number. But when you have some data, the probabilities change. For example, if you have observed 50 spins and #13 has hit 3 times, the probability that it will hit on the next spin is:

[math]P(13|3) = \frac{3+1}{50+37} = \frac{4}{87} = 0.046 [/math]

It's important that the categories are mutually exclusive and exhaustive, i.e., that only one event can occur and that one of the events must occur. For example, the categories red and even are not mutually exclusive because the event could be red and even, and the events red and black, although mutually exclusive, are not exhaustive because the event zero has not been included.

If the event doesn't make any difference in terms of payoffs, it can be ignored. e.g. the tie bet in baccarat, so there would be just two categories in this case: P & B.

An important assumption for this formula is that nothing regarding the cause of the events is known about. Of course, in casino games we do know in a sense what causes the outcomes, but not in any detailed way. So most of the time in roulette we don't know why the ball falls into a particular number, because the train of events leading up to that outcome is too complex, although in principle it could be known, given enough information regarding the "initial conditions" (ball speed, wheel characteristics, ball type, environmental conditions, etc).

So if we do have some evidence that the wheel is biased, say, then we ought to take this into account, but we cannot use the above formula because it will give misleading results (the full machinery of Bayesian statistics is required, which is much more complex). So this formula only applies when we have data in terms of the frequencies of the events, nothing more - as far as we're concerned, the events are "random" and all we have are the frequency counts.

If you haven't realised yet, the formula is basically a trending system. It always gives those events which are "hotter" a higher probability. Obviously, this is controversial in regard to casino games, which are designed to have a fixed probability at all times. But are we justified in going along with this? even if we are, even if it's reasonable to conclude that casino games are deliberately set up to be as "fair" as possible (meaning no bias), variance is undeniable. So even if "in the long run" probabilities do converge to that predicted by the standard theory, the conventional probability formulas are of no help. What's needed is a way of tracking the short-term fluctuations (biases, if you prefer).

In the long run, the formula does predict what the other formulas say will happen in the long term. For example, if after 1000 spins you have 31 zeros, 496 reds, and 477 blacks, m = 3, n = 1000, n_i = 477, so the probability of the next spin being black is:

[math]P(black|477) = \frac{477+1}{1000+3} = \frac{478}{1003} = 0.477[/math]

which is not far off what it "should" be.

For some more background reading, see:

https://archive.org/stream/theprinciplesof00jevoiala#page/256/mode/2up

In particular I like the passage on page 260:

[attachimg=2]

There is a more mathematical account of the formula here:

http://en.wikipedia.org/wiki/Rule_of_succession

Hi RoulettePlayer,

I find it instructive to compare bets like these to doubling up on the even chances. So what is the equivalent of your system in terms of a martingale (which it effectively is)?

Your trigger is 9 unique streets. What are the chances?

On the first spin any of the 12 could hit, and in subsequent spins it's required that previous streets are removed, since no repeats are allowed, so on the 2nd spin there are only 11 streets "available", as it were, and on the 3rd spin 10 streets, and so on. So the number of ways (permutations) that 9 unique streets can hit is

12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 = 79,833,600

The number of ways that 9 streets can arrive without any restrictions (i.e. repeats are allowed) is

12 x 12 x 12 x 12 x 12 x 12 x 12 x 12 x 12 = 12⁹ = 5,159,780,352

So the probability of 9 unique streets is 79,833,600/5,159,780,352 = 0.01547 or 1 in 64 approximately. In this case, it's easy to see that this is equivalent to a run of 7 on the EC's.

You can work out the probability of getting 12 unique streets in 12 spins in a similar way, which is 1 in 18,614, as you have posted.

In terms of an even chance run length for this probability, you can use trial and error, or logs:

For any even chance, the probability of a run of length n is

p = (1/2)ⁿ

log(p) = n x log(1/2)

n = log(p) / log(1/2) = log(1 / 18,614) / log(1/2) = -4.2698/-0.3010 = 14.184

i.e., a run of length 15 (since we need to round up).

So your system is equivalent to waiting for a trigger of 7 reds (or perhaps Bankers, in terms of Baccarat) in a row, and then doubling up, hoping that the run will not exceed 15.