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Ralph Scalitzia · Junior Member Joined: 14 Mar 2003 Posts: 3

Dear Sir,
In a single zero roulette game, the PROBABILTY of winning increases if you place a portion of your money on fewer numbers for more spins versus covering more numbers per spin, an example:
If you are willing to risk 500$ in order to win 250$ then you could:
Option (A): place 250$ on any of two dozen and should you be a winner you will win 250$. The probabilty of that happening is 24/37=(.648648).
Option (B): Place 125 on any one dozen and should you be a winner you will win 250$ and walk away. However, should you lose you can now bet187.5$ on the same dozen and should you be a winner you will win 375$ which will get you the 250$ and the 125$ you lost on the previous spin. Now should you lose on both spins you still have 187.5$ to play with and you can place 20.833333$ on any nine numbers and should you be a winner you will get 750$ which is equal to your 500 original capital plus 250$ in winning which was your goal. The probability of that happening meaning either hitting a dozen OR nine numbers at LEAST once in three spin is equal to[1-(25/37)x(25/37)x(28/37)]=0.65451.
Hence, for the SAME capital and for the SAME payoff you are able to increase your PROBABILITY of success as in option (B) if you play fewer numbers with less money but for MAYBE more spins.(As you might win on the first spin)
You can even improve your probability further if you play only six numberes at a time and try to win 250$.
Any Explaination??!!!!

casinosurfer · Junior Member Joined: 01 May 2003 Posts: 26

Has anyone tried this system? Any successful examples?

Might want to apply this to my game play.

Desparate to win... (^_^)