Single number betting

BlueAngel · Aug 22, 2023, 08:31 PM

The procedure is relatively simple, chart the last 37 spins to find a number which has NOT hit during the last 19 spins, but HAS hit only once the previous 18 spins.

In case there are 2 or more qualifiers pick the 1 which its right AND left adjacent numbers have not hit OR have 2+ hits.
Example:
Number 2 has its right adjacent number, 25, with no balls received and its left adjacent neighbor, 21, with 1 hit, this would be no bet because we don't want to have 2 adjacent numbers with only 1 hit.

Number 1 has its right adjacent, 20, with no hits and its left, 33, with 2 hits, thus this would qualify to bet number 1 because from one side has an empty pocket and from the other a number with 2 balls already got.

Number 3 has its right adjacent,26, with 1 hit and its left, 35, also with 1 hit, this would be no bet for any of those 3 numbers.

Number 4 has its right adjacent, 21, with 2 hits and its left, 19, with 3 hits, in such case the number 4 would be qualified for betting.

If there are not any qualifying numbers then wait until there are, when we eventually qualify only one number for betting then we will proceed with the specific number for a maximum of 18 consecutive bets (or till win, whatever happens first).

When we win repeat the process, when lost qualify another number and this time we will bet it for 19 consecutive spins (maximum), but this time, since you have arrived here after 18 lost bets, make the first seventeen bets with 1 unit and the last two with 2 units, like this:

1st set/number with up to 18 bets } 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
2nd set/number with up to 19 bets } 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2
3rd set/number with up to 18 bets } 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3
4th set/number with up to 19 bets } 3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4

A total risk of something more than 140 units.
Each set qualifies another number, never continue with a losing number beyond the 18th/19th bet!