Example:
Roulette Number: 6 Hits = 24 times Skips = 51 43 104 1 2 29 37 77 79 15 130 15 12 7 11 131 11 129 8 16 20 14 21 12 * Sorted Skips = 1 2 7 8 11 11 12 12 14 15 15 16 20 21 29 37 43 51 77 79 104 129 130 131 * Median Skip: 16
So, roulette number 6 hit 24 times.
The first row called Skips shows (from right to left) the skips of number 6. After 12 spins, hits again, after 21 spins hits again, after 14 spins number 6 comes up again etc. The last three skips of number 6 are 104, 43, 51.
The second row called Sorted Skips shows all the skips for a particular number in ascending order.
The third row called Median Skip show the median value of the sorted skips. The median is the middle value of a string of numbers. In this example the median value of skips is 16. It should not be confused with the average.
Therefore 50% of the values are within the median or less, and 50% are within the median or more. In my example of number 6, the skip median is 16. That means that 50% of the time number 6 hits within a skip of 16 or less.
Now if in the string of skips of number 6 I mark:
I obtain this:
51 43 104 1 2 29 37 77 79 15 130 15 12 7 11 131 11 129 8 16 20 14 21 12 + + + - - + + + + - + - - - - + - + - - - - - -
THE STRATEGY
We bet on roulette numbers which have a 3 consecutive streak of skips greater than the median value of skips. (That means three consecutive +)
In our example, number 6 is qualified because the last three skips (consecutive) are 104, 43, 51. We have + + + (three consecutive value of skips, greater than median value of skips: 16)
(104 > 16), (43 > 16), (51 > 16)
It is possible to code this roulette gambling system with these options:
I also use this strategy with success in Bookmakers Lottery (Irish, German, Spanish, 6 of 49 lotteries where I can bet on one number only.)
Rarely will a number have three consecutive skips in ascending order.
Example: 0 12 13 22 15 17 0 12 30 12 15 24 22 12 4 0 22 34 17 20 25 36 35 14 19 19 21 22 14 0
In above case the skips for roulette number 0 (IN ORDER) are: 5 8 13.
So in this case we must bet on number 0. When 0 hit we remove the number.
The system is simple.
We bet all the numbers which have three CONSECUTIVE SKIPS IN ASCENDING ORDER.
In this case for roulette number 0 = 5 < 8 < 13.
Roulette number 12 has these consecutive skips: 5 1 3.
This number is not qualified because the skips do not have three values in ascending order.
You can code this system (method) with these settings:
When a number hits, remove it and continue in this manner.
I do NOT recommend skip systems above the FFG median individual or the game constant. The skips above FFG median can go on for very long periods of time sometimes. There are so-called roulette sleepers that skip over 200 spins each when over 1000 roulette spins are analyzed! The systems above median have a higher COW (cost-of-winning) and their winning frequency is usually lower.
In the Apache Roulette System I, it is preferable to apply the system in reverse. Look for 23 skips under the median (the minus sign ). They do assure a lower COW. I usually see that 23 skips under the median have a better frequency too. You see in Apache's report strings of (minus) 3 or 4 skips in length. And the waiting between the signs was far shorter than in the + (above median) situations.
Let's take the 3 consecutives case. The string of 6 minuses (under median) at the end would have resulted in 4 successes: | 14 21 12 |, | 20 14 21 |, | 16 20 14 |, | 8 16 20 |. The string of that followed would have yielded 2 hits: | 12 7 11 |, | 15 12 7 |. Total successes: 6.
There was only one qualifying string of pluses (the one at the start is still running!) The + (above median) string would have resulted in 2 successes: | 37 77 79 |, |29 37 77 |.
The (under median) method wins by the score of 6 2. Not to mention that the waiting periods for the (under median) strategy were far shorter, therefore the COW a lot cheaper. Analyzing 7000 spins from the Hamburg Casino (Spielbank), Germany, 25 of the 37 roulette numbers had skips of over 300 spins at least once.
Axiomatisch, here is data from the same casino showing clearly a better performance by the under median roulette strategies.
FFG-1: 2 skips SUM-UP TO 52 = 2823 hits in 7000 (40%) = 16 numbers (to play) FFG-2: 2 skips UNDER/EQUAL 26 = 1806 hits in 7000 (26%) = 12 numbers FFG-3: 2 skips ABOVE 26 = 1681 hits in 7000 (24%) = 11 numbers FFG-4: 3 skips UNDER/EQUAL 26 = 927 hits in 7000 (13%) = 7 numbers FFG-5: 3 skips ABOVE 26 = 859 hits in 7000 (12%) = 8 numbers FFG-6: MEDIAN & SUM <= 52 = 2309 hits in 7000 (33%) = 11 numbers FFG-7: 1st Skip <= 26 = 3505 hits in 7000 (50%) = 17 numbers FFG-8: 1st Skip > 26 = 3384 hits in 7000 (48%) = 20 numbers SYS-: Your System <= 20 20 = 1244 hits in 7000 (18%) = 7 numbers
The under median systems offer more hits, while resulting in fewer roulette numbers to play. It is a two-edge sword against the odds. Furthermore, the 3-skip systems can be totally replaced by Your System with the last 2 skips set to less than or equal to 20 (median -25%, or minus approximately one FFG deviation).
And remember to always keep track of the hits and misses of every system.
The Apache System II has no mathematical foundation that I can think of. I'll never play such a roulette or lottery system. What I am saying is, what Liru Crocodilu noticed may occur periodically, but I don't figure out a formula to back his observations.
Read my theory of skips in gambling and lottery here (a repost from my now-closed forums, 2006):
The only software that plots the skips, creates gambling systems from skips, and generates combinations is SkipSystem. The presentation of the program with illustrative screenshots: