Suppose we are betting randomly the pattern #2, #5 , #8 and #15 of the Big Road. (two time pattern #15 hadn't come out):
S-S-A-A
A-A-A-A
S-A-A-A
A-A-A-A
A-A-A-A
A-A-S-A
A-A-A-A
A-A-A-A
S-S-S
A-A-S-A
S-A-S-A
A-A-A-S
A-A-A-S
A-S-A-A
S-A-A-A
S-A-A-A
S-A-A-A
A-A-A-S
A-A-A-A
A-A-S-A
A-A-A-S
S-A-S
What and when to bet at these successions?
If S= -3 and A= +1, before vig any line will get:
-4
+4
0
+4
+4
0
+4
+4
-9
0
-4
0
0
0
0
0
0
0
+4
0
0
-5
Total= +2
If adopting the strategy to play A-A one time and A after S one time we'll get:
(-3)(+1) = -2
+1
+2
+1
+1
+2
+1
+1
-3
+2
(+1)(-3)(+1) = -2
+1
+1
(-3)(+1)(+1) = -1
+2
+2
+2
+1
+1
+2
+1
(+1)(-2) = -1
Total= +15
Therefore if we'd assume a A=0.75 p and S=0.25 p, the expected A/S ratio is 3:1. So it's the average more likely ratio while considering four A/S decisions (when applicable).
Thus when an average ratio shows up no possible permutation will deny us to make a +1 or more probable a +2 profit.
In fact a single S among three As cannot produce any loss.
Within sets of 4 resolved hands, losing streaks can only come out when two or more S happens.
Anyway 4 S are just a loss of -3
3 S produce a loss of -6, -6, -3, -2.
2 S produce a loss of -2, -5, -6, -1, -2; and a win of +2.
0 S are always a +1 win.
Paradoxically we are in less worse shape when 4 S are showing up than when 3 S are coming out.
2 S are really hurting us just in two out of six possible permutations; in the remaining cases we'll get a -2 or -1 controllable loss and even a win of 2 units.
Run this situation infinitely (that here were taken randomly even if some positive variance happened) and let's see how many 4-decision sets are getting the negative 3 S or, at a lesser degree, the 2 S negative enemy.
A more aggressive plan needing a very large bankroll would be to double the A-A bet and the S-A bet after two or three losses in a row with the addition of betting the A patterns until they'll stop and until the deficit is recovered.
A plan at least 50x fold better than betting Banker in whatever sauce.
as.
S-S-A-A
A-A-A-A
S-A-A-A
A-A-A-A
A-A-A-A
A-A-S-A
A-A-A-A
A-A-A-A
S-S-S
A-A-S-A
S-A-S-A
A-A-A-S
A-A-A-S
A-S-A-A
S-A-A-A
S-A-A-A
S-A-A-A
A-A-A-S
A-A-A-A
A-A-S-A
A-A-A-S
S-A-S
What and when to bet at these successions?
If S= -3 and A= +1, before vig any line will get:
-4
+4
0
+4
+4
0
+4
+4
-9
0
-4
0
0
0
0
0
0
0
+4
0
0
-5
Total= +2
If adopting the strategy to play A-A one time and A after S one time we'll get:
(-3)(+1) = -2
+1
+2
+1
+1
+2
+1
+1
-3
+2
(+1)(-3)(+1) = -2
+1
+1
(-3)(+1)(+1) = -1
+2
+2
+2
+1
+1
+2
+1
(+1)(-2) = -1
Total= +15
Therefore if we'd assume a A=0.75 p and S=0.25 p, the expected A/S ratio is 3:1. So it's the average more likely ratio while considering four A/S decisions (when applicable).
Thus when an average ratio shows up no possible permutation will deny us to make a +1 or more probable a +2 profit.
In fact a single S among three As cannot produce any loss.
Within sets of 4 resolved hands, losing streaks can only come out when two or more S happens.
Anyway 4 S are just a loss of -3
3 S produce a loss of -6, -6, -3, -2.
2 S produce a loss of -2, -5, -6, -1, -2; and a win of +2.
0 S are always a +1 win.
Paradoxically we are in less worse shape when 4 S are showing up than when 3 S are coming out.
2 S are really hurting us just in two out of six possible permutations; in the remaining cases we'll get a -2 or -1 controllable loss and even a win of 2 units.
Run this situation infinitely (that here were taken randomly even if some positive variance happened) and let's see how many 4-decision sets are getting the negative 3 S or, at a lesser degree, the 2 S negative enemy.
A more aggressive plan needing a very large bankroll would be to double the A-A bet and the S-A bet after two or three losses in a row with the addition of betting the A patterns until they'll stop and until the deficit is recovered.
A plan at least 50x fold better than betting Banker in whatever sauce.
as.