The previous example as well as the variance test showing independent waves got me thinking...

To return to the variance problem as well as whether we should stitch an outcome or not and other concepts discussed herein, I am starting to think there might be a riddle to solve here:

1) The outcome of the dozen cycle is not officially set in stone until the cycle has closed.

2) The cycle outcome can occur across 3 different spins and we have various ways of playing Same and Different based on stitching or missing out spins.

3) The repeat depends on the uniques; Same/Diff also depends on the spin number. For example, dozen 1 has more chance of repeating if it's in the starting partition; Diff has more chance of resulting if the cycle is on spin 2 as opposed to spin 1 or spin 3; Same has more chance on spin 1 followed by spin 2, etc.

4) Spin 1/CL1 can only be missed out and not stitched, so spin 2-3 are probably our main weapons.

5) Each cycle outcome is independent of the previous cycle besides the carrying over of the defining element; Same/Diff probability/maths expectation depends on several cycles as opposed to a single cycle.

6) Same is not necessarily better than Different because each bet is a break even bet in the long run (notwithstanding MLE) - but the dependencies differ in different situations.

7) Different cannot happen unless the cycle remains open beyond spin 1; Same on spin 3 cannot happen unless the cycle remains open beyond spin 2, etc.

8 ) We don't yet know why we should stitch, not stitch, play a spin or miss out a spin during an open cycle for the purpose of exploiting the variance - only becomes official when the cycle is closed.

Since finding out that waves of variance between cycles are completely independent of previous waves, and the wheel has no memory, it then follows that solving this riddle must go something like this...

We need to find a way to bet for multiple things to happen over the next X cycles - more specifically so that Same ends up totaling more than Different via each individual spin result (also tallying up to maths expectation, respectively). If we try to bet Same only then we reach the house limits when using progression or we get eaten by dispersion; if we try to bet Same on first spin only the same thing also happens, so we cannot force just a single part of the system. Perhaps we need to find some way to alternate a "betting wave" and make it dynamic somehow.

To test this we really need to think outside the box: I'm guessing we need to measure spin 1 vs. spin 2 vs. spin 3 tally when, say, Same is 5 and Different is 3; something like that might provide us the right betting template?