Theorem on friends and strangers

[falkor2k15]

Does anyone know any examples of how we might apply this to Roulette:
link:s://en.wikipedia.org/wiki/Theorem_on_friends_and_strangers

[falkor2k15]

Does it perhaps resemble 3 parallel games using ECs? PA said we can win with the official ones: he said clusters exist when moving from one EC stream to another.

[falkor2k15]

[falkor2k15]

How about in 6 lines there will be at least 3 results same or 3 results different?
123444

[falkor2k15]

My first attempt at testing this... (mechanically)

[falkor2k15]

See the following possibilities:
W - 58%
LL - 35%
LW - 7%

If we play all combinations the way the example suggests, in terms of individual outcomes, we will get 58% wins and break even. Unless you can find a way to make this tilt towards one side for a set of all the possible combinations, this doesn’t have an edge on its own and it’s a failure. Some runs will give you profit, some will give you loss and if you add Zero to the mix you will get -2.7% equating to a single zero house edge.

Unless you are able to increase the Win% which is currently standing at 58% in the above set , whatever variance based methodology or progression that you use will drive you down.  Good luck 

[3Nine]

link:s://:.getyarn.io/yarn-clip/a44193b1-f650-4507-8ea6-a65c4199f532

[maestro]

link:s://:.getyarn.io/yarn-clip/e717676d-d60c-467e-b324-fd11aee75bea           



lol

[falkor2k15]

Edgey wedgie... (right-click > save target)



Up next: VdW!

[RouletteGhost]

What the hell are you talking about

[falkor2k15]

It's conquered!

When we get a triple+ connection, like 3 lines all greater than 1 there's more chance that we can finish the 6 spins with an additional connection: in this case it ends in 5 lines all greater than 1 compared to L6 on only 1 appearance/show.



The above has a 13% edge:

avg	45563	29%
	31656	20%
	80451	51%
	157670
s3	742	12%
	1954	33%
	3247	55%
	5943