I've gone back to trying to solve the riddle.
Excluding zero, either dozen 1,2 or 3 MUST show once within 1 spin. We have a 33% chance and payout of 2:1 costing 1 unit, hence breaking even in the long run.
Excluding zero, either dozen 1,2 or 3 MUST show twice (repeat) within 4 spins. We have a 62% chance it will be defined same as the starting dozen, but the cost could be up to 3 units, hence breaking even in the long run.
1... 62% chance repeat will be on 1, i.e.:
11
121
1231
131, etc.
121... 71% chance 2nd repeat will be on 1, i.e.:
1 2 1 3 1
1 2 1 1
1 2 1 1
1 2 1 1
1 2 1 2 3 1
1 2 1 1
1 2 1 3 2 3 1
1... 44% chance repeat will be on Cycle Length 2, i.e.:
121
122
CL2... 76% chance repeating cycle length will be CL2, i.e.:
CL2 CL2
CL2 CL1 CL2
CL2 CL1 CL3 CL2
Are those the "imbalances" - with increasing stats - that we are meant to be "riding on" since there's always more chance of a repeat in the short term (against brick design) compared to all uniques showing?
Over the course of 2+ spins we can also bet several events at the same time, i.e.:
*Dozen Cycle Length 2
AND
*Defined by same dozen as starting partition
AND
*Defined by different line as starting partition
Is there more chance of a dozen repeating with a line? Should we bet:
2(L3)... bet 1+3 and if win:
2(L3)3(L6)... bet 2+3+L1+L2+L4+L5+L6
OR
2(L3)... bet 1+3 and if win:
2(L3)3(L6)... bet L4+L5+L6
?
Am I on the right track or do I need to dig deeper?
Gilius-Falkor,
Welcome back !
Just when you thought you had dug yourself to the bottom of the stupid hole, you look across and see that the guy next to you is still digging.....
you'll be able to chat with Steve then