The search for the perfect roulette system

10 September 2017

By John Grochowski

The first time I ever played roulette, I kept it simple. It was a $5 minimum table, and I did nothing but bet $5 on red or $5 on black for the hour or so I played.

I was alone in my one-bet method. Everyone else at the full table was spreading chips all over the layout, looking for that perfect combination.

That search is both eternal and futile. There is no such combination that will swing the odds in favor of the player.

However, a reader found something I once wrote saying it was easy to design a combination that wins more than it loses.

“Would you please explain a little more,” he wrote. “I’d like the feeling of winning more than I lose.”

With a caution that losses are larger than wins in such systems, you win more often than you lose if you cover 20 or more numbers with bets proportioned so that any win is larger than your total bet.

One simple way 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% of your wagers, the normal house edge at 00 roulette.

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% 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 5.2%.

At Michael Shackleford’s wizardofodds.com, a dealer asked about something that looks like the above system on steroids.

The dealer said he had seen players betting $75 on the first 18, $50 on the third dozen and $10 on the 0-00 split, the player profits on 32 of the 38 numbers – there are only six losing numbers.

Wagers on each spin total $135. The first 18 pays even money, the third dozen pays 2-1 and the split pays 17-1.

So when any of the first 18 hit, you win $75 and keep your $75 wager for a total of $150, making a $15 profit. If any of the third dozen hits, you win $100 and keep your $50 bet, again showing a $15 profit. When 0 or 00 hit, you win $170 and keep the $10 bet for a $45 profit.

That all sounds good, but when reality hits, the six losers each cost you $135.

Per 38 spins, you risk $5,130. Your wins plus the bets you keep total $4,860. The house has $270, or 5.26% of wagers.

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.


Look for John Grochowski on Facebook (http://tinyurl.com/7lzdt44) and Twitter (@GrochowskiJ).