@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???
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.
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....
Quote from: BEAT-THE-WHEEL on September 20, 2018, 04:35:16 AM
and since we can't avoid them, then the edge of 2.7 green, and 1/27possibilities, still remain....
No doubt, these will occur
Thanks Stringbeanpc,
Seems that there no way to beat them.