Players float roulette systems past me several times a year. The variety of wagers available opens the door to creating combination bets, with players hoping different bets will cancel out each other’s weaknesses.
It doesn’t really work that way. The house edge remains intact, though there’s no real harm in trying as long as you stay within a budget you can afford.
One simple combination was forwarded to me in April:
“I bet $10 on black and $1 each on five different red numbers,” the player wrote. “That way, I have 23 numbers that win and only 15 numbers that lose. I win $10 on each of the 18 black winners and $35 on each of my reds. That’s $255. I lose $15 on each of the 15 losers, and that’s only $225.
“It looks to me like I should win $30 more than I lose.”
What the player was overlooking was that on each winner, the other bet loses. On 18 black winners, the $10 bet is paid $10, but there’s a $5 loss on red bets. That leaves the profit on each of those spins at $5, not $10. Similarly, on red winners, the black bets lose, so the profit on those spins is $25, not $35.
Let’s look at 38 spins of a double-zero wheel in which each of the 38 numbers comes up once. You risk $15 per spin, so you have a total of $570 in wagers.
On each of the black winners, you keep the $10 bets plus you’re paid $10. For 18 winners, that puts $360 on your side of the table.
On each of the red winners, you keep $1 and are paid $35, and that leaves another $180 with you.
So at the end of the trial, the $360 from your black winners and the $180 from the red single numbers leave you with a total of $540. Since your wagers totaled $570, that leaves $30 with the house.
Divide that $30 by the $570 risked, then multiply by 100 to convert to percent, and you get 5.26 percent, the normal house edge on double-zero roulette. The combination wins on more spins than it loses, but the house still collects its normal percentage.