I just got back from Atlantic City and I have a nice roulette strategy.
The tables were a $15 minimum there, so this actually helps minimize losses.
First off, you wait until there are 4 consecutive colors in a row: 4 black/4 red.
In this example, let's say 4 blacks. 3 black, 0/00 works too, which is preferable. You place $20 on red, $15 on black.
If you win, great! You got $5. If you lose, you only lose $5.
If you lost, you then bet $25 on red, $15 on black. If you win, you get $10, which cancels out with the 5 you just lost, giving you $5 profit.
If you LOSE, you flat out bet $15 on red. If you win, you break even, if you lose, which is unlikely but still possible, you lose $30 instead of $45.
Now the reason this works is because with each consecutive result, let's say 4/5 blacks, mathematically, the odds are FOR it being red. To give a bit of insight: try flipping a coin and getting heads more than 5 times in a row.
I tried this and actually won $50 over the span of an hour.
Funny thing, and proof that it doesn't ALWAYS work: I lost $100 because of 14 blacks in a row. I did my steps above, took my $30 loss and walked away.
I came back 15 minutes later and the board was lined up with black numbers, so I thought "guaranteed red.." I put $30 on red, lost it. Next time, put all I had on red (kinda stupid, but then again, mathematically, the odds of it being black again are VERY slim to none).. black again.
After that, there were 3 more black balls.. so people were losing left and right. It was crazy.
In any case, my method really does get some nice results
