I have a question that seems obvious on the face of it but I don't know how to approach what should be an easy answer.
We all know that during the course of 37 spins, there will be a given number of repeats (I have heard minimum of 4, but the actual number is irrelevant for this question unless the minimum is 0). At the end of the 37 spins, we will have 4 or 6 or 10 spaces that haven't shown. Looking back at what is now history, we could argue that those spaces were never possible options. We didn't know what the numbers were that wouldn't hit, but we don't have to if we are playing for repeats.
If the minimum is actually 4, are not our odds of getting a hit on say 9 spaces covered 9/33 rather than 9/37? It doesn't matter which 4 don't hit, they simply won't hit. It becomes a discussion of misses being any of the 24 losing spots instead of the 28 "possible" spots, as we know they won't all hit. The implications of this are fairly drastic over just a few spins. In 1 spin, it is 27.27% vs. 24.32%. However, over the course of 3 spins, it moves to 62.71% instead of 56.66%. 5 spins, our odds of hitting move to 88.52% against 75.18%.
If our "real" odds are not 9/37 and 28/37 respectively, but rather 9/33 and 24/33 in the worst case scenario of only 4 repeats, am I incorrect that there would be a positive expectation?
Oops
I know it is easy to dismiss this question. I ask it in all seriousness.
For instance, while we know that we can theoretically get 50 reds in a row, after the spins have come out, we know that they will fall into the long term statistics that act exactly how they should. We just can't anticipate where and when they will fall.
My question is - does it matter where they fall? We know that 4 (or more) won't hit. It doesn't matter which numbers won't hit, all that matters is that some won't hit. Why can't we use this to our advantage?
Sorry Colbster. My oops didn't mean that I thought your idea was and oops, it was that I made a post that after reading it I realized that it was illogical so all I could do was to delete the whole message which left a blank post. The system doesn't allow that so I added oops so I could post the deletion.
I think it is an interesting idea. Capitalizing on it is the problem.
GLC
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.
AMK started link:://:.rouletteforum.cc/index.php?topic=13544.0 (link:://:.rouletteforum.cc/index.php?topic=13544.0). It got overshadowed by one of the occasional hot threads but remains the most important thread in my mind on this forum in terms of bet selection. I have played around with the concept I laid out there with much success. It isn't anything definitive but that is my go-to brain teaser concept. I think this might be related but I don't quite know how.
Colbster, I don't believe you can use it. If for instance you have average N repeats within M spins we are talking about laws of very large numbers. To get what you want you need so many spins played it does not make any sense. You don't want to have a couple of units won after playing tens thousands of spins. This game becomes unrealistic.
Just IMHO
i checked it many times in practice, stiff patterns don't work. If you play repeaters only-- finally you will lose at the end. Even if you win for long time, at the end you will lose. There gonna be times when repeaters will go to sleep so many times you can't believe it will happen before you experience it. You will get feeling that the wheel works especially against you. That what roulette does. Any stiff patterns lose at the end.
I agree that you can't use my observation that the number of possibilities might be fewer than 37, although I do think it points towards repeats instead of sleepers. That is the only take away I can see from that avenue of discussion.
Regarding my reference to AMK's thread, if that is what you are speaking of being unusable, I strongly disagree. I point to a comment in the thread by Ken where he expresses his liking of the fact it is "event betting" - what will happen vs. what might happen or what should happen or what could happen. My post in that thread represent a definite way of getting a repeat during a series of spins. The repeat is absolutely guaranteed, the trouble comes from the progression that is needed. Money management issues plague the idea, but the premise is unassailable.
Colbster
Only a few people will be truely interested in this and I am one.
Number Six did reams on this in VLS. I read them all.
I think the problem lies in the fact that group's of 37 lie nested within each other and the amount hit slowly changes over time....hovering away from mean (24 hits in 37)and back.
To ignore 4 because statistically it is negligible is what quantum physicists do.
There are more likely events....however still rare ( like 29 hits in 37) that can be considered.
I'm with colbster and Turner here. I loved that AMK thread and did quite a bit of testing on that bet selection. You do need a filter , I only attacked for a certain amount of spins / chips as it definitely needs a stop loss and overall it was profitable. You'll see the sweet spot emerge as you test it. The only reason I didn't carry on with it was, as with all inside number systems the variance can be frustrating and large drawdowns and sitting through many spins til it becomes profitable again is just not me.
Just read AMKs post in which I feature as a mad scientist who talks too much. When they make the film I will be played by Christopher Lloyd.
I have no memory of typing any of that.
Im not sure how I make the assumption that all these observations in inside numbers will correllate all nice and even into dozens.
Its one over simplification too many...diluting an already tentative premise.
Turner
In theory your premise kills the "cold numbers will hit" theory.
Since we know that usually all 37/38 numbers will not hit in
37/38 spins chances are the unique numbers are most likely to hit.
Paradox is each spin is independent, so
there shouldn't be any bias with hitting numbers.
I agree that there is no bias hitting numbers. What I wonder is if there is a bias away from the W,X,Y and Z numbers that won't hit in a stretch. I understand that the spins are independent of the past and the future. Where I get interested is that they are only hypothetical as long as they haven't been spun. It is something like Schrodinger's Cat - each number is both hit and unhit until the spin is complete. We treat it as both, but in the end the cat is actually either alive or dead. The spaces either get a hit or don't in 37. Once we look back at a set of 37/38 spins, a specific number didn't hit and, in retrospect, could not have hit. I might be a lunatic but I think its a nod towards continuing my (our) pursuit towards repeats rather than sleepers.
Quote from: Turner on Sep 16, 02:52 AM 2014
There are more likely events....however still rare ( like 29 hits in 37) that can be considered.
There are more likely events and those help our cause. If at least 4 hit but 8 actually do, we have more opportunities to win our bets and maximize profits while lessening drawdown. It is fortunate that 4 won't happen because it makes the numbers more manageable. Like klw said, the variance and bankroll considerations can be painful as I was playing it. I never (ever, ever, ever) didn't get the hit. It was just a matter of how do we handle our BR until it inevitably comes. Playing to a profit on one hit was expensive and ran afoul of table limits. Somehow, I think there is a Lanky's 6-leveler progression or something that could be very well suited to this. That isn't my area of expertise, though, and I struggle to optimize the money part of this.
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.
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)
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.
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 ;)
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.
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
Thanks for a great post Colbster,
It sounds logical! Where are the terms and conditions? :D
Definately something worth looking into here.
Quote from: Colbster on Sep 15, 07:25 PM 2014
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.
Quote from: Colbster on Oct 06, 08:49 PM 2014
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.
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
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...
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?
Quote from: vladir on Dec 15, 10:43 AM 2014
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.
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.
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
Turner, good stats to have. Thank you.
That's why people who like to chase sleepers have always been told by the wise never bet sleepers before the 5 deviation.
But no no one wants to listen.
I am repeating this again, sorry.
My discussion with Winkel was the same, if you are smart and know and can see whats happening you start a permutation of combination , union and difference to come up with some lower number to bet.
You might not always be right, but statistics never lie, you with flat betting as Winkel said will come out ahead.
Turner I have one question for you, not that it matters the numbers would still be the same.
Did you take 37 and stop or did you keep looping as in 1-37, 2-38, 3-39 or did you do 1-37 , 38 to 74?
Using Droidman's track4 software is great for seeing how all this rule of the third stuff works over the 37 spin cycles.
Quote from: Azim on Dec 15, 05:10 PM 2014
Turner I have one question for you, not that it matters the numbers would still be the same.
Did you take 37 and stop or did you keep looping as in 1-37, 2-38, 3-39 or did you do 1-37 , 38 to 74?
Azim,
I said in bold that they arnt my stats. They are from a poster called Teorulte
The sum of the 37 stats are 499964, and its over 500K spins, so its looped
13,14 and 33,34 would show over higher spins, perhaps 2,3 million
Quote from: Turner on Dec 15, 05:26 PM 2014
Azim,
I said in bold that they arnt my stats. They are from a poster called Teorulte
The sum of the 37 stats are 499964, and its over 500K spins, so its looped
Yeah. Sorry, I realised it after I posted the question. Sorry.
Hi,
I am Teorulte, and I did the stats test for the Law of the third and they are looped, ie: 1-37, 2-38 etc
I was thinking of doing some similar testing but also with evens, and seeing if there was a possibility in cutting down numbers to bet on somehow.
I feel like there is something I am missing here, like there is a solution but I can not find it.
Quote from: LuckoftheIrish on Dec 15, 05:58 PM 2014
Hi,
I am Teorulte, and I did the stats test for the Law of the third and they are looped, ie: 1-37, 2-38 etc
I was thinking of doing some similar testing but also with evens, and seeing if there was a possibility in cutting down numbers to bet on somehow.
I feel like there is something I am missing here, like there is a solution but I can not find it.
You are correct. With the stats just provided over the last 3-4 posts. There is something out there.
All that needs to be done is put the pieces of the puzzle together.
It's not easy it's trying to understand what you want from the stats.
Bottom line is, we have to sit down and work our way back. Meaning start from the final destination and go back to the source.
What do I mean by that is you know the result as an example:
You want to go from point A to point B.
You Google it and there are 5 ways to get there.
There is no right or wrong way, its a preference of how you want to get there.
Some will choose scenic route and some will choose road full of potholes so they can get there faster.
Its all in preference as to what you want.
I think Colbster is absolutely correct with this thread. It's all about having a system that concentrates on repeaters... I reached a similar conclusion in a thought experiment a few nights ago based on Law of the Third.
Quote from: falkor on Dec 15, 06:25 PM 2014
I think Colbster is absolutely correct with this thread. It's all about having a system that concentrates on repeaters..
It's possible. Here is a kind of reverse rule of the third stat which I discovered when working with Droidman's tracker.
Take two cycles of 37 spins (74 spins) and you get roughly 24 numbers appearing 2 times or less and the other 13 numbers appearing 3 times or more. Combining a few of these different stats together may help in figuring out a point of attack.
Quote from: wiggy on Dec 15, 07:39 PM 2014
It's possible. Here is a kind of reverse rule of the third stat which I discovered when working with Droidman's tracker.
Take two cycles of 37 spins (74 spins) and you get roughly 24 numbers appearing 2 times or less and the other 13 numbers appearing 3 times or more. Combining a few of these different stats together may help in figuring out a point of attack.
Where do you get this droidman's tracker?
Vladir, I can't post an external link for the software. You can find it in the downloads section on the VLS roulette forum. Or just google 'droidman track 4'
It's intended for play with Winkel's G.U.T. Just ignore the crossings at the bottom and concentrate on how the numbers develop over 37 and 74 spins.
Ok, here is something interesting. Using excel, I randomized 100.010 "spins" and analyzed the sequences. I did it more 10 times (so a total of 1.000.000), results are similar in all. I analyzed them for the appearence of the first repeater in a sequence of 37. (see attached image).
Not sure yet of how- or if -this info could be used properly...
In 1.000.000 "spins", the maximum for a repeater to appear was at "spin" 26, or in other words, a maximum of 25 uniques before a repeater.
You wouldn't be betting the sleepers waiting for the first repeat. The best system would bet on the repeats until they appear twice or more.
So if you keep betting the streets as they appear with a progression then sooner or later the street has to repeat. Worst case scenario is that you are betting on 11 streets and the 12th street comes in (if that's possible even?).
The problem is that once you have to bet many numbers or streets simultaneously then the BR starts getting exhausted quickly.
An interesting thing I found is playing the Warriors system on Double Streets: when that system loses in a set from hell - the situation is perfect for betting individual number repeats since whilst you were playing the lines there would have been no number repeats greater than 1 (or only single appearances). Therefore a combination of systems might be the best way to go - based on Law of the Third.
Here is an idea based on what I shared above:
When we have 4 unique numbers, start betting them, flat up until 8 uniques maximum, risking only a maximum of 30 units... We will win about 47,5% of the time one of this values:
32
27
21
14
6
Then track again and keep flat beting or for example.... raise base bet by 1 unit on a loss, reduce by 1 on a win, reset when in plus.
This is a test I did in November. I tested over 1 million single zero RNG spins and got this:
2 3210
3 6251
4 9017
5 10906
6 12398
7 12852
8 12454
9 11478
10 10007
11 8619
12 6845
13 5243
14 3857
15 2730
16 1816
17 1128
18 687
19 355
20 246
21 96
22 61
23 29
24 16
25 5
26 3
27 0
28 0
29 0
30 0
31 0
32 0
33 0
34 0
35 0
36 0
The most unique numbers in a row was 25 (The 26 means it hit on the 26th spin, so 25 unique numbers in a row) and this happened only 3 times.
Thanks luck of the irish. Similar to what I got too.
Quote from: vladir on Dec 16, 05:24 PM 2014
Thanks luck of the irish. Similar to what I got too.
Thank you for posting your stats. And our results are very close ;)
Does anyone know if all splits or streets could hit without a repeat?
Quote from: falkor on Dec 18, 06:10 PM 2014
Does anyone know if all splits or streets could hit without a repeat?
Yes all splits and all streets can hit without a repeat. I tested splits a while back.
ok thanks
Quote from: LuckoftheIrish on Dec 16, 02:16 PM 2014
..
23 29
24 16
25 5
26 3
..
The most unique numbers in a row was 25 (The 26 means it hit on the 26th spin, so 25 unique numbers in a row) and this happened only 3 times.
They say order isn't important, but would be interesting to check out these sets to see how far away the repeating number was from it's initial appearance. For example, was it the most recent number (or 2nd most recent number) that hit? Or was the repeating number one of the ones that first appeared much earlier?
Quote from: falkor on Dec 19, 07:52 AM 2014
They say order isn't important, but would be interesting to check out these sets to see how far away the repeating number was from it's initial appearance.
read the most simple statistic of the binomial distribution table and you will have the answer!
Quote from: falkor on Dec 19, 07:52 AM 2014
For example, was it the most recent number (or 2nd most recent number) that hit? Or was the repeating number one of the ones that first appeared much earlier?
You have to learn a lot about the distribution
I don't wish to see tables based on probability, but actual test results based on live roulette data. You don't learn by somebody else's theory - but by your own testing and observation.
Quote from: falkor on Dec 19, 08:16 AM 2014
I don't wish to see tables based on probability, but actual test results based on live roulette data. You don't learn by somebody else's theory - but by your own testing and observation.
Do you know that results have to be the same? The only diffrence is that your datas are too less to be exact. If you find something is in front after a million spins, it has to be in the back in the next 1 million spins.And that will eat your money.
It would be much smarter to declare the appearances that came less as favorit for the next spins.
Well, according to the statisticians roulette cannot be beaten due to gambler's fallacy, yet I already proved it can be beaten by playing the waiting game, so why study their work... 1 million spins is more than enough to test a system for viable play during our lifetimes. And if my 500 BR got eaten once any typical winning system that passes such a test would guarantee several K before or after that event.
I agree with less/more vs. order, but the strategy I have in mind for numbers would win on the first repeat, so that kind of data (less/more) doesn't apply to what we are testing here unless you are tracking hundreds of spins prior, which again I haven't got time for.
Quote from: falkor on Dec 19, 09:32 AM 2014
..., yet I already proved it can be beaten by playing the waiting game,
You are seriously telling us, that waiting is the way to win at Roulette? Of course you wouldn´t lose during Waiting, that´s correct. But if you don´t bet you will win neither.
Quote from: Turner on Sep 17, 05:46 AM 2014
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)
Turner,
Did you end up testing this at all.. The nested 37's?
This has caught my attention for a while now.
I have been thinking and can't figure a way out, but it sure has merit. However it's tracking will be worse than GUT. Colbster, were you thinking on the same line as I am now:
If we track 1000 spins in reality that's 964 spin cycle of 37 (1000 - 36). (Turner is my number right?)
Unless there is a bias RNG or Wheel whichever you want to look at. There will be 24 number's hit and 10 not hit in each cycle?
The key to your thinking is to catch the 24 number's and play them to repeat up to a certain cycle?
Is that right?
Every spin occupies a unique place in each of 37 different spins (actually, it's infinite but that doesn't help us). For instance, the next spin is the first spin of a 37-spin cycle, but it is also the 12th spin of a cycle that began 11 spins ago, the 24th spin of a cycle that began 23 spins, and the last spin in a 37 spin cycle that started 36 spins ago. Thus, tracking is constant. That said, we know that a certain number of spins will not show and as repeats occur, we can quantify how many won't hit this cycle. As we approach that number, the odds can be understood to turn in our favor as there cannot possibly be 8 unique numbers hit over the next 4 spins (as an example). It is just a matter of grabbing those bets that give us positive expected returns.
I get it.
What you saying is if we had people joining in at the table after every spin. what could be your spin 50 could be someone else's 30th spin.
Which would make it a hell lot of tracking at the tables, would it not?
Can any math person correct me on this.
I an running a test against exact numbers are 51788484. They have just been downloaded from Random.org
If my programming is correct I should have a result set of 517788448.
That should give me 517788448 cycles of 37 spins.
Is that a correct number? I think it is, just by using a small example if we had 40 spins recorded and the table got closed. We would only have 4 people who would be able to complete the cycle of 37 spins?
Am I right?
Quote from: Azim on Jan 04, 09:17 PM 2015
Which would make it a hell lot of tracking at the tables, would it not?
Not to my way of seeing it. We don't have to play EVERY cycle, just be aware that we are well into cycles no matter where we enter or end our current tracking cycle. I would play until a win and then back track by deleting just those spins previously that cause the current cycle to end. To illustrate this, let's say we're tracking for a first repeat. At spin 13, we get a repeat of a number that initially occurred at spin 5. Rather than starting all over to track, I would just delete the spin 5 and all previous and my new cycle is at spin 8 of a cycle that began with the old spin 6. You have already paid house edge for that information so why would we want to forfeit that useful information?
True.
Lets look at a cycle from hell, if you going to do that.
Your first repeat could be the last of the series of that number and it will be the last time for it to hit till the next 3 cycles.
Isn't that possible, if we just recycle the 37 spins?
I must have mis-explained. I'm using a for instance of getting ANY repeat, not necessarily the one that just repeated. By removing the first appearance and the previous spins, you will have remaining the most possible available information where you have not had a repeat yet. By taking away the first appearance, you have removed the circumstances by which any repeats exist in your cycle but you already have a certain amount of info already tracked. It will save you 2 or 7 or 15 spins of waiting that way by allowing you to utilize the information you have already "paid" for, either in time or chips.
I will re-do this again.
Spin 1 to 16 no repeats.
Spin 17 we get a repeat that was from spin 5.
We are at spin 17 at the table but out cycle is now at spin 12.
Is that correct?
Yes, that is how I see it. :thumbsup:
Ok.
Here is our first repeat.
This can go up to an average of 14 numbers without a repeat. Worse case situation would be 18 - 20 number's.
The repeats in the previous cycle happened at the very start of the old cycle and now it's catching up with the new hit number's.
If you really want to do it this way, you will have to keep track of 3 -4 37spin cycles.
29 22 20 3 16 4 34 2 33 12 31 11 32 22 29 0 18 19 21 10 9 10
|
These are actual number's from an auto roulette machine.
If you missed the 29 and 22 there is 20 number's before a repeat. Unless I have missed a number.
That's one of the reasons, playing auto roulette is awesome. You not forced to make a bet to get it to spin.
You wait for a situation like this and now wait. once we have repeat starting to come. They will come for about 10 - 12 spins gap.
I was just using repeats as an example of how I saw cycles. I wasn't actually implying anything about tracking for a repeat. I'm sorry if it seemed like I was.
I have a feeling you have a merit point as long as we not just going after repeats but there is quite a bit of tracking involved.
Colbster, can I hijack your thread or do you want me to start a new one?
If you are good with it, let's try this with someone posting 30 spins. Minimum start will have to be at 30 spins and play this out till our first hit.
Have at it, Azim! ;D
Id on't want to use my own number's anyone willing to have a go with 30 numbers?
Quote from: Azim on Jan 06, 10:36 PM 2015
Id on't want to use my own number's anyone willing to have a go with 30 numbers?
Here you are:
7
2
12
7
22
10
14
2
35
18
29
24
26
1
23
19
34
36
35
22
23
33
25
19
34
18
14
6
14
36
They are from random.org, hope it doesn't matter. :-)
It doesn't matter. As long as you are honest about the results I predict.
I would normally start with $50.00 on a 0.10 cent table which would give me 500 units.
30 number's tell me that in the first 37 cycle we have 18 number's that are sleeping, 9 numbers that have hit once and 9 numbers that have hit twice and and 1 number that has hit 3 times.
The same number's from 15th spin on has a 37 cycle were the number's are 26 number's that are sleeping, 6 number's have hit once and 5 number's have hit twice and no number's have hit 3 times.
Which means if i were to play the next spin the number's I would play are : 1, 2, 10 ,12, 24,26, 29 and 7.
Playing 8 numbers brings my bankroll to 492.
When you ready. Let me know the next number.
If you like 30 numbers from60spins rng william hill inspired gaming
5
7
20
30
20
7
14
8
15
16
11
27
16
11
30
6
26
15
5
6
25
28
0
26
16
24
27
24
2
8
30 numbers from60spins rng william hill inspired gaming
This 30 cycle numbers are->This one has 20 no hit number's 5 numbers hit once 11 numbers hit twice and 1 number hit 3 times.
At the same time 20 cycle numbers are 23 no hit numbers 8 numbers hit once 6 numbers hit twice and no numbers 3 and over.
With this i would bet 4 numbers which are 5. 7. 20 and 30.
Let me have the next number when you ready.
15
30 numbers from60spins rng william hill inspired gaming
15 Last number : Bankroll is 496.
Its now current cycle is at 31 and cycle we picked of 20 is at 21 spins
spin 31: has 20 unhit 5 hit once 12 hit twice an d2 hit 3 times
spin 21: has 23 unhit 7 hit once and 7 hit twice none hit 3 times.
Number's I would bet now are 8 15 16 26 27 7 11 14 20 and 30.
If I missed I will take 1 number if I hit I will take 5 number's.
BankRoll dropped to 486.
17
Jimmie B can verify these numbers
Quote from: Azim on Jan 07, 06:17 AM 2015
It doesn't matter. As long as you are honest about the results I predict.
I would normally start with $50.00 on a 0.10 cent table which would give me 500 units.
30 number's tell me that in the first 37 cycle we have 18 number's that are sleeping, 9 numbers that have hit once and 9 numbers that have hit twice and and 1 number that has hit 3 times.
The same number's from 15th spin on has a 37 cycle were the number's are 26 number's that are sleeping, 6 number's have hit once and 5 number's have hit twice and no number's have hit 3 times.
Which means if i were to play the next spin the number's I would play are : 1, 2, 10 ,12, 24,26, 29 and 7.
Playing 8 numbers brings my bankroll to 492.
When you ready. Let me know the next number.
Sorry, I just generated those 30 numbers, not more. Of course I could generate another number right now, but it will not be from the same series. Thought this shall not matter.
I would repeat the same numbers for the next spin : 8 15 16 26 27 7 11 14 20 and 30.
If I missed I will take 1 number if I hit I will take 5 number's.
BankRoll dropped to 476.
14
29
31
6
6
29
With the hit on 14 Bankroll 512.
Next numbers to bet from the same cycle: 0 5 7 11 16 20 25 26 30
Bankroll down to 503.
4
We have just finished a full cycle of 37 and are now into the next cycle.
Balance was 503 less 10 numbers... New balance 493.
New 10 numbers: 0 2 4 5 7 17 25 29 30 31
12
Hi Azim,
NTP is out for most of the day, however, he has sent me over the remaining numbers and I can send them to you when you are ready for them.
Regards
Jim
NP.
Balance was 493 less 10 numbers... New balance 483.
New 10 numbers: 0 2 4 5 12 17 25 29 30 31
Hi Azim,
Next number form NTP's session is 18
Jim
TY.
Balance was 483 less 9 numbers... New balance 474.
New 9 numbers: 0 2 4 12 17 25 29 30 31.
....6
TY.
Balance was 474 less 9 numbers... New balance 465.
New 9 numbers: 0 2 4 12 17 25 29 30 31.
...19
TY.
Balance was 465 less 12 numbers... New balance 443.
New 12 numbers: 0 2 4 5 7 12 17 18 19 25 29 31 .
18
TY.
Balance is now 479 less 5 unit .. New balance 474.
New 4 numbers + 1 extra unit on 30 only : 7 11 20 30 .
19 again
TY.
Balance is now 474 less 5 unit .. New balance 469.
New 4 numbers + 1 extra unit on 30 only : 7 11 20 30 .
34
TY.
Balance is now 469 less 6 unit .. New balance 463 + 10 Error in reply 93 math. 473
New 5 numbers + 1 extra unit on 30 only : 30 5 7 11 20 .
25
TY.
Balance is now 473 less 7 unit .. New balance 466.
New 6 numbers + 1 extra unit on 30 only : 30 5 7 11 20 25.
35
Balance is now 466 less 8 unit .. New balance 458.
New 7 numbers + 1 extra unit on 30 only : 0 30 5 7 11 20 25.
4
Balance is now 466 less 16 unit .. New balance 450.
New 15 numbers + 1 extra unit on 30 only : 30 0 11 20 1 3 9 10 13 21 22 23 32 33 36.
16
I guess the 37 cycle isn't going to work... Or I have misunderstood from the way it was explained.
dont give up heres the rest
29,22,26,17,18,23,9,24,8,1
better to see the last of the numbers you might find a tweek for next time. would you like another 60 spins these would be global draw unless they changed their name. the previous were inspired gaming, find there rng hard ,very hard
Ok... Lets try this one more time..
Lets start with 10 number's first...
7,28,34,5,2,2,16,14,36,8
1 unit on 2.
500 less 1 unit balance 499.
32
1 unit on 2.
Balance 498:
29
Know this off topic but look at even odd reply 12 permanenzen table 3 Jimmie B
1 unit on 2.
Balance 497:
31
again off topic 82 units Jim
1 unit on 2.
Balance 496:
Jimmie once 26th # come 12 spins for 27th stakes sheet would need to wait atleast 6 before betting
25
1 unit on 2.
Balance 495:
18
1 unit on 2, 16, 17, 18 19 and 20.
Balance 489:
1
1 unit on 2, 36 0 3 2 1.
Balance 483:
23
Azim is 2 dupli
yes..
1 unit on 2. 21 25 22 24 23
Balance 477:
29
2 units on 29
1 unit on 2. 27 31 28 30
Balance 470:
25
2 units on 25
1 unit on 2. 29 23 27 26 24
Balance 472:
16
2 units on 16
1 unit on 2. 14 15 17 18 25 29
Balance 463:
20
No bet for next 4 spins...
Balance 472:
10
1
9
11
2 units on 16, 25 29 2
Balance 455:
26
same bet:
2 units on 16, 25 29 2
Balance 447:
9
3 units on 16, 25 29 2
Balance 435:
4
3 units on 16, 25 29 2
Balance 423:
8
If these are real numbers by now i would be home... Create another session.. zip it and password protect it.. and send it over...
Any one playing the session you gave out the numbers with would have been up 100. by now...
real numbers but rng
23
no need to make them up what would be the point your testing and us, I would like you to succeed.
Create another session.. zip it and password protect it.. and send it over...
haven't a clue how to do this.
William hill?
Are these number's like last one from william hill rng?
no ladbrokes rng, Global draw if they still go under that name
can give some old smartlive from aug 2013
4 units on 16, 25 29 2
Balance 407:
4
4 units on 16, 25 29 2
Balance 391:
34
think your've got 5 more spins and i think its going to get nasty
Looks like my kind of spins...
6 units on 16, 25 29 2
Balance 367:
7 units on 35 33 30 27 24 22 21 19 17 15 13 12 6 3 0
Balance 262
think your watching J B,
with had 22 gone and last 4 have been repeats, remember 20th to 26th usually in in 3 spins
30 turned out ok
J B
30 was 23rd
yeah...
6 units on 16, 25 29 2
new balance 490...
30 do you believe it
azim
you bet the 15 that had not come
I Know, I have a madness to my calculations...
If I was at a real table.. I was just shocked why no hit for so long...
I now know why I have all the bots playing, to take away my emotions from the game. Have faith in my methods and strategies.
I guess now people know why I don't gamble..
I would have doubled my bankroll by now if it was a bot playing the way the numbers were coming..
We still betting:
6 units on 16, 25 29 2
new balance 490...
reply 164 is the next number 30 repeated
7 units on 30 23 4 34
the last 4 spins were 4,34,30,30
0
did i hit any of those number's or have i lost track?
no last number zero
I am showing we gone through 36 spins is that right?
no zero is spin 35 i believe
no i've just checked zero 35th
i guess i am 1 spin ahead.. and from 4 30 34 and 23 i only hit 30 once? that's when I played the 7 units?
yes 30 hit reply 159
yes got you back to 514
27
35 and good night
Good Night.
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
So 500,000 37 spin cycles produced 102,814 sets of 24 unique. Is it okay to say 20%.
If another set of 500,000 was done, will it be 20% again? Will it always be 20%
In GUT its said at spin 13 we could see 26-9-2, well i've been at spin 13 many times seeing 24-13-0, so would you not now bet the 1X's. Look how many times 15 unique appear.
Quote from: nottophammer on Dec 11, 12:37 PM 2015
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
So 500,000 37 spin cycles produced 102,814 sets of 24 unique. Is it okay to say 20%.
If another set of 500,000 was done, will it be 20% again? Will it always be 20%
In GUT its said at spin 13 we could see 26-9-2, well i've been at spin 13 many times seeing 24-13-0, so would you not now bet the 1X's. Look how many times 15 unique appear.
Hi nottop my results:
see attachments
Does this mean 13 unhits in 37 spins is most common?
Thanks
Quote from: RouletteGhost on Dec 11, 02:02 PM 2015
Does this mean 13 unhits in 37 spins is most common?
Thanks
Yes but 14 is not far away, because the theoretical average is at 13.5 numbers
Quote from: winkel on Dec 11, 02:06 PM 2015
Yes but 14 is not far away, because the theoretical average is at 13.5 numbers
Thanks
So how do we take advantage of the repeaters.... :yawn:
Quote from: RouletteGhost on Dec 11, 02:08 PM 2015
Thanks
So how do we take advantage of the repeaters.... :yawn:
that is the wrong question!
How can we take advantage of the development of the trot of 37 spins?
So we can classify:
- The trot is slow: means many repeaters
- The trot is fast: more unhit hit than repeaters appear
- The trot is an average trot
Quote from: winkel on Dec 11, 02:24 PM 2015
that is the wrong question!
How can we take advantage of the development of the trot of 37 spins?
So we can classify:
- The trot is slow: means many repeaters
- The trot is fast: more unhit hit than repeaters appear
- The trot is an average trot
Beyond my thinking. Need a math wizard.....you? Lol
Rg
You don't have to be a maths wizard at all.
You just have to understand a few things.
The bell curve above shows 25, 24, 23 as being expected.
That's cos they fall within 1SD of the mean. 2SD less frequent but not never. The start and end of the bell curve are deviating from the mean (SD) so much its near on never going to happen
Like 32 hit and 5 don't or 15 hit and 22 don't
Thing is.....even if my mother (no disrespect mum) saw 30 hit in 37 she would say "what's wrong with that" but even she would spot all the repeats if 15 hit.
But they have the same probability
This stat sheet above and the bell curve is all you need to know really.
You have to know a bit of maths to produce those two charts but don't need any maths to look at them.
Quote from: winkel on Dec 11, 02:24 PM 2015
that is the wrong question!
How can we take advantage of the development of the trot of 37 spins?
So we can classify:
- The trot is slow: means many repeaters
- The trot is fast: more unhit hit than repeaters appear
- The trot is an average trot
Yes watch the trot, keep in mind LOTT
QuoteNow you all know when playing KTF I say use 37 #’s, why because they are all due, but not necessarily in 37 spins.
Luck of the Irish/Winkel did some work on 37 spin cycles.
Let’s just work with the 24 non-hit that come in 37 spins. Of that 24, 13 repeat, 13 don’t come.
So KTF would have us bet for those 13 remaining non-hit, if not made the win you want. At 60 spins on KTF sheets we see it average 30.5 non-hits, so 6 more non-hit in 23 spins could be possible.
Would it be better to in the next 23 spins, bet the 24 that have come in the 37 spins?
On today’s Jackpot247.com we get 26 non-hits in 37 spins, so you can see on the sheet I’ve marked the 26 on the mat.
At the 12th spin its +£40 #14, so do we stop?
I carried on to see what happens in the 23 spins using +1/-1 and not betting the #’s hit, again bets get cheaper.
Yes it is possible 37 outcomes, notto you are so close to hg, just play chaos theory together with law of the third and the magic door will open.
Mumbo
Can you shed some more light on chaos theroy, if a butterfly flaps its wings and may be the cause of a tornado,i'll need some help to open HG's door
Just some quotes from others and i tend to agree with them,
Quote from one mathematician , statistician :
***
The roulette wheel is not random. It is chaotic! Future spins are determined by past spins. Patterns therefore occur. Find them. Exploit them.
Orderly disorder exists ***
quotes from some scientists
*** the presence of chaos may actually produce ordered structures and patterns ... ***
***Patterns will appear out of the chaos. And this, in its essence, is chaos theory: finding order in the chaos. ***.
Okay on todays J247.com sheet you can see at the 37th spin theres 22 1x's.
So we bet the 22,1X's,like below, use +1/-1 +£58
- 38-22,w
39-21,w
40-20,L
41-40,w
42-19,w
Quote from: Colbster on Sep 15, 03:19 PM 2014
I have a question that seems obvious on the face of it but I don't know how to approach what should be an easy answer.
We all know that during the course of 37 spins, there will be a given number of repeats (I have heard minimum of 4, but the actual number is irrelevant for this question unless the minimum is 0). At the end of the 37 spins, we will have 4 or 6 or 10 spaces that haven't shown. Looking back at what is now history, we could argue that those spaces were never possible options. We didn't know what the numbers were that wouldn't hit, but we don't have to if we are playing for repeats.
If the minimum is actually 4, are not our odds of getting a hit on say 9 spaces covered 9/33 rather than 9/37? It doesn't matter which 4 don't hit, they simply won't hit. It becomes a discussion of misses being any of the 24 losing spots instead of the 28 "possible" spots, as we know they won't all hit. The implications of this are fairly drastic over just a few spins. In 1 spin, it is 27.27% vs. 24.32%. However, over the course of 3 spins, it moves to 62.71% instead of 56.66%. 5 spins, our odds of hitting move to 88.52% against 75.18%.
If our "real" odds are not 9/37 and 28/37 respectively, but rather 9/33 and 24/33 in the worst case scenario of only 4 repeats, am I incorrect that there would be a positive expectation?
I give colbster the thumbs up here.
Mogul
the line in red, works the same for non-hit,the more spins you wait for a particular non-hit past your known average to hit, the better the chance you'll win, of course it might hit by its known average to hit.
Now if you was to collect the non-hits where you play in the good old USA, after collecting 100 games of at least 60 spins, you would have an average to start working with.
I'd bet your dollar that you'd see for up to the 19th non-hit they would average to hit in just 2 spins, there max to hit ? but the more games you collect the max will appear.
See if you can work with this, its not been updated for months, but these averages are the same as others i have.
Quote from: Colbster on Oct 05, 07:27 PM 2014
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.[/color]
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.
The above in maroon, is when you will win, with flat bet, the only part is, what winnings you are going to take? if using £1 units, i'd stop as soon as +£24 is reached.
Don't believe, i'd suggest you test and soon as 12non-hit are left start flat betting.
Remember Nottophammer told you this :thumbsup:
Just played on MPR 25th came
35-R
34-26th
27-27th
Now this is betting non-hit, but with 30 seconds would not get on,but when live table is busy it can be done, so theres your answer, why they like to spin fast, so you cant spread the large non-hits at the start.
But like Colbster says whether inadvertently bet the last 12 non-hit.
From me flat bet works, you just have to keep your faith.
(link:://:.pichost.org/images/2016/12/25/temp_678299.png) (link:://:.pichost.org/image/eFD5) |
This should be in KTF but look at the trot, 7,+2, 12,+2, the old average so next 10 spins could see 3 0x's, countback showing how fast and look at spin 37, Luck of Irish shows 24 at 37. countback saying could see 21, so you know its fast.
After 25th, flat bet the remaining 12,0x's, made the +24 so stop.
WTF, do you reckon Celtic would have reset after 29,29, looking further, i think he'd have reset at spin 10
Quote from: Colbster on Oct 06, 08:49 PM 2014After 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.
I agree that there are not anymore enough spins for all 25 unhit numbers to hit so there will be some left unhit in the 37-spin cycle. BUT I think the problem still remains that in the remaining 22 spins we don't know WHEN will a repeater be or when a new unhit number will occur. aka we don't know in a specific spin what the outcome will be. We do know that in the end some will stay unhit but we do not know in a specific hit the outcome and HOW/in which way this sure knowledge of us for the end result will eventually form.
So what I think is that even knowing that 3 will stay unhit for a specific spin the numbers that make us lose are still 25 and not 22. Which would lead to:
We have 25 chances to lose 12 units (25 * -12 = -300). We have 12 spaces where we can win 24 units if they hit (12 * 24 = 288). Net of 288-300 = -12units. we have an expected return of -12 units which would be spread over the 22 spins, or -0.545 units expected return per spin negative.
I HOPE I'm wrong but what you guys think? :/
Quote from: nottophammer on Dec 25, 03:12 PM 2016Don't believe, i'd suggest you test and soon as 12non-hit are left start flat betting.
Remember Nottophammer told you this
So from further testing we see it should be when only 9 non-hit remain of the starting 37.
What you think guys on my reply#201? Can there really be any positive expectation per spin the way suggested? I have good results so far but it could be just luck.
PS: Wishing you a very good and productive year as well in every field of your life. And with health to you and your loved ones :)
Quote from: BellagioOwner on Jan 01, 06:31 PM 2017
What you think guys on my reply#201? Can there really be any positive expectation per spin the way suggested? I have good results so far but it could be just luck.
PS: Wishing you a very good and productive year as well in every field of your life. And with health to you and your loved ones :)
I would like to know who is right, but I'm not a math boy. But I tested it some time ago and posted the results somewhere in this thread.