Different bet selection ideas for the 11-33-99 progression.

[wiggy]

I didn't want to put this idea in the grassroots thread because it involves betting streets and not dozens. Feel free to share any bet selection ideas you may have for this progression.

My idea is to keep a rolling count of the 8 latest streets and then you will have 4 sleeping streets. Then I would wait for one of the 4 sleeping streets to hit twice in a row and then bet the 8 latest streets to hit on one of the next three spins using the 11-33-99 progression.

A total loss of the 11-33-99 progression would mean that the 4 sleeping streets hit 5 consecutive times.

Here is a quick summary of how it all looks. There were two bets in this example which both won on the first attempt.

11 (1)x
8 (2)x
7 (3)x
12 (4)discard
11 (4)x
11 (4)discard
4 (5)x
7 (5)x
2 (6)x
8 (6)x
10 (7)x
2 (7)discard
7 (7)x
8 (7)x
6 (8)discard the 4 missing are 1,3,5,9.
10 (8)x
5 (8)discard the 4 missing are 1,3,9,12.
3 (8)x the 4 missing are 1,9,11,12. After 2 losses bet the 8 streets 2,3,4,5,6,7,8,10.
4 (8)x
7 (8)x
7 (8)discard
10 (8)x
8 (8)x
12 (8)x the 4 missing are 1,2,9,11.
3 (8)x
10 (8)x
10 (8)x
3 (8}
11 (8} the 4 missing are 1,2,6,9.
8 (8}
10 (8}
12 (8}
4 (8}
9 (8} the 4 missing are 1,2,5,6.
1 (8}x the 4 missing are 2,5,6,7. After 2 losses bet the 8 streets 1,3,4,8,9,10,11,12.
1 (8}

To keep the rolling count up to date, I put an x on the furthest back street when it repeated.
Then when I eventually had 8 unique streets and hit a sleeper I had to discard the furthest back street in the 8 otherwise I would have 9 streets.

The idea behind this was how it seems quite rare to find back to back dozens like the 123-123 repeating.

So maybe even waiting for 3 losses would be better although more time consuming or use a 11-33-99-2727 progression after two consecutive sleeper streets appear.