Not sure if you have the right understanding? Every 9 spins 3 of the same colour will form an arithmetic sequence.
1 2 3 4 5 6 7 8 9
So we can guarantee that at least one of these 3 will be of the same colour (all red or all black), but we don't know which one:
1,2,3
2,3,4
3,4,5
4,5,6
5,6,7
6,7,8
7,8,9
1,3,5
2,4,6
3,5,7
4,6,8
5,7,9
1,4,7
2,5,8
3,6,9
1,5,9
So we start looking for 2 matching colours: if spin 1 and 2 is red then spin 3 could be red; if spin 1 is black and spin 5 is black then spin 9 could be black, etc. So we would bet the 3rd spin to be red or the 9th spin to be black. But actually we would never bet on the 9th and final spin - instead forfeit the set - since red and black would both have an arithmetic progression. Priyanka is better at explaining it than me, so maybe re-read first 2 pages.
Falkor, I understand perfectly whats going on with the simple explanation, thanks. In fact my numbers match yours except I forgot 7-8-9 in mine
When I say has to win on ODD I mean the last place holder in the arithmetic sequence is ODD
123
135
147
= all these can only "win" on an ODD spot (3-5-7)
I hope this explains what I was trying to say a bit more clear
When there is multiple sequences for red and black, do you cut the 9 spin window at the no bet spin and start over? Is that how it is done in your programming?
Oh, and 1 and 2 are just representing any even chance (red/black,player/banker,etc)