On the first part I've proved with calculations how a progression can turn a negative expectation in to positive.

This time, on the second part of the mini series, I'm going to show how a bet selection without any progression can result in positive results in contrast with the conventional negative expectation.

Take a close look at the following groups of numbers

ROL 1 3 5 7 9 BEH 20 22 24 26 28

ROH 19 21 23 25 27 BEL 2 4 6 8 10

REL 12 14 16 18 BOH 29 31 33 35

REH 30 32 34 36 BOL 11 13 15 17

These are the groups of all 3 EC's combinations, each group contains 4 up to 5 numbers.

Let's say I bet simultaneously Black,Even,High what would make me lose all 3 bets?

The numbers 1,3,5,7 and 9, a total of five numbers which all of them belong to 1st dozen...

The 1st dozen contains also numbers which I could win with my EC bets, such as: 2,4,6,8,10

By betting 3 EC's simultaneously there are more chances to lose or win 1 unit, rather than win or lose 3 units and that's because more numbers are resulting in 2 out of 3 EC's to change or remain the same.

Except from the

**Even,Black,High** there is also equally good the

**Red,High,Odd** in combination with

**1st dozen**.

A total of 2 winning combinations, each one contains 4 flat bets with 1 unit each.There is 1 number resulting in -4 (0) -4

There are 5 numbers resulting in -1 (1,3,5,7,9 for BEH or 2,4,6,8,10 for ROH) -5

There are 5 numbers resulting in +3 (1,3,5,7,9 for ROH or 2,4,6,8,10 for BEH) +15

There are 5 numbers resulting in +2 (20,22,24,26,28 for BEH or 19,21,23,25,27 for ROH) +10

There are 4 numbers resulting in 0 (30,32,34,36 for BEH or 29,31,33,35 for ROH) 0

There are 2 numbers resulting in +1 (11,12 for both BEH and ROH) +2

There are 6 numbers resulting in -2 (13,14,15,16,17,18 for both BEH and ROH) -12

Calculating all totals we find: (15+10+2)-(4+5+12)= 27 - 21 = **+6 grand total**

This is what I call positive expectation and if you are not convinced so far, then nothing and nobody could ever convinced you!