Are there really 37 possible outcomes?

[GLC]

Just because my original post was akin to insanity and hopefully I got it off before too many read it, I really do like this general concept.

I like Colbster's suggestion re: Lanky's 6 point divisor.  I don't see right off the top of my head how to use it with such large bets required, but I think we do need to consider multiple wins to come out ahead.  Of course a win on the 1st or 2nd attempt would be nice.

I wonder if we can determine with a very high degree of surety how many numbers will hit in a 37 spin cycle.  For example let's say at least 24 numbers will always hit in a 37 spin cycle.  If we are tracking and reach spin 27 and only 18 numbers have hit, we know that we have 10 spins left in the cycle and can expect to get 6 wins and 4 losses.  By flat betting on those 18 numbers for the 10 spins or until we are ahead by 1 hit, we can realistically expect to come out ahead every time.  Unless we can't know for sure how many numbers always hit in 37 spins.  We're not talking about being 100% sure and the number may be so close to 37 that we can't capitalize on it.

[Turner]

George...I read it. I think "oops" was better (only pullin yer leg)
To have any idea if your 18 was of any use you would of have to of been tracking nested 37s for a while.  The trend could of been (h
it) 27,27,26,25,26,25,24,23,24,23,24,23,22,21
or 21,21,20,21,22,22,23,24,25,24,23
Its a lot of tracking with little certainty.
As I have said...number six in his 2011 era in VLS goes into great depth. You do have to tolerate endless interuptions from Herb (Caleb/snowman/xander/real)

[Colbster]

There will come a time (if this observation proves true) where the expected return turns negative if we shift the ratios.  For instance, if after 15 spins, we have 3 repeats and 9 singles, our bet is 12 units for a possible +24 (+35 - 11 losers).  If we shift the number of potential numbers left to hit down by the number of repeats, there are really only 34 possible outcomes.  We take 12/34 * 24 for our expected wins and get 8.471.  We have 22/34 * -12 for our expected losses and get -7.765, netting us 0.71.  We have better than even odds that we will be positive at the end of all this.  This would be our cue to start betting on all 12 of the units for the remainder of the 37-spin series.

Prior to this spin (which I am claiming as a repeat for the sake of this argument), we had 2 repeats and 10 singles.  With only 2 repeats, there are the potential for 35 unique numbers.  12/35 * 24 gives us 8.229 and 23/35 * -12 gives us -7.886 for a net +0.34.  We could also have played beginning with this spin and would have had a win as indicated above.

If the spin before this was also a repeat and we had 1 repeat and 11 singles to this point, our math changes to 12/36 * 24 and 24/36 * -12, or 0.  At this point, we were neutral and there was no justification for a bet.  Anything before this would have been neutral or negative prior to the repeat.   I think that with some basic tracking, we can identify the exact spin where our expectations move from negative to neutral to positive.  I'm working a spreadsheet now to test by hand, but I bet there is a simple mathematical equation that could solve the dilemma for us with less hoopla to just indicate if we are positive or negative on expectation.

[wyldegibson]

I'm very interested in this subject Colbster. The Law of Third is very steady in us hitting around 24 numbers in a 37 spin cycle. I too have been trying to figure out when to be betting in those 37 spins and when to be waiting for the best time based on the amount of repeats and amount of singles. I'm sure there is an easy mathematical way to figure this out if you are going to be assuming 24 unique numbers will be coming out but I unfortunately am no good at this kind of math myself

[Colbster]

Well, lesson learned: Always type important posts in Word and then copy them to the editor on the forum so your hard work doesn’t go down the pipes!

In response to the last post, it simplifies the question a little too far.  We cannot assume that we will have exactly 12 repeats in a cycle, just that we will average around there.  We can have as few as 4 and as many as 14 or 15 or more.  The math needs to adjust to the dynamic situation that is roulette, and I think I have managed to get it in place.

To jump straight to the point, the formula for when we have a mathematical advantage by playing those numbers which have already shown during the current 37 number cycle (I use European examples but it works for no-zero and American as well) is (36-C)/(N-S) where any result greater than 1.00 is to our advantage.  C=Total number of spaces covered by our bets, N=Number of spots on the wheel (36, 37 or 38 depending) and S=Total number of spins so far in this cycle.

As an example, on a European wheel where we have had 10 singles and 2 repeats during the 14 spins so far, our formula is (36-12)/(37-14) or 24/23.  Since this is greater than 1.00, we can now safely bet for the remaining spins of this 37-spin cycle with an expectation of advantage by betting 1 unit on each of the numbers that have shown so far and adding a chip onto each new number that shows going forward.

To further illustrate, on our first spin of a cycle, the formula is (36-1)/(37-1) or 35/36.  This is less than 1.00 and the house still has the advantage.  Later in the cycle, if we are at 8 singles and 1 repeat that have shown in the first 10 spins, we have a formula that has changed to (36-9)/(37-10) or 27/27.  This is 1.00 and we are neither at an advantage nor a disadvantage to the house.

This is the same situation we find ourselves in on a no-zero table before the first repeat.  If we start a session with 5 singles, we have (36-5)/(36-5) or 31/31.  At 1.00, we have the expected no-advantage situation that makes these tables nice to play when available.

To make this much, much simpler, just know that as soon as you have 1 repeat more than there are zeroes on the board, you have the edge for the remainder of the cycle.  A no-zero table turns to your favor after the first repeat, a European table after two and an American table after the third repeat within the cycle.  You can bet on all remaining spins in the cycle by placing 1 unit on each number that has hit and each number that shows within the cycle with a positive expectation.

Why is this the case?  After 15 spins (9 singles and 3 doubles), there are 25 spaces left unhit but only 22 chances to hit them.  The odds of any of the unhit numbers falling are now 22/25. 
Although we don’t know which 3, there are at least 3 numbers that simply CANNOT fall during the course of 37-spin cycle.

The advantage we have in such a situation can be quantified.  We have 22 chances to lose 12 units (22 * -12 = -264).  We have 12 spaces where we can win 24 units if they hit (12 * 24 = 288).  Net of 288-264, we have an expected return of 24 units which would be spread over the 22 spins, or 1.091 units expected return per spin positive.

It is important to note that this expectation only holds true for this particular spin.  The numbers change upon the next result.  If we have a loss of 12 units by having an unhit number fall, our next bet shows us with 21 chances to lose 13 units (21 * -13 = -273) and 13 chances to win 23 (13 * 23 = 299), a net expectation of 26 spread over 21 spins or 1.238 units expected positive return on the next spin.

If we had had a win on the previous spin, we now have 21 chances to lose 12 (21 * -12 = -252) and 12 chances to win 24 (12 * 24 = 288), a net of 36 spread over 21 spins or 1.714 units positive expectation.  Win or lose, the number of chances decreasing increases our positive expectation.

It is critical at this juncture to remind everyone that positive expectation does not mean that you will have a positive result at the end of every 37 spins.  Look at the casinos we go into.  They have a 2.7% advantage on us as soon as we hit the table.  Despite that advantage, the variance that is at the core of roulette allows some players to go home with much more money in their pockets at the end of the night than when they walked through the doors.  However, over the long-term, casinos know their advantage will eventually gobble up huge profits.  Likewise, we will have session where variance kicks our butts and hammers our bankroll.  Then we will get those golden sessions where repeats can’t come quick enough and we walk away from the cycle with a pile of profit.  However, in the long-term (and I can’t define what long-term is), our advantage will show us with the winning conclusion guaranteed.

Because of that large variance, this will require a hefty bankroll that can withstand big hits.  It also needs a long-term perspective and solid money management (handling money, not progressions).  On a related note, there is absolutely no need for a progression here, simply flat betting on the numbers that have shown.  However,  I won't attempt to offer suggestions on session bankroll, win goals or loss limits.

[sniper]

Hello Colbster,

Thank you very much for your interesting post.

This is something different from normal mechanical system we have seen so far.

I will work on this.

Regards and Best Wishes

[ugly bob]

Thanks for a great post Colbster,

It sounds logical! Where are the terms and conditions? 

Definately something worth looking into here.

[nottophammer]

I think that it can only be applied in that it gives credence to repeaters and not chasing those that haven't hit.  We will have numbers V,W,X,Y,and Z that won't hit as an example.  We cannot guess which of the unhit numbers those variables will ultimately be.  However, the numbers that have already hit are not variables, they are now established fact.  We know which set of numbers therefore are possible repeaters.  I'm not certain it leads to an application, it just seems to confirm my bias towards repeats in the grand scheme of things.

Hi colbster
In the above first line you talk of repeaters, well repeats are the way.
If you look in notepad (trying something Smartlive) think its on page 4 at the moment.

each post has 60 spins in groups of ten spins,
Now if you look at the first 10 spins, in the next 10 spins you usually get a repeat from the previous 10. Well each ten gets a repeat, needs a progression thou.

Think the only 10 spins that does not repeat in the next 10 is in reply 5, spins 31-40 numbers 9,7,27,18,14,36,12,14,16,30, but the missed 10 come in in 14 spins possible with a progression.

But as we know in roulette the first 10 might miss for 20spins.

[MrG]

Well, lesson learned: Always type important posts in Word and then copy them to the editor on the forum so your hard work doesn’t go down the pipes!

In response to the last post, it simplifies the question a little too far.  We cannot assume that we will have exactly 12 repeats in a cycle, just that we will average around there.  We can have as few as 4 and as many as 14 or 15 or more.  The math needs to adjust to the dynamic situation that is roulette, and I think I have managed to get it in place.

To jump straight to the point, the formula for when we have a mathematical advantage by playing those numbers which have already shown during the current 37 number cycle (I use European examples but it works for no-zero and American as well) is (36-C)/(N-S) where any result greater than 1.00 is to our advantage.  C=Total number of spaces covered by our bets, N=Number of spots on the wheel (36, 37 or 38 depending) and S=Total number of spins so far in this cycle.

As an example, on a European wheel where we have had 10 singles and 2 repeats during the 14 spins so far, our formula is (36-12)/(37-14) or 24/23.  Since this is greater than 1.00, we can now safely bet for the remaining spins of this 37-spin cycle with an expectation of advantage by betting 1 unit on each of the numbers that have shown so far and adding a chip onto each new number that shows going forward.

To further illustrate, on our first spin of a cycle, the formula is (36-1)/(37-1) or 35/36.  This is less than 1.00 and the house still has the advantage.  Later in the cycle, if we are at 8 singles and 1 repeat that have shown in the first 10 spins, we have a formula that has changed to (36-9)/(37-10) or 27/27.  This is 1.00 and we are neither at an advantage nor a disadvantage to the house.

This is the same situation we find ourselves in on a no-zero table before the first repeat.  If we start a session with 5 singles, we have (36-5)/(36-5) or 31/31.  At 1.00, we have the expected no-advantage situation that makes these tables nice to play when available.

To make this much, much simpler, just know that as soon as you have 1 repeat more than there are zeroes on the board, you have the edge for the remainder of the cycle.  A no-zero table turns to your favor after the first repeat, a European table after two and an American table after the third repeat within the cycle.  You can bet on all remaining spins in the cycle by placing 1 unit on each number that has hit and each number that shows within the cycle with a positive expectation.

Why is this the case?  After 15 spins (9 singles and 3 doubles), there are 25 spaces left unhit but only 22 chances to hit them.  The odds of any of the unhit numbers falling are now 22/25. 
Although we don’t know which 3, there are at least 3 numbers that simply CANNOT fall during the course of 37-spin cycle.

The advantage we have in such a situation can be quantified.  We have 22 chances to lose 12 units (22 * -12 = -264).  We have 12 spaces where we can win 24 units if they hit (12 * 24 = 288).  Net of 288-264, we have an expected return of 24 units which would be spread over the 22 spins, or 1.091 units expected return per spin positive.

It is important to note that this expectation only holds true for this particular spin.  The numbers change upon the next result.  If we have a loss of 12 units by having an unhit number fall, our next bet shows us with 21 chances to lose 13 units (21 * -13 = -273) and 13 chances to win 23 (13 * 23 = 299), a net expectation of 26 spread over 21 spins or 1.238 units expected positive return on the next spin.

If we had had a win on the previous spin, we now have 21 chances to lose 12 (21 * -12 = -252) and 12 chances to win 24 (12 * 24 = 288), a net of 36 spread over 21 spins or 1.714 units positive expectation.  Win or lose, the number of chances decreasing increases our positive expectation.

It is critical at this juncture to remind everyone that positive expectation does not mean that you will have a positive result at the end of every 37 spins.  Look at the casinos we go into.  They have a 2.7% advantage on us as soon as we hit the table.  Despite that advantage, the variance that is at the core of roulette allows some players to go home with much more money in their pockets at the end of the night than when they walked through the doors.  However, over the long-term, casinos know their advantage will eventually gobble up huge profits.  Likewise, we will have session where variance kicks our butts and hammers our bankroll.  Then we will get those golden sessions where repeats can’t come quick enough and we walk away from the cycle with a pile of profit.  However, in the long-term (and I can’t define what long-term is), our advantage will show us with the winning conclusion guaranteed.

Because of that large variance, this will require a hefty bankroll that can withstand big hits.  It also needs a long-term perspective and solid money management (handling money, not progressions).  On a related note, there is absolutely no need for a progression here, simply flat betting on the numbers that have shown.  However,  I won't attempt to offer suggestions on session bankroll, win goals or loss limits.

This caught my attention. So I simulated it for quite a lot of spins for no zero roulette. The cycle has 36 spins, betting starts after first repeater and the whole cycle is played. After that a new cycle begins, waiting for trigger and again whole cycle is played and so on. Tested on a few million spins. Results can be seen in attached pictures. No progression used, only flat bet.

[nottophammer]

mr G  your reply
is it regarding  repeats  .
now is it i've been out eating and drinking so looking at replys as i'm posting and i see something from colbster this is hard keys are blurry tryin to make between (36-1) and so on . what is colbster on about. personally dont care. this hard work repeats are the way ask the seagull a repeat comes in the next ten or what is its aaaaaavg

[vladir]

This is extremly interesting. Some time ago I was arround this same thinking of repeaters. I generated some billion sets of numbers in excel and analyzed for 37 spin cycles. Only 1 time I got only 5 repeaters in 37 spins, and very, very rarely only 6 repeaters.  Most of the time at least 7 repeaters appeared in a 37 spin cycle, and many times more then that.

So, taking for granted we will always get 6 repeaters in a 37 spin cycle (and crossing fingers to never have the session from hell in my lifetime - only 5 repeaters), I made a formula to calculate how much to bet to ensure a win in the end of a 37 spin cycle, based on how many repeaters had shown up and the total of plays we still had to reach the end of the cycle. Problem was bets could still get substantially large and it would require a bankroll of more then 3000 units if I'm remembered...  I think some early stop losses (like not having a repeater for the first 20 spins) could allow to overcome this, but to figure it out, It would have to be tested and analyzed... but I never got it coded to do that properly...

Anyway, I like this approach more, of betting flat after 1 hit repeater. But still not sure if its good enough...

[vladir]

Let's raise this topic from the dead.

I'll make this question in another way. How many spins can we have maximum withouth a repeat?  Since 4 is the minimum of  repeats we have in 37 spins, we will at least have 1 repeat at spin 33. But I believe even in the worst cases, it should be much sooner. Anyone has done tests on this and has any data to share? Any ideas if this can be exploited in some way?

[Azim]

Let's raise this topic from the dead.

I'll make this question in another way. How many spins can we have maximum withouth a repeat?  Since 4 is the minimum of  repeats we have in 37 spins, we will at least have 1 repeat at spin 33. But I believe even in the worst cases, it should be much sooner. Anyone has done tests on this and has any data to share? Any ideas if this can be exploited in some way?

The data you looking for is already provided by Winkel in his original GUT theory.

[Azim]

Here you go:

37 spin cycle:

24 numbers show  13 no shows.

24 Shows 10 numbers repeat 14 numbers appear once

13 numbers no show.

Spin 1 to 13 10 numbers show.

Spin 14 to 25 8 new numbers show

Spin 26 to 37 6 new numbers show.


Here it's summed up easy to understand I hope.

[Turner]

Here are some great stats from a poster called Teorulte

I thought they were relevant to Colbsters post

I ran about 500 000 cycles of 37 spins and came up with the following:

(Number on left is number of unique numbers and next number is how many times a spin cycle had exactly that many unique numbers in the test)

37 spin cycle:

15   1
16   12
17   202
18   1417
19   6020
20   18309
21   41304
22   73951
23   98780
24   102814
25   80164
26   47276
27   20934
28   6706
29   1685
30   349
31   35
32   5

74 spin cycle:

25   46
26   304
27   1898
28   7536
29   22123
30   50675
31   88059
32   114050
33   107055
34   69801
35   29634
36   7735
37   1011

111 spin cycle:

29   31
30   181
31   1407
32   7570
33   29508
34   83849
35   151889
36   155781
37   69674

148 spin cycle:

32   15
33   828
34   8540
35   52625
36   181344
37   256499

Mean and median were approximately:

24 for 37 spin cycle
32 for 74 spin cycle
35 for 111 spin cycle
37 for 148 spin cycle