One for the Mega Minds on Standard Deviation/Normal Distribution

[Mr. Investable]

Hi Guys,

A quick question for the math genius' of the VLS forum;

The dozens or columns have 1:3 odds when played individually. When played as "Double Dozens" the odds are quite obviously 2:3. We can expect to see, approximately, 2 wins in 3 spins when applying the "Double Dozen" bet. One would naturally expect, approximately, 60 wins in 100 spins or 600 in 1000 spins. However, thanks to the beautiful random nature of Roulette the results simply are not that predictable.
According to the odds what is the range of Standard Deviation/Normal Distribution when playing "Double Dozen" bets over
A) 100 spins
B) 1000 spins
C) 10,000 spins
Even though I understand the process of Standard Deviation/Normal Distribution I am unable to do the math - F for flunk!
Thanks for the help guys. I have a very promising system on the drawing board and it has returned some attractive results. As soon as its ready I will more than gladly share. 
   

[Mr. Investable]

*EDIT*

66 in 100
666 in 1000

[Bayes]

Hi Mr I,

This is the formula -

W = np + z*sqrt(np(1 - p))

W =  wins
p = 0.67
n = 100, 1000, or 10,000

If you take z to be -3 (or +3 if you want to know the upper boundary), then this covers 99.7% of cases. But it can get better or worse than this, I would go to a worst/best-case scenario of +/- 5, although you may not ever see it.

Plug those numbers into the formula to get your answers. 

[Mr. Investable]

Hi Bayes,
     Thank you for the formula!    I plugged the numbers over a 1000 spin cycle and got 715    Is this correct and does this mean we can expect a "Double Dozen" win/loss hit rate to deviate by only 49 losses in 1000 spins (715-666)?
If this is true its very promising but I have a feeling I have just Flunked my Mr. Bayes math class 
If I have made a mistake, could you kindly point out my error and give me the three examples over 100/1000 and 10,000 spins?
My workings were:

n= 1000
p= .67
z= 3
np= 670
1-p= .33
(np(1-p))= 221.1
sqrt(np(1-p))= 14.86943173090351

So:

670+3*14.86943173090351 = 714.6082951927105 or 715

Thank you once again, 

[Mr. Investable]

 

...does this mean we can expect a "Double Dozen" win/loss hit rate to deviate by only 49 losses in 1000 spins (715-666)?...

I think I ment "can we expect the win hit rate to deviate from the mean by 49"

[Bayes]

670+3*14.86943173090351 = 714.6082951927105 or 715

That's ok as far as it goes, but you've only done half the calculation. To get the lower number (the least number of hits assuming a standard deviation of -3) you need to subtract from the average (all you've done is add to it). Remember, the average is in the middle of the bell curve, which is symmetrical.

670 - 3*14.86943173090351 = 625.

BUT, it can get worse!

Do the same calculations with z = +/- 4.0. That will give you a more realistic idea of what to expect.

[Mr. Investable]

Thanks for your help Bayes! I have been working on a system and its producing some exciting results. Its not perfect and needs more work but as you can see from the graph it certainly has potential! I managed to programme the basic mechanics into Roulette Xtreme which allowes me to test the variables over thousands of spins at the click of a button.
Its a math based system that can be applied in Land Based Casino's but its better suited to Online Casino's especially RNG. It exploits the mathematically probability a random sequence, restricted by parameters, can not avoid 
Keep in touch and I'll keep you updated!
Peace

[Mr. Investable]

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