Progression from Bayes that would appear to survive all tests so far:

http://www.rouletteforum.cc/index.php?topic=8165.0

In its usual form, the Labouchere progression starts with string of numbers and aims to make the target by cancelling 2 losses with each win. It can work well, but the stakes  get out of hand when you hit losing streaks. One way around this is to break down the strings of numbers (which represent your debt) into smaller "substrings" when the stakes get too high, then work on clearing the substrings. Another way is to use a "divisor" which keeps stakes from getting too high in the first place. The disadvantage is that you need more patience!


The basic idea is to choose a divisor, which is a number greater than or equal to 2 (a divisor of 1 corresponds to the standard labby). Higher divisor = more safety. Using this method involves some work because you need to calculate the current stake from the standard labby string. But it does keep the stakes lower and maintains the same advantage of cancelling 2 losses with each win.


First, a bit of terminology. You need to use what programmers call the "Ceiling" function. In plain English, this just means after dividing one number by another, you round the result UP to the nearest whole number.


e.g. Ceiling(4/3) = 2   because 4/3 = 1.3333 but when rounded up gives 2.


The divisor, call it X, always goes in the bottom part of fraction, and the top of the fraction is the stake which is generated by the standard labby.


The best way to explain is by an example using a divisor of X = 3



1 L 11                        stake = 1
1 L 112                      stake = 1
1 L 1123                    stake = 2 (ceiling(4/3) = 2)
2 L 11234                  stake = 2
2 L 112345                stake = 2
2 L 1123456              stake = 3 (ceiling(7/3) = 3)
3 L 11234567            stake = 3
3 L 112345678          stake = 3
3 L 1123456789        stake = 4 (ceiling(10/3) = 4)
4 W 12345678           stake = 3 (ceiling(9/3) = 3)
3 L 123456789          stake = 4 (ceiling(10/3) = 4)
4 W 2345678             stake = 4 (ceiling(10/3) = 4)
4 W 34567                 stake = 4 (ceiling(10/3) = 4)
4 L 3456710              stake = 5 (ceiling(13/3) = 5)
5 W 4567                   stake = 4 (ceiling(11/3) = 4)
4 W 56                       stake = 4 (ceiling(11/3) = 4
4 W                           *string cleared*


As you see, you only needed a third of the bets won + 1 in order to clear the debt, but with a divisor of 3 your highest stake needed was only 5, instead of 13 if you'd used the standard labby.
If you play around with this, you'll notice that with larger divisors you're left with some units over, and that these accumulate if your bets don't rise above 1 unit. This happens if, for example, you're using a divisor of 5 and you get a series of decisions which don't include 5 or more losses in a row, but the overall number of losses is more than the wins.