SIXAINS ROULETTE SYSTEM
I'd like to put the following roulette system playing sixains up to general consideration.
Empirical testing of 3 months of records have shown an average gain of 2,6% of turnover. It is designed to be played on "French" tables with the Maximum pay-out usually being 1200 times minimum (i.e. the maximum stake on sixains is 200 times minimum.) Only "natural" sixains (or as Americans call them: streets) are considered. I.e.: I: 1-6, II: 7-12, III: 13-18 etc.
The aim is to win at the first repeat of ANY (not a particular!) sixaine. Watch the coups (taking notes of the sixains drawn). If within 4 coups ("throws" for my American readers) none has repeated start betting on the last drawn sixaine (step I). If lost bet additionally on the second one drawn, (stepII). If again no hit, spread your bet on all three of those drawn last (stepIII), according to the progression shown below. (Additionally "buy" insurance on Zero). A theoretical step IV can be imagined but as can be seen below is definitely to expensive; required capital would increase fourfold). (This is eliminated by 3spaces, hope it displays properly; all numbers refer to units):
Step Numbers bet size bet 0 total bet/step cumulative bets
I 7 1*5 1 6 -6
II 13 2*5 1 11 -17
III 19 3*10 2 32 -49
(IV) (25) (4*28) 5 (117) (-165)
If hit profits will be as follows (all previous losses recovered): Step hit on sixaine hit 0
I +24 +29
II +13 +18
III +12 +19
(IV) (+2) (+4)
If no hit by step III, cumulative losses will be -49. The day's play should be abandoned if three losses occur (day's maximum loss -147 units).
Wait with first actual bet, until first theoretical loss has occurred.
Wait one additional spin between signal and placing bet.
If a loss has occurred, follow the game until a hit would have occurred, then immediately start betting again following the pattern above.
Capital requirements: In my opinion three-day's of losses, to be on the safe side (i.e. almost 1500 units). Not cheap I know.
Mathematical expectations according to binomial distribution (50%\expectation)
Step I: 16,8%
Step II: 34%
Step III: 56%
Empirical testing showed significantly better results for steps II and III.
Reason for this seems to be a; waiting period until first theoretical loss has occurred, b; the fact that betting doesn’t start until at least three different sixains have appeared. For those mathematically inclined the general formula for calculating expectations for series (a series of two sixains is used for winning in the above system).
Whereby: p=expectation, q=counter expectation (1-p), s=length of series, Cv=size of covered area (i.e. 37/size), m=numbers covered, n=lenght of spins played, W=probability, #=average, E=exponent (scientific notation, easier to display)
For any chance:
solitary series: qE2*pEs*Cv (e.g. for plein: (36/37)E2*(1/37)Es*37
soziable series: q*pEs*Cv (e.g. for plein: (36/37)*(1/37)Es*37
sliding scale: pEs*Cv
reciprocal expectation (1/W) thus,
in step I: soziable determined sixaine: 1,464, any sixaine 8,784
in step II: soziable determined m=13: 6,76, any m=13 19,25
(Note soziable series are usually only useful if a immediate gain, "paroli" play is planned. Length of series in most cases will be 2 or 3, which is done in this system). I can't go further into stochastic explanations, whole books have been written on this... Anyway, I hope to have given you an idea, and hope the explanations are sufficient. For those of you playing a 38 number table, the zero insurance in step I and III should be bet a cheval on 0/00, in step II it should be increased to one unit each. This will reduce pay-outs on 0, but not significantly overall. Anybody playing on European tables in step II, if all covered sixains equal manqué or passe should play this chance instead to profit from the "enprison"-rule.
Crucial to winning is the waiting periods!