Sep

27

2017

# How Do You Win More Than You Lose in Roulette?

There are so many possibilities in roulette, it’s easy to design a system that wins more often than it loses.

Problem is, when the losses come, they’re much larger than the wins and wipe out any profit – and then some.

One simple method in American roulette with both 0 and 00 is to bet \$1 each on 20 single numbers. Any winner pays 35-1. So you risk \$20 and if any of your numbers hits, you have \$36, since you keep your \$1 wager along with the winnings.

That’s a profit of \$16 per win and you’ll average 20 wins per 18 losses. However on each of the 18 losses, you lose the full \$20.

Per 38 spins, you’ll risk \$760 and at the end you’ll have \$720. The house will have \$40 – 5.26 percent of your wagers, the normal house edge at 00 roulette. The losses wipe out your wins and give the house its percentage

Another easy way is to bet \$6 on the first 18 and \$5 on the last dozen. Now you have 30 numbers covered and you’ll make a profit on 79 percent of spins.

Your risk is \$11 per spin. If the ball lands on any of the first 18, you win \$6 and keep the \$6 bet for a total of \$12 and a \$1 profit. If the number is anything from 25 through 36, you’re paid 2-1, so you have a \$10 payoff and a \$5 bet for \$15 and a \$3 profits.

The house takes your full \$11 on numbers you don’t have covered. At the end of the trial, you have \$396 of your \$418 in wagers, and the house has \$22 – or 5.26 percent.

The possibilities are practically boundless. There are many, many possible systems in which you win more often than you lose. But because the losses are bigger than the wins, the systems don’t really help you.

You can run from the house edge. You might even be able to hide from it for a little while. But in the end, it will claim its share.