Roulette Systems

[BrianK]

Originally Posted by Pacman

Yes I realize it works in some respect, but it doesn't because eventually black will hit ten times in a row and you'll loose how ever many grand you suggested which would be ridiculous because you'll be betting $1,000 to try to win $10...

I suggested $2550, because 10 + 20 + 40 + 80 + 160 + 320 + 640 + 1280 = 2550. At that point if you happened to have another $2560 on you, could could double up one more time. But then if you lost, you couldn't double up anymore because table limit is $3000. That's where I came up with that number. It's doubling up.....anyway see below.

Originally Posted by guybrush

For every dollar you bet you'll lose around 5% of it. You can't change that by betting double the next time. The only thing you can change with a betting strategy is your variance but never your odds.

I realize you can't change the odds. The odds are always in the house's favor, this strategy is essentially making a new game out of it. If you play over a bunch of spins of the wheel and the board reads this:

B
R
R
R
B
R
R
B
R
B
B
B
R
B
R
B
R
R
R
B

You simply count up the "B's", and you multiply that number by 10, and that is your profit. In the series above you just made $90.

The only time you lose, is when there are 9 "R's" in a row, and in that case you lose it all.

It's not changing the odds in your favor, no. It's just a different way of playing the game. Everytime black hits, you win $10 more dollars, everytime red (or 0 or 00) hits it doesn't matter. Only 9 losses in a row is an actual loss.