The sweet spot

[TwoUp]

One the things we have to do as a player is decide what events we are going to lose on vs the drawdown and profit potential.

Where does a smart bettor play where making  a profit is reasonably stable and the downside is nicely constrained relative to the level of profit?

The answer needs to be that 3 to 5 profitable sessions more than pay for a losing session and that losing sessions are infrequent (you need a  failure rate less than 1 in 50 for bankroll compounding to be effective).

There has been plenty of discussion about the law of the third (LOTT) which I have explained elsewhere on this forum how it comes about with some very simple math.

Many have sought to exploit the LOTR statistical characteristics of 23 to 24 uniques and 13 repeats / 13 unhits in a 37 spin cycle. This is a very stable statistic, and certainly worth exploring, if only to get a more stable result. Stability can be compounded, volatility cannot and in fact it ensures our ruin.

If we are looking for a statistical characteristic that is quite stable from cycle to cycle then the LOTT is a good place to start and firmly anchored in math, not mysticism or wishful thinking.

Further analysis shows that every spin cycle we expect only 13 to 14 of the previous 24 uniques to  be part of the new cycle and 9 numbers to changeover, meaning 9 of the previous 13 to 14 unhits will now be hit and 9 of the 24 of the previous cycle will now go 'cold' and be part of the 2nd cycles 13 to 14 unhits.


This is happening all the time, not just on the spins you decide to call a cycle.

The interesting thing is that nine comes up quite a bit, as we expect a single repeat within the previous 9 numbers on average, and 9 new numbers per cycle.

When we start looking at cycles the wheel is quite constrained and it is rare to see the averages deviate much from 24 unique numbers per cycle, it may go as high as 28, but given on average it takes 155 spins to witness all numbers (and often many more as numbers do go missing for hundreds of apins), and that even after 74 spins the wheel only pushes as far as 34 unique numbers, those last few numbers can take a significant number of spins to close out, often many hundreds of spins. This is interesting from a house edge perspective which is based on an extra  pocket, when right here from simple observation we have 3 missing for many cycles.

Fundamentals are important, the worst the wheel can do is 37 uniques in 37 spins which is a 1 in 766 trillion event (1 in 766,879,127,067,900).

Getting 30 uniques in 30 spins is a 1 in 40 million event (1 in 40,714,151).

Getting 24 uniques in 24 spins is a 1 in 19,605 event.

Getting 18 uniques in 18 spins is a 1 in 150 event.

Getting 9 uniques in 9 spins is approximately a 1 in 3 event (1 in 2.89).

As we can see the difference in difficulty for the wheel to achieve more and more uniques fades to practical impossibility.

We also see that 24 uniques by spin 24 is still right up there in difficulty for the wheel to achieve, as it climbs from 1 in 150 event for 18 uniques in 18 spins to a 1 in 20k event for 24 uniques in 24 spins. So we expect at least one repeat the 19,604 times we play and to miss once.

We also see that we don't have to wait long for a betting opportunity entering after witnessing 9 uniques, whilst the probability of getting a repeat  is skyrocketing and the outlay in terms of numbers bet is modest.

So now we have looked at what the math dictates over a cycle on average, quite clearly there is an optimal and limited range to bet as we need to control our drawdown.

Also keep in mind that every spin is part of 37 other cycles, it is simultaneously the start of a new cycle, the end of a previous cycle and part of 35 cycles not yet completed, there is nothing special about the cycle you define when you walk up to the table or decide to bet. This is good as we can enter any time the trigger condition is met.

The math says by spin 9 we expect a repeat and by spin 24 we expect 18 uniques and 6 repeats. The sweet spot in terms of numbers covered vs expected hits is when there has been 9 spins with at most 1 repeat in those 9 numbers. Enter on spin 10 on the last 8-9 unique numbers for 10 spins (which is typically around spin 24). Don't ever bet more than 18 numbers (which you can't if you're only betting 10 spins) and instead start over.

The payouts are favourable in this range, betting 9 numbers nets 27 units, up to 18 uniques (covering 18 numbers) the return is 18u (which should be around spin 24-28).


Your maximum possible outlay over 10 spins is 9+10+11+12+13+14+15+16+17+18 = 135 units

It will be less with any repeats present.

You are expected to get 6 hits within the first 24 spins of a cycle (18 uniques, 6 repeats is 24 spins)  but you will still be ahead if you only get 4 hits. 6 hits provides 81 units of profit, so even with an outlier that gets not a single hit, you can recover with 27 units of profit in hand in just 2 coups.

The advantage of this approach is the relative stability in the statical characteristics. Enter when you observe any window of 9 non repeats, which by definition means if you just lost the last nine bets on the back of a previous window of 9 uniques, then you have a situation where the wheel has done 18 uniques in 18 spins (which will happen about 1 in 150 attempts).

This is also a good time to enter again, the statistics are well and truly on your side as two back to back 1 in 150 events is a 1 in 22k event.

The trigger is not that important, you need to enter on spin 10 and around nine numbers to bet so you may have observed 1 repeat already when you enter. This area needs further exploration..


Even if you do get an outlier and lose the series twice back to back (1 in 22,283 event), it is still only a total net loss of 270 units which is more than recovered with 54 units of profit in just 4 coups, given the average expectation is to have won 6 repeats by the time you are at spin 24.

This is playing flat, no progressions, just accept the losses and let the statistics play out.

There are many variations and adjustments that are possible and many finer points to make a reliable system..

Wishing you all a Happy and prosperous New Year!

[Martingale]

https://www.roulettelife.com/index.php/topic,3656.0.html

Hello everyone. I was interested in this topic. Especially after the message of Mr. Kramskoy, where he conducted testing. Unfortunately, I can't write on that forum, because they can't confirm my registration, nonsense.. So I would ask Mr. Kramsky personally. I suggest discussing this topic here. First of all, I didn't understand how Kramski plays. I can assume that he waits for 14 numbers without repeating, then starts betting on these 14 numbers on the 15th spin and closes a new number for 9 spins with each spin. Moreover, it is unclear why it is 9 spins, and not 10, because we are playing in a 24-spin cycle. For example. We got 14 numbers without a repeat: 14, 26, 20, 30, 21, 6, 18, 22, 13, 28, 23, 25, 35, 10. We're starting the game. And on the 15th spin we bet on these 14 numbers, $ 1 for each number, only $ 14. We will get number 3. Now we close our 14 numbers + number 3 again. In total, on the 16th spin, our bet is $ 15, etc. And so we play 9 spins, our total bet (risk) = 14+15+16+17+18+19+20+21+22= 162$ Mr. Kramski claims with his research that playing this way we have an advantage over the casino. To be honest, I hardly believe it. He says he has conducted his test for 117,200 spins and 61,890 bets. Again, I did not understand this. What does 117,200 spins and 61,890 bets mean? According to his schedule, he made 6,139 bets, which 61,890? to wait for 14 numbers without a repeat, you need a lot of spins, this is a fairly rare event, if he means waiting for the fact that we are waiting for 14 numbers without a repeat, and then we play, then there cannot be 117,000 spins and 61,890 bets. This is complete nonsense. I say right away, I absolutely do not know English and I use a translator to communicate. I would like to know exactly how Mr. Kramski plays. Secondly, on how many spins (excluding waiting for 14 numbers without repeating) he conducted this test. If it is 6139 spins, it is not enough for objective testing. And how often the event of 14 numbers without repetitions occurs. What do you think about all this?

[Martingale]

It turns out that we play a fixed 9 spins, regardless of how many repetitions we have. The maximum number of numbers we bet on is 22 numbers. in the absence of repetitions (hits) 14+15+16+17+18+19+20+21+22
I have played 40 games and the average number of replays during 9 spins is 3.375, which to put it mildly is not very profitable. Maybe I don't understand something?? How did Kramsky get a plus in the end?

[Martingale]

TwoUp, do you understand Kramsky system? You can ask him about her?

[Martingale]

Even theoretically, from 15 to 24 spins, there should be 4 repetitions. And 4 repetitions do not give us profit. And why exactly 9 spins, and not 10? After all, Kramski is talking about the 24-spin cycle, not about the 23-spin cycle.

[TwoUp]

TwoUp, do you understand Kramsky system? You can ask him about her?

I have not been in contract with Kramsky, his last post basically said that you have to abstain / stand aside when there is a period of "negative deviation" to ensure profitability.

He correctly quoted that in 14 spins on average we expect 2 hits. I have attached my table so you can review the expected uniques, spins and repeats based on the probabilities.



 
Kramsky used an absense of 2 repeats in 14 spins as an indication to abstain/stand aside and avoid cycles that tend be low on repeats. He also mentioned cycles being overlapped.

I have not tested the validity of this idea but his profit line looked quite healthy making 15.16% profit per bet (all flat) with his test of 117,200 spins and 61,890 bets. It is a reasonable statistic which on it's face warrants further investigation as if you were using $100 units flat betting that is just shy of $1 million in profit and there was zero drawdown in his chart.


This means following Kramsky's filter you would be entering on spin 15 with at most 12 uniques and a minimum of 2 repeats betting for 9 spins and thus ending on spin 24. This is why he only bets for 9 spins.

Kramsky also mentioned that he stood aside 47% of the time using this filter so it obviously does weed out a lot of games.

The maximum drawdown with no hits whatsoever would be 12+13+14+15+16+17+18+19+20+21 = 165 units

Additionally some stats provided by Janusz on that thread for 136,032 spins, making a total 5668 games of 24 spins indicated there were only 38 games that had just 2 repeats, and only 7 with just 1 repeat and zero without any repeats.

Here is the bar chart of the stats to show the distribution of repeat counts. As you can see many games are 6 repeats and above.



So any filter that avoids betting on sequences with low repeat counts may really help if it doesn't also filter out too many profitable games.

This is the next step.

[Martingale]

You've made the situation a little clearer to me now. It turns out that we are not waiting for 14 numbers without repetition, but 12 numbers and 2 repetitions. Then we start betting. I agree with the table that you have attached, but even according to it, the average number of repetitions from 15 to 24 spins is 4. It is not profitable for us. Again, you say you need to bet on 9 spins, but: The maximum drawdown without heads will be 12+13+14+15+16+17+18+19+20+21 = 165 units. this is 10 spins, not 9.. so are we playing 9 or 10 spins?

[Martingale]

minimum of 2 repetitions

[TwoUp]

I just reread Kramskys post and it appears he only plays nine spins. His maximum drawdown would be reduced 21 units to 144 units.

However he did make a final statement:

How many numbers to play for the next 9 spins is the result of other connections.

Which could mean anything, he may only play 9 numbers, or a specific range.


I have done some preliminary analysis to attempt to identify the best balance of numbers to cover over vs repeats over a set number of spins with the aid of the binomial mass function.

It gets a bit tricky as the total numbers to bet each spin is changing and probability distributions need the probability to be fixed and not changing.

Still it does indicate that there is a peak value for the number of spins for each repeat count vs numbers covered and diminishing probabilities beyond that point.

Obviously we want the sweet spot, not too many numbers as that hurts profitability and not too few as that always reduces overall probability of success.

It is a multidimensional optimisation problem so I will need to do some further analysis before I can share anything definitive.

There are some things we can also test, including:

• should we have stopping points similar to the table I have provided where we abandon the series if we don't get X repeats by spin X +1 or +2
• or stop at the spin where we expect the following repeat to occur by on average? (logic being we have gone through 2 waypoints and still no repeat, so the average is clearly not holding on this game, so cut our losses).

[Martingale]

TwoUp, let's take your system. For example, we waited for 9 numbers without repetitions. We are starting to place bets. Do we finish either when we have a bet on 18 numbers or wait for 24 spins? Did I understand correctly? At the beginning I thought we were doing 9 spins fixed, now I understand that something is probably wrong. And our number of spins depends on the number repeats.
for example.
our 9 numbers without repetition: 0,1,4,6,8,13,19,25,27.
spin:
10.bet 9; number 7 fell out
11.bet 10; number 18 fell out
12.bet 11; number 7 fell out
13.bet 11; number 1 fell out
14.bet 11; number 10 fell out
15.bet 12; number 9 fell out
16.bet 13; number 1 fell out
17.bet 13; number 29 fell out
18.bet 14; number 23 fell out
19.bet 15; number 11 fell out
20.bet 16; number 27 fell out
21.bet 16; number 25 fell out
22.bet 16; number 17 fell out
23.bet 17; number 4 fell out
24.bet 18; number 10 fell out (should I play this spin?)

number of repetitions: 6
In this example, are we playing 24 spins? or are we playing 23 spins?
total: -184+216=+32 (if 23 spin)
total: -202+216=+14 (if 24 spin)

[Martingale]

2 example.
our 9 numbers without repetition: 6,7,17,19,20,25,29,30,33.
spin:
10. bet 9; number 13 fell out
11. bet 10; number 35 fell out
12. bet 11; number 16 fell out
13. bet 12; number 14 fell out
14. bet 13; number 31 fell out
15. bet 14; number 0 fell out
16. bet 15; number 30 fell out
17. bet 15; number 13 fell out
18. bet 15; number 34 fell out
19. bet 16; number 23 fell out
20. bet 17; number 11 fell out
21. bet 18; number 3 fell out
We played 18 numbers, so we finished the game.

number of repetitions: 2
total: -165+72=-93
is that right?

[Martingale]

3 example.
our 9 numbers without repetition: 3,11,21,27,29,31,32,35,36.
spin:
10. bet 9; number 10 fell out
11. bet 10; number 19 fell out
12. bet 11; number 14 fell out
13. bet 12; number 14 fell out
14. bet 12; number 11 fell out
15. bet 12; number 3 fell out
16. bet 12; number 32 fell out
17. bet 12; number 31 fell out
18. bet 12; number 22 fell out
19. bet 13; number 24 fell out
20. bet 14; number 26 fell out
21. bet 15; number 36 fell out
22. bet 15; number 30 fell out
23. bet 16; number 22 fell out
24. bet 16; number 24 fell out
We have reached 24 spins, we are completing the game. (Let's say we play 24 spins, 24 inclusive)

number of repetitions: 8
total: -191 + 288 = +97

[Martingale]

I do not know how the Kramsky plays. I want to find out. At the expense of your assumptions, I'm still not sure that he plays that way.

[TwoUp]

@Martingale you have played the basic approach I laid out correctly. Play 10 bets starting at spin 10.

My original post is not a system, it is an idea for developing a full system based on 6 repeats in 24 spins and the way the numbers drop on average.

I also suggested that having a repeat in the first 9 spins is still ok and based on Kramsky's results, statistically there may be validity in only betting series that conform to the expectations and avoid those that do not.

I have not had the opportunity to validate  Kramsky's results and experiment with his filter rule yet as I have been away enjoying the summer weather where I live.

I also want to investigate possible stopping rules that I listed above. The problem with many rules in general is they cut both ways as they eliminate both losers and winners and you're no better off.

The other thing is that you need many thousands of games to even begin to draw  conclusions. Kramsky bet 61,890 thousand times in 117,200 spins all flat betting so it appears something is working. While he had no drawdown below the zero line (starting with a balance of zero) there was some oscillation of 1k to 3k and then 4k to 2k (about 2k unit window) before it steadily climbed to the next tier and peaked at about 10.5k units just before finishing at 9,382 units.

Even with this many spins it only provides a moderate degree of confidence and you can still have favourable runs so I will be simulating and testing for millions of spins which is easy for me to do.

I am also assuming Kramsky may have varied the units he played at different points, he basically said how many units to play is the result of other connections.

The statistical analysis of closed cycles based on the binomial probability distribution of the sortie of numbers is fundamental to understand abstention and connection with arbitrarily defined subcycles.
In 24 spins we have "on average" 18 numbers once and 6 repetitions:
17.83 numbers 1 time;
5.0502 numbers sortied 2 times;
0.9676 numbers sortied 3 times;
0.1360 numbers sortied 4 times;
0.0147 numbers 5 times;

The above is basic binomial probability calculations and hinting that once a repeat (aka "sortie") occurs the balance of probabilty vs the cost favours not betting beyond 1 repeat and at most 2 repeats.

So you can safely drop a number once it hits without too much of a drop in probability of overall success.

This is the further analysis I was talking about to precisely identify number of spins to bet a number for vs the number of repeats we can expect to get and expect to miss out on doing so. Aiming for the sweet spot.

[Martingale]

As for Kramsky's research. What do the numbers at the bottom of the graph mean? What is it if not the number of spins?
100,000 spins is certainly not a small test for a flat bet, but it is better to test on 500,000 spins or a million, for the objectivity of the results. So that it was impossible to refer to an accident. Yes, he did not have a drawdown, but there are downturns on his chart, and if you go into such a downturn at first, you can lose the whole bank. Having a $1000 bankroll is dangerous to play $1 per bet. According to his schedule, you need to have at least $4,500 to survive the pit. I think Kramski didn't change his $1 bet during the whole game. The binomial distribution is more likely to be a different Bernoulli distribution. I have heard a lot about it, but I do not know a single person who has been able to use it to create an advantage over roulette. I would really like to listen to Kramskoy and watch the test on more spins.
I reread Kramskoy's posts, but he says you don't need to wait for situations without repetitions, for example, when numbers don't repeat for 14 spins. You don't have to do that, but it still doesn't say much about how he plays. If he appears on the forum again, please ask him more about how he plays.