Mar

30

2015

# Placing Bets In Roulette

There are so many different wagering options at roulette that it’s only natural that players would try to come up with combinations to try to overcome the house edge.

The reality is that no combination can change the math of the game. You can have winning streaks and winning sessions. Without them, we wouldn’t play the game. But over time, the odds will drive the overall results in the house’s favor.

One system that came to my attention recently was from a reader who likes to bet the columns. Those familiar with the roulette layout know that other than the 0 and 00 at the top, it’s a grid with 12 horizontal rows of three numbers each, which can also be read vertically as three columns of 12 numbers each.

The reader’s system was to bet two columns at once. That covers 24 numbers at once, and if any one of them comes up, you get a 2-1 payoff on one bet. For example, if you make two \$5 bets and the ball lands in one of your numbers, you lose one \$5 bet, but you keep the other and get \$5 in winnings besides. That’s a \$5 net profit for your \$10 in wagers.

The house edge on betting columns is 5.26 percent on a double-zero wheel. That’s the same house edge as you face on single numbers, two-number splits, dozens, red or black, or almost any other wager. The exception is the five-number bet on 0, 00, 1, 2 and 3, where the house edge is 7.89 percent.

There are 38 numbers on the wheel, including 0 and 00, so while you’re covering 24 numbers, there still are 14 on which you lose both of your column bets. Winners pay 2-1, but when you win, you lose your bets on the opposite column. The bottom line is that for every \$38 you wager, you lose \$2, or 5.26 percent.

Betting so many numbers at once smooths out the rough spots. You’ll have fewer long losing streaks than if you bet fewer numbers. It also tamps down the high points, since you’re guaranteed that at least one bet is a loser on every spin. It doesn’t hurt you in terms of the house edge, but it doesn’t help, either.