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!
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?
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?
TwoUp, do you understand Kramsky system? You can ask him about her?
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.
Quote from: Martingale on Jan 07, 2023, 11:13 AMTwoUp, 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.
Screenshot_20221228_090418.jpg
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.
Screenshot_20230107_214103.jpg
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.
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?
minimum of 2 repetitions
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:
QuoteHow 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).
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)
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?
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
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.
@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.
QuoteThe 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.
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.
It would certainly be great if Kramski shared his way of playing) but don't count on it. It is necessary to collect everything bit by bit. Do you think his filter for entering the game is: entry for 15 spin with a maximum of 12 unique numbers and at least 2 repetitions? Farther. Is he guaranteed to stop betting when he reaches 24 spins? Or is he betting on 24 spins inclusive? Is 9 spins always played fixed? Perhaps Kramski also excludes the numbers that fell out twice and removes the bet from them.
judging by his words that there are only 9 spins, it turns out that he does not play 24 spins, but only 23
He did explain his approach.
Only bet on sequences that have 2 hits by spin 14, abstain from betting otherwise.
Using the binomial distribution allows us to determine the profitable window/sub cycle to bet a number for as it becomes diminishing return to keep betting on a number.
For example, one could bet on a number for hundreds of spins and as you continue to infinity the probability of seeing a hit keeps approaching but never reaches 1. But we can't do that profitably.
The binomial distribution helps answer questions like exactly how many occurances of an event can I expect in N spins.
The binomial mass function:
(https://en.m.wikipedia.org/wiki/File:Binomial_distribution_pmf.svg)
The binomial cummulative distribution function:
(https://en.m.wikipedia.org/wiki/File:Binomial_distribution_cdf.svg)
Imagine a coin flip 3 times with all possible outcomes listed in the following table:
TTT
TTH
THT
THH
HTT
HTH
HHT
HHH
The binomial distribution can give us exact results for questions about that table. Such as in 3 flips what is the probability of witnessing a head? Count the rows and we 7 out of 8 rows (7/8 is the probability).
How about how many times do we see exactly 1 head and not more than 1? Count the rows with only 1 head and we see 3/7 rows.
How about 2 or more heads? Count the rows and we see 4/8 or 1/2.
How about 2 or fewer heads? Count the rows and we see 6/8 or 3/4.
What about 1 or more heads in just two flips of a coin?
So we can cut down the table to just:
TT
TH
HT
HH
Here we see that 3/4 have at least 1 head and with 3 flips it was 7/8 so our probability has reduced from 0.875 with 3 flips to 0.75 with just 2 flips.
This may help identify the best range where we can potentially apply a cutoff as the probability advantage on average may not be worth the cost.
If we do 4 flips the probability increases to 0.9375, and 5 flips is 0.96875, 6 flips is 0.9843, 7 flips is 0.99218. (I used the binomial functions in excel to compute these).
So as you can see the gain in probability of success for each additional trial is diminishing quite a lot while the cost is constant.
There is a crossover point where it makes no sense to keep going for the added probability increase.
Obviously with roulette we have 37 outcomes at each step instead of just 2 for a coin but hopefully that explains the binomial distribution in a more intuitive fashion.
I understand what you're talking about, but we still don't know exactly how Kamsky bets. Do you have the opportunity to test the system on a large number of spins? I suggest this approach: we start betting on the 15th spin if there were a maximum of 12 unique and at least 2 duplicate numbers. But we don't bet on numbers that have been repeated! That is, if we have a situation of 12 unique numbers and 2 repetitions, then we bet on 10 numbers that have not been repeated, we do not touch the repeating numbers. We play 9 spins. We do not place a bet on numbers that are repeated, and if our number is repeated, then we withdraw the bet from it. Does this approach make sense?
maximum minus 10+11+12+13+14+15+16+17+18 = -126 (with 12 unique numbers and 2 repeats)
In 24 since 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;
there is also a reverse approach: we only bet 9 spins on repeating numbers, in the hope that the number will repeat 3 times. Since theoretically for 24 spins, one number should be repeated 3 times.
is there any other stats that can help you here two up?
Some basic rules to test:
- We expect 2 repeats and stand aside otherwise. (entry trigger)
- We expect 3 repeats by spin 17 (potential early exit trigger).
- We expect 4 repeats by spin 19 (potential midway exit trigger)
- We expect 5 repeats by spin 21 (potential late exit trigger)
- Betting plan: 12+13+14+15+16+17+18+18+18 = 141 units.
Note we do not bet the first 2 numbers that have already had a repeat and only go as far as covering 18 numbers (we will never get this high if using the exit triggers). We will also be betting less units as repeats reduce the numbers that need to be covered for the remaining spins and additionally we can also remove any number that repeats, which reduces drawdown further. I would expect maximum numbers bet per spin to be around 16 with 2 hits by spin 21, and less than this if using the exit triggers. - Payout based on 4 hits is 36×4 = 144 units, so the worst case betting plan with no exit triggers nets profit with 4 hits.
- The worst the profit could be with 4 hits and taking away repeats is 12+13+14+15+16+17+16+15+14 = 132u, leaving a profit of 12u
- The average profit with 4 hits and taking away repeats based on the averages is 12+13+14+13+14+13+14+13+14 = 120u, leaving a profit of 24u
- if we do not take away repeats (even on the first 2 we didn't bet on) we get a loss even when the averages play out as we are starting with more numbers covered, 14+15+16+16+17+17+18+18+18 = 149u, leaving a loss of 5u
- if we only take away repeats on the first 2 we didn't bet on (as per the betting plan) we get a profit when the averages play out 12+13+14+14+15+15+16+16+17 = 132u, leaving a profit of 9u
Based on the above, we do need to remove numbers, and those that have already repeated once are the obvious candidates, however this is not the only way.
A better rule might be to only bet a number for X spins max. Every spin we drop the oldest number that has not hit, so we instead cap our betting to X numbers.
If X is say 9, then we can limit the numbers bet to 9 most recent numbers, but perhaps priotising those that have occured twice (given stats show almost 1 3-peat per 24 spin cycle), and always dropping those that have appeared 3 times (given the stats show it's a bad bet).
So testing the numbers to cover for how long may be productive. If you think about it, using a window covering 9 numbers is not surprising as we expect a repeat by spin 9.
The window may not be fixed, maybe it grows from 9 to 14, which is where testing over statistically significant datasets can provide this insight.
Another mathematical method I will also explore is what the Poisson distribution indicates in terms of expected hit rates over the spin interval vs numbers covered (which provides the expected mean hit rate and variance for the Poisson distribution). The stats should ultimately conform to the math, although with varying numbers covered (changing average hit rate) the math becomes more difficult as we would need to compute a combined distribution.
Anyway that's more than enough for folks to chew on, so roll up your sleeves and test with a large data sample.
I will be setting up my test rig to do some further exploration on both rng and large historical datasets.
A correction to my post above:
Quoteif we only take away repeats on the first 2 we didn't bet on (as per the betting plan) we get a profit when the averages play out 12+13+14+14+15+15+16+16+17 = 132u, leaving a profit of 12u
Also keep in mind Kramsky had an average profit of about 15% per bet all said and done. He bet for 61,890 spins and made 9,382 units. I note that 61,890 doesn't divide by 9 bets evenly which indicates he may use exit triggers when the averages don't hold.
Even so, 9,382÷(61,890÷9) = 1.364u per bet made using the average of 9 bets per game.
The above shows an average profit per bet of 12/9 = 1.34
Whilst the other scenarios had double the profit. Accounting for the inevitable losses we may be in the ball park with the adjustments I have indicated in my post above.
The histogram I previously posted shows the bulk of games have 5 repeats or more. If the entry and exit rules weeds low hitting games we save a ton of bankroll and on the average lose less per game than the winning games.
A game can go max 12+13+14=39 unit loss at spin 17 before abandoning. If we got a hit (3 repeats total) in the first 3 bets (+36) we can continue 2 spins more to spin 19 with a balance +36-(12+13+14+14+15) = -32u debt before abandoning. If we get a 2nd hit (4 repeats total) our profit is 36×2=72u with a net profit of +4u. So we bet until spin 21, 15+16=31u. If no 3rd hit (5 repeats total) our debt is -27u, but if we get a hit our net profit is +9u minimum, so we continue until spin 23 16+17=33u. If no forth hit (6 repeats or more) our debt is 24u, but 1 hit we get +12u profit worst case. We are not removing numbers here, only not betting 2 from the first 14 spins.
Given the stats show most games have more than 5 repeats. We avoid losing a lot of units on games with low hit counts, but are open to games with higher hit counts. The filtering will remove some games that may have turned in our favour with very late repeats, but even then the profit is very marginal as you would have to bet more numbers for longer.
I have attached some stats and payouts based on the worst case (no removal of numbers other than 2 oldest non repeats from the first 14 spins):
Screenshot_20230112-132111_Excel.jpg
Note that I cannot fully account for the filtering effects, many games that have late repeats will not be bet on at all, we will miss out on but their profit is marginal/break even in general, so the effect of these is effectively washed out.
Games that have early repeats and peater out are handled by the exit triggers. There may be high repeat count games prior to spin 14, and I cannot account for these other than refining the trigger to stand aside if the repeats are more than 2 in the first 14 spins.
Where there may be an overestimation is high repeat count games that have none in the first 14 spins, which is almost a contradiction, as a high repeat count game will generally have repeats by spin 14 given only 9 spins remain. This is why the triggers may actually be valid for eliminating marginal or losing games whilst not removing too many profitable games.
This needs testing which I plan to do, so don't start risking cash until you have validated for yourself.
Apologies for my error below.
QuoteEven so, 9,382÷(61,890÷9) = 1.364u per bet made using the average of 9 bets per game.
The above shows an average profit per bet of 12/9 = 1.34
What I meant to say is Kramsky has only 1.364 units of profit
per game.
Whereas the crude profit per bet based on estimates against the betting plan as per my table in the previous post is 25u. The filtering effects will alter this of course so I do expect the actual profit per game to be lower due to a percentage of profitable 6, 6 & 8 repeat games being filtered out.
Kramsky said he eliminates.about 44% of games. If I assume we eliminate 44% of the winning games only
and still keep all the losing games (games with 5 or less repeats) we still have 10.64u per game vs Kramsky's number 1.34u per game, or a 118% profit per bet made vs Kramsky's 15%.
Even taking 2/3 (66%) of the profitable games off the table and keeping 100% of all the losing ones we still have 2.67u per game or 30% return per bet.
Even inverting all the profit from 6 repeat games and treating it as a loss and eliminating 44% of the profitable 7 and above repeat games, the average profit per game is still 6.13u per game or 68% per bet. What is clearly evident is that higher repeat games have a lot of profit and easily compensate lower repeat games.
This all points to testing, testing and more testing as no doubt betting less numbers will lead to reduced repeats, however even discounting the winning games by 66% it still shows profit potential.
you should use my roulette numbersandneighbours application tool for this two up..or the very least the ayk tracker
Quote from: TwoUp on Jan 12, 2023, 02:58 AMSo testing the numbers to cover for how long may be productive. If you think about it, using a window covering 9 numbers is not surprising as we expect a repeat by spin 9.
first pic is last 9 numbers which is shortened by a repeat or repeats...2nd pic is true 9 numbers..
I am a developer of some 40 years, and have commercial experience in bigdata, data analysis pipelines and numerical processing.
I'll probably share some example workbooks for the members here as you can use convenient and free cloud services for this kind of stuff. I recommend checking out Google colab.
the max for 1st pic to hit a repeat in the 9 was 31 spins..you can see it below the neighbours tab
the true 9 spins was a max of 27 before a repeat hit following the last 9....
if you have windows two up you can gladly use this software..I,ll just have to activate you on my server when you first try to run it..it,ll give a pc code...you can play about with anything on it its fully adjustable.....thats what it was designed for..crystal 2000 made it to my specs...its a useful tool...specifically designed to place bets quickly on the downloadable williamhill casino...
Thanks for the offer.
For myself I will stick the tools I am using as I need speed of thought and flexibility to explore different ideas and angles.
My problem is getting screen time, with family, and summer holidays here it's a bit difficult at the moment.
no problem ..I,ll probably put it in the roulette software section for members to use...
Quote from: TwoUp on Jan 12, 2023, 12:29 PMI need speed of thought and flexibility to explore different ideas and angles.
maybe I can help you with this with some basic ideas if you like....just ask..I don,t usually do it but its basic stuff ...
I have to think about it, thanks anyway for sharing this, even if it doesn't work.
Yes, with such a system, our losses will be small, but at the same time they will occur very often.
me need to test 100-200 games on paper, see how it goes.
https://rouletteprofs.com/threads/topic-the-sweet-spot-roulettelife-forum.26/
What do you think about testing Amro? I think he has errors in calculations.
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@Martingale Take a closer look at the suggestions for betting I detailed a few posts back. Really study it.
The entry trigger is that you must see 2 repeats in the 14 spins
otherwise it means stand aside. You keep watching until you see 2 repeats in a 14 spin window. Given this is the mathematical expectation you don't need to wait very long at all.
At that point you start betting on spin 15. How many numbers certainly is the frontier to explore. I have proposed some things to investigate which may or may not turn out to be optimal.
I proposed starting on spin 15 betting 12+13+14. This is just a suggestion and I am encouraging anyone reading this topic to explore for yourself.
If you don't get a hit then that is a potential exit trigger. Also if you don't get a hit then your future profitability is also reduced, so it is a reasonable approach to stand aside when you don't meet or exceed the average early in the game, however you may elect to continue to the next waypoint at spin 19 (again I recommend exploring other exit points).
Your next bets on spin 18 and 19, are 14u,15u if you had one hit, or 15u,16u if you had no hit. If you had two hits then it's 13u,14u and for the trifecta of 3 hits it's 12u,13u.
At this point if you haven't had a hit by spin 19
then it's a very good exit trigger.There are 4 spins remaining, so if you bet the next two spins then spin 21 is the last exit trigger. Maybe you had 1 hit but not 2 as expected so you decide to indulge two more spins to spin 21. Principle here is to keep the leash short but not too short.
At spin 21 there are only 2 spins left betting to spin 23 (as per Kramsky) and you can only get 2 hits. At this point your betting level would be two bets at 18u,18u if you had no hits and continued betting like a fool, which is not going to provide a good return or even make up for the losses, hence why the exit triggers make sense.
Don't end up here with no hits and then complain your losses are high.Spin 18-19 seems to be a reasonable go-no-go point and a mandatory exit trigger if you haven't had a hit (or two) and it gives some room for variance.
The other aspect to investigate to increase profitability (at some reduction in probability) is dropping numbers, which will require some in depth testing to determine what the sweet spot (if any) is.
When I posted the stats and profit estimate it is clear that the 7 and 8 repeaters are very dominant in the histogram (moreso than 6) and the profit skyrockets, easily eclipsing the losses on the low yielding games, even when we take away 2/3 of them!
The thing about the profitable hands is that on average they must be hitting through the 24 spin cycle, a rare few will be the last 5-6 spins and the point is who cares! as those are not as profitable games in any event as you would need to bet more numbers over more spins and in effect stick your neck way out to get there. So don't do that.
What I am trying to say here is that the triggers should be keeping you out of games where the hits don't come until late in the cycle. Sure you will have some quick losses abandoning games and miss out on some late bloomers but the gains on games with early repeats given the stats should make up for it. I've discounted the wins by 2/3 and negates the profits on games for 6 repeats to make them losers and the overall profit seems to be carried by the 7 and 8 and 9 repeat games.
My gut tells me that provided you do get some early wins in the cylce it may be beneficial to expand the numbers bet to all that appeared since spin 1 rather than reducing, so potentially it's start with slightly fewer (eg 12) and either exit if no hits or expand based on the early repeats occuring. Again testing will validate the ideas.
Even just using the entry and exit triggers alone may be enough to ensure profitability betting all the numbers that have appeared. Testing will help to confirm what the best points are to cut losses. I would still not bet a number that has repeated twice given the stats show it's quite rare to see a 3rd repeat vs the outlay.
What we have is a probability distribution with the bulk of the weight in the higher repeat counts, which is why it's the sweet spot.
We also already know that 2,3,4 and 5 repeat games are few, 3-repeat (3.7%), 4-repeat (10.4%) 5-repeat (18.5%). The bulk of the games are higher repeating 6-repeat (24%), 7-repeat (21%) and 8-repeat (13%).
All one has to do is ensure the average cost of the losing games is less than the average profit on the winning games. As we can see, when we get more than 6 repeats the profit skyrockets. We have 33.6% of games being losers (many we avoid altogether) and 65.7% are winners. Quite clearly our average per game loss can even be larger than average winning game profit and we still net a profit as we win 2/3 of the time and only lose 1/3 of the time.
But as I have shown the losing game losses are far less than the winning game profits, which is just what we want.
The other characteristic in our favour is that the high repeat count games mostly (but not all) have to have repeats occuring throughout the cycle, so the exit triggers should work reasonably well to cut losses on losers and marginal games.
You only need to bet 4 spins from spin 15 to 19 and then you know absolutely if it is a high repeat (7 or above) game as there are not enough spins left barring an outlier. I would say betting to spin 17-18 is enough to call it a marginal play to continue if you didn't get a hit.We need to elimate the losing games that we don't really know are losers and evict them for not paying rent on time, before it starts to cost too much. The entry trigger of 2 repeats by spin 14 keeps us out of most poor performing games and the exit triggers get us out of the games that fade on us. Sure we may evict some games that may have turned out good, but we can afford to throw away more than half of the winning games and still be profitable as per the chart I included previously.