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Main => Full Roulette Systems => Topic started by: Norman Bates on January 08, 2012, 04:41:40 PM

Title: 2 to 3/3 to 2. A $1,500. Dollar Domination System
Post by: Norman Bates on January 08, 2012, 04:41:40 PM
Flat Bet American Roulette System
:skull:
Using The Same Spins Each Time, And Work Your Way Through Them  Using The, "Buy-In" Listed Under Each Example As A Guide.  For Example; When You Reach The, "The Buy-In" Amount For The Example Below The One You Are Working On, Keep Going Until You've Won However Many Chips You Think Is Possible.  When Your Chip Count Starts To Back Down Again, You Can Either Stop There, Or Continue On To The Next Example Below The One You Are Working On, And Use That Example To Wager With.

:skull:
1st Example:
3 chips on numbers 0, and 00
5 chips on line (1-00), and corner (2:00)
16 chips on numbers 5, and 8
20 chips on line (4-9)
36 chips on line (10-15)
104 chips per spin 1,600 chip buy-in.
:skull:
2nd Example:
6 chips on numbers 0, and 00
8 chips on line (1-00), and corner (2:00)
28 chips on numbers 5, and 8
38 chips on the the 1st dozen
66 chips on street (10-12)
188 chips per spin, 2,850 chip buy-in.

10 chips on numbers 0, and 00
15 chips on line (1-00), and corner (2:00)
50 chips on numbers 5, and 8
67 chips on the 1st dozen
117 chips on line (10-15)
334 chips per spin, 5,000 chip buy-in.
Thank You For Your Patience.
Title: Re: 2 to 3/3 to 2
Post by: Mr J on January 08, 2012, 06:09:22 PM
I like your 4th example. So even at $1 per chip, its a $5,800 buy in.  :suicide:

Ken
Title: Re: 2 to 3/3 to 2
Post by: pins on January 08, 2012, 06:22:41 PM
you are only backing the first dozen.  and 0-00; whats the point.
Title: Re: 2 to 3/3 to 2
Post by: Norman Bates on January 08, 2012, 06:39:23 PM
I just thought that I should allow 18 spins, so I just multiplied the number of chips per spin by 18.  Is that A good Idea, or do you think I should divide it by 2, and multiply that by 3?
Title: Re: 2 to 3/3 to 2
Post by: Mr J on January 08, 2012, 06:42:30 PM
I like your method but this is what I would do >>


Since P(Win)  =  1/19 and a win results in 1700, we multiply

        (1/19)(1700)  =  1700/19

To find the probability of losing, just subtract the probability of winning from 1 to get

        P(Lose)  =  1 - P(Win)  =  1 - 1/19  =  18/19.

Losing results in winning -100, so we multiply

        (18/19)(-100)  =  -1800/19

finally add these partial results to get

        1700/19 + (-1800/19)  =  -100/19  =  -5.26

Title: Re: 2 to 3/3 to 2
Post by: bombus on January 08, 2012, 06:53:28 PM
Quote from: Mr J on January 08, 2012, 06:42:30 PM
I like your method but this is what I would do >>


Since P(Win)  =  1/19 and a win results in 1700, we multiply

        (1/19)(1700)  =  1700/19

To find the probability of losing, just subtract the probability of winning from 1 to get

        P(Lose)  =  1 - P(Win)  =  1 - 1/19  =  18/19.

Losing results in winning -100, so we multiply

        (18/19)(-100)  =  -1800/19

finally add these partial results to get

        1700/19 + (-1800/19)  =  -100/19  =  -5.26

lol... Bloody mathboy!
Title: Re: 2 to 3/3 to 2
Post by: insidebet on January 08, 2012, 11:53:35 PM
Mr. J

I understand perfectly  where you re going with this professorial post.
But this guy needs a prescription renewal, not a math lesson.

Insider
Title: Re: 2 to 3/3 to 2
Post by: Proofreaders2K on January 09, 2012, 10:57:38 AM
Crazy enough to work?

I wonder what would happen if all of Norman Bates' systems were put together in an Ophis MST style tracker (we could choose the chip amounts) with a positive progression of +1 on a win/less 2 on a loss?