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[BEAT-THE-WHEEL]

@stringbeanpc,
Thanks for your posting..
111   211   311
112   212   312
113   213   313

121   221   321
122   222   322
123   223   323

131   231   331
132   232   332
133   233   333

Three Same Dozens 3 out of 27
111
222
333

Three different dozens 6 out of 27

123   213   312
132   231   321

Two dozens 18 out of 27
&&&&&&&&&&&&&&&&&&&&&&&

As we know, its always 33% vs66% minus green probability.

But as the only2dz in matrix aka pinwheel,
If we forego the 111, 222, 333 matrix,
And only bet the formation of matrix consist of "only 2dz in 3spins"
Then the probability is 24probability,

Then 18/24=75%
Why 24, as we left out the 111, 222, 333.

6/24=25%...?

doesn't this sort of distorted edge, since we not bet the 111, 222, 333,
Instead of 66/33%....if we bet from the begining of 3spins, but 75% for 2dz/3spins???

[stringbeanpc]

Beat-the-wheel

I had never thought of dozens as you describe above.

Just to be clear of your post

DO NOT BET
111
222
333

123   213   312
132   231   321

Only bet there could be 2 dozens in a set of 3 spins

112
121
122

131
133

etc, there is a total of 18 of these combinations

I do not see how this could improve the odds because combinations like 111 & 312 will still appear.

The only control we have is our bet, not on what appears.

[BEAT-THE-WHEEL]

Thanks Stringbeanpc,

Do you mean if someone could avoid betting the matrix (as you described),
then the edge will be distorted?
(To 75%....)


and since we can't avoid them, then the edge of 2.7 green, and 1/27possibilities,  still remain....

[stringbeanpc]

and since we can't avoid them, then the edge of 2.7 green, and 1/27possibilities,  still remain....

No doubt, these will occur

[BEAT-THE-WHEEL]

Thanks Stringbeanpc,
Seems that there no way to beat them.