144 Number Roulette

[falkor2k15]

These are not my points of view. I am letting the evidence speak so to speak! And the above is what it's telling us...  so we'll just have to wait and see if this method yields any fruits of the kind we want.

[Firefox]

I think you're correct.  Nothing wrong with presenting the results from some abstract ideas. Far better to give stuff to others that they may be able to develop. That's how progress is made. It's far more constructive than some of the other members here , full of riddles, and pretending one has things when in fact they have nothing of any practical use.

I personally think there's nothing in the non-random  approach, but certainly the more basic cycles should offer some crumbs if the concept is going to be of any kind of use at all.

[falkor2k15]

Am getting ready for the first practical test now having optimized the simulation code:

[falkor2k15]

I'm running some more stats first on the unique streets that the number cycle depends on - will be interesting to see if they keep their ratios since the streets themselves aren't using a cycles framework.

[Firefox]

How are you betting if an advantage found. Presumably 1 unit, win 11 and then 12 on second street, win 132 + stake = 144 units?

143 units profit, a priori odds 0.00657.
1 chance in 152.

[falkor2k15]

I made a fundamental mistake in my earlier thought experiment: we cannot cover all the sleepers early on in this game because all 12 streets are still required for other dual spin combinations, so if we bet 11 streets then we aren't covering all the sleepers. However, the repeat is not of immediate concern anyway, as it takes quite a long time to arrive. It just feels like we are betting random streets with little correlation to the 144 number cycle that is churning out it's own sleepers. I tried the perfect loser strategy and went down to -2,000 before returning back to zero, so is essentially an independent break even game.

I think we need to concentrate on the other idea I was discussing with regards to negative progression and creating dependency using 2 spins as 1 spin as 1 outcome. There's another way we can do it using Six Dozen Options or Quadruplets without using 2 spins:

11 = Option 1
12... Option 1 cannot repeat

Perhaps that is also dependency-related.