The sweet spot

[Martingale]

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.

[Martingale]

judging by his words that there are only 9 spins, it turns out that he does not play 24 spins, but only 23

[TwoUp]

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:


The binomial cummulative distribution function:


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.

[Martingale]

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;

[Martingale]

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.

[6th-sense]

is there any other stats that can help you here two up?


 

[TwoUp]

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.

[TwoUp]

A correction to my post above:

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 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):



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.

[TwoUp]

Apologies for my error below.

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

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.

[6th-sense]

you should use my roulette numbersandneighbours application tool for this two up..or the very least the ayk tracker

[6th-sense]

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.

first pic is last 9 numbers which is shortened by a repeat or repeats...2nd pic is true 9 numbers..

[TwoUp]

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.

[6th-sense]

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....

[6th-sense]

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...

[TwoUp]

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.