Non-Random combinations - revisited

[falkor2k15]

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?

[DoctorSudoku]

Gilius-Falkor,
Welcome back !

[Taotie]

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.....

[nottophammer]

you'll be able to chat with Steve then