Correct?

[Turner]

Now I agree...as I said in my first post....setting a goal, say at 3 wins in the next 7 would be 28% chance...but that is worked out in binomial distribution, and the formula uses the odds of winning and losing...see?

7C3 x 18/37^3  x  19/37^4
=28.02%

This is "how many 3's in 7" x win probability x loss probability

No special odds for each spin. Just the odds of winning (18/37) and losing (19/37)

You can clearly see them in the formula. Win 18/37 and loss 19/37

do you see my point?

My point being, yes, a prediction has a probability which changes depending on the prediction, but each spin is same probability.

[Mr J]

Thanks Turner, much appreciated !!

Ken

[Edgars776]

playing roulette will really challenge your patience. So good luck!

[Nathan Detroit]

This system   playing for  a change ( e.g. after 3 Reds  bet  on Black   a certain   number of times) is being recognized by many seasoned  gamblers as  being  what   card counting  is in Black Jack.

This method is highly recommended   by 2  of the top US  professional gamblers. J.P and  F,S.

AMATEURS  don`t need to apply.
.

MrJ  thanks again for  having started this thread.


Nathan Detroit



'Posted August 14th, 2014  1700 hrs EDST ( Filed)

[palestis]

This system   playing for  a change ( e.g. after 3 Reds  bet  on Black   a certain   number of times) is being recognized by many seasoned  gamblers as  being  what   card counting  is in Black Jack.

This method is highly recommended   by 2  of the top US  professional gamblers. J.P and  F,S.

AMATEURS  don`t need to apply.
.

Amazing. The best systems turn out to be so simple and less complicated.
I agree with this system too. Though I play a different version. (I don't like progressions that shoot up too fast).
Instead of black or red missing for 3 spins, I Play a dozen missing for 6 or more spins (whenever I see it on a score board). And I play it for about 3 or 4 times, sometimes 5.  And stop.  The logic behind it is this. If a dozen is missing 6 times, will it always go on and become missing 10 times? Cycle after cycle after cycle? I doubt it. Sometimes you will see missing 20 + times, but since you only play it 4 times, you can't lose your shirt. However 6 misses turning out to become 10 misses each and every time, it's very unlikely.
Just as if red was missing for 3 spins, turning out missing for 6. Every time.

[Nathan Detroit]

If you were playing at a 0/00 wheel then you  could  disregard the fixed table  layout for the dozens by creating your own dozens.

Just check  on google WASHOO2  2 dozen   roulette system.

You find it also  in the archives at one of Steve`s  boards.






8/16/2014 Post  on File

[Turner]

Summary (if you are interested  )

Q: Why are the odds of getting at least 1 black in 5 spins around 97%?

1) The odds of getting 1 and 1 only black in 5 spins

how many 1's in 5 = 5C1 = 5!/1!/4! =5 (! is factorial. 5! is 5x4x3x2x1)

win = 18/38 and loss = 20/38 or 0.472 and 0.526

5C1 x 0.472 x 0.526 x 0.526 x 0.525 x 0.526 (or 0.526^4)

=0.1806 = 18%

so the odds of getting just 1 black in 5 spins is 18%

2) odds of getting 2 blacks only

5C2 x 0.472^2 x 0.526^3

=  10  x 0.03242 = 0.3242 =32.42%

3) odds of getting 3 blacks only

=29%

4) odds of getting 4 blacks only

=13%

5) odds of getting 5 blacks

=0.472^5 = 2.41%

So....18%+32.42%+29%+14%+2.41% = 96%

BUT.....for a very close estimate, forget zero and use win 0.5 and loss 0.5 = 97%

[Nathan Detroit]

Mr Turner,
For simplification purposes:


Afeter 3  REDS  bet the next spin on Black. If you win that one  then the series is over.

If you had lost that first bet on nBlack you now bet  once more on Black. If you win then the series is over.
I
If you lost that bet you bet the  third time  on Black. Win or lose   the    series   is over..

The odds that a  6th Red could appear is 63  : 1.

Just  play it at  your  live local casino.

[Turner]

Thanks Nate..

I just saw that stat.....97%

Well...I believe nothing unless I see it for my self.

So I broke it down

[Turner]

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