hi folks,

i have a question and I hope someone here can help:

what is the probability that within 3 spins all 3 different dozens hit (in which order doesn't matter). and how is the calculation done?

thx a lot already and cheers

hans

33%, minus HE.....as in pinwheel.

Single zero wheel- 3 spins- 3 different dozens- any order

(36/37)*(24/37)*(12/37)=.20468     or 20.468%


explanation of calculation-  First spin can be any of 3 dozens which is 36 numbers,  the second spin to be   one of remaining  2 dozens which is 24 numbers,  and last spin remaining 1 dozen which is 12 numbers.

thanks guys for your instant replies. 

We could only bet the third spin, after the first and 2nd spin produced 2 different dozens.

eg.
1st =dz 1, 2nd=dz2, then we bet the 3rd spin=dz3.
Thus the probability= 33%, minus HE.


The reverse also true,
If we bet there only 2 different dozens in 3spins, =66%...minus HE.

The bad news is , extreme variance,  and it may not rtm in short run.  Though it quite STABLE  in 300+- spins, imho.

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

Three dozens with back to back repeats of same dozen 15 out of 27

111   211   311
112
113

   221
122   222   322
   223

      331
      332
133   233   333

Thanks Stringbeanpc.



Say, what the probability , if

123, then 123 hit again,

thus 123, 123..
And then, other permutations hit for next tens of spins,

Now 123 hit again....will next three spins, be 123?again?


123
123....
×××
×××
×××
×××
123...now, will 123 hit again?...
=============

Other example...
Say,

312,then...
111.....
Xxx
Xxx...
312  hit,again,... (will 111 hit again, after 312 hit, repeating the previous 312, 111? Back to back?

We can bet the highly unlikely coincidence,  though the probability remain the same...

Say,
If you see a repeat of matrix,

123, then..
123 again,
After a while,
123  hit, will the next three spin be...
123 again?


Or,
123, follow by..
111....
Then after a while...
123 hit, will the next three spins be...
111....again?

If you see..
111, follow by...
222, do you think that...it be follow by...
333...?

Three different dz in three spins,


223
(123)...a single
233...

That a single of a permutation, among , (only two dz in three spins matrix)


Say,
123...follow by another three different dz in next three spins=33% minus green.

eg
123,
213....as a SERIES,  of TWO...(of three different dz matrix)

321,
123
231....=series of three

231,
321,
213,
231..=series of four.

And you will see series of five, six, seven  ...too!

Probability 33% for single..
66%..for series (of 2 and more)

Thus when you see a single, you may bet, it may go to series, or becoming a single...

eg..

You see..
123....bet next matrix may three different dz ....=33% chances. minus green


You see..
123...bet next matrix= only two, or only one dz..=66%....minus green

Thus, if you see...
123...there be three possibilities....

If you bet any ,two dozen, then if it "only two different dz in 3spins,",.. you win...and
If it "3 different dz in 3spins"...you win...

Lose when only one dz hit, that you choose not to bet....

Any combination of three dozens in three spins are equally probable, be it 123 or 111 or 231 or any other combination of 1,2 or all three dozens, you  can make or expect. That will be 3.6% in European Roulette. Now, you can multiply this % with number of combinations you can have.Three different could come as 123,231,321,132,312,213, i.e. six ways. Hence, you have a probability of 3.6 x 6=21.6% to get three different dozens in three consecutive spin.  Similarly, chances of any combo of ECs in three spins is 11.5%.

Thanks Albalaha for your post.

Though the probability for 3different-dz/ 3spin, will produce no edge whatsoever,
I think the coincidence of a combo to repeat itself, may due to another theory,
Probability-of - coincidences,  perhaps?

If we see a rather rare coincidence,
Say,
123, follow by...
111....then after a while...

123, followed again, by,
111....then after a while...

123...hit, will the next three spins, be
111....again?

Though the probability remain unchance, and has no edge, but the repeat of coincidence, remain highly unlikely.


The same , as to ds,
If we see, six different ds in six spins,
Will the next six spins has six different ds...again?
Though the probability remain the same , but the coincidence highly unlikely....

say,
If we see five different ds in 6spins,
123455...will the next 6spins produce...
123455....again?

People ask,
How to bet SINGLE DOZEN IN MATRIX OF 3DZ.

we bet when first and second matrix are TWO DIFFERENT DOZENS, and bet the third as SINGLE BET.

say,
Permutation of 3different-dz, may SINGLE VERSUS SERIES-OF 2, 3, 4, 5, 6..INFINITE

Thus, when a single permutation  hit, we bet it may produce a series...

123, bet next 3spins, will 3different dz , again
321, bet the third spin=1dz....hit...
The probability=33%, minus green, thus a long day wait to bet.