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You can't escape the house edge

29 May 2012

When casinos offer bets on opposite outcomes of a play, they always leave a clear path to a house profit.

In craps, pass and don't pass don't quite offset, because a loser 12 on the pass line is just a push rather than a winner in don't pass. In baccarat, player wins less often than its opposite, banker, and the house preserves an edge by collecting a commission on winning banker bets. In roulette, betting red and black at the same time isn't a breakeven method because both bets lose whenever a green 0 or 00 turn up.

That doesn't stop players from scheming to find that one opposites combination that will turn the tables.

The latest comes from a reader who signed himself Eric: "For roulette, let's say you bet \$5 (table minimum) each time on both red and black and \$1 each time on green. Obviously, the red and black bets are washes and the dollar on green is your real bet. Is this allowed?"

Short answer: No. The \$5 minimum applies separately to outside bets, such as red or black, and inside bets, such as the zeroes and other numbers. If you bet \$1 on the zeroes, you still have bet another \$4 on inside numbers.

Even if you could make that combination, it doesn't help you. Red and black aren't really washes, and the house edge is the same 5.26% it is on each bet in the combination.

Imagine a sequence of 38 possible spins in which each number comes up once. Eric's system risks \$11 on each spin -- \$5 on red, \$5 on black and a \$1 split bet that covers both zeroes. The total risk for 38 spins is \$418.

On each of the 18 black numbers, he keeps his \$5 wager on black, and gets \$5 on winnings for a total of \$10.

It's the same deal on each of the 18 red numbers. Chalk up another \$180 in his stack.

That leaves the two green numbers, 0 and 00. Two-number splits pay at 17-1 odds, so on each of the greens he keeps his \$1 bet and gets \$17 in winnings for an \$18 total. Multiply that by the two winners, and that's \$36.

Add all that up, and Eric has \$396 of his original \$418, and the house has the other \$22. And that \$22 is 5.26% of his wagers, the normal house edge at double-zero roulette.

If Eric wants the zeroes as his "real" bet, he's at least as well off to bet them while skipping the red and black. The house edge isn't any lower, but at least he can bet less money per spin and limit his losses.

He specified a table minimum of \$5, so let's say he made a \$5 split on 0 and 00 for each spin, for a total risk of \$190 per 38 spins. That's less than half the amount Eric wagers in his system.

Now he loses his full \$5 on each of the 18 red numbers, and on each of the 18 blacks. On the 0, he keeps his \$5 bet and wins \$85 on the 17-1 payoff. Same thing on the 00: He keeps his \$5 bet and wins \$85. Total that up, and at the end of the 38-spin trial he has \$180, and the house has kept \$10.

Divide \$10 in losses by \$190 risked, then multiply by 100 to convert to percent, and we have that familiar figure, the 5.26% house edge on double-zero roulette.

In terms of the house edge, it doesn't really matter if Eric bets red and black to go with his wagers on the green zeroes. What does matter is that he's betting more money his way, and that leads to larger long-term losses.

ANOTHER OPTION: Bets with even-money payoffs are less volatile than split bets such as 0/00. The 17-1 payoff on split bets can lead to some nice wins in a short streak where they show up a little more often than normal. However, they win an average of only once per 19 spins, and a small shortfall in frequency can deplete the bankroll rapidly.

Eric's system would leave a less volatile game, since most of his money would be on even-money bets. However, his \$1 bet on the greens can't stand on its own on a table with \$5 minimums.

How could he get the desired effect? He could place six \$1 double streets, with \$1 on 1-2-3-4-5-6, \$1 on the next six numbers, and so on. Along with his split bet on the 0 and 00, he'd have a total of \$7 in action on each wager.

The double streets pay 5-1, so after each red or black number, Eric would still have \$6 of his \$7 wagered. On each 0 or 00, he'd have \$18 for a \$11 profit.

That would leave a less volatile game with nice wins on the zeroes. The house edge? In no surprise, it's 5.26%.

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Best of John Grochowski
John Grochowski

John Grochowski is the best-selling author of The Craps Answer Book, The Slot Machine Answer Book and The Video Poker Answer Book. His weekly column is syndicated to newspapers and Web sites, and he contributes to many of the major magazines and newspapers in the gaming field, including Midwest Gaming and Travel, Slot Manager, Casino Journal, Strictly Slots and Casino Player.

Listen to John Grochowski's "Casino Answer Man" tips Tuesday through Friday at 5:18 p.m. on WLS-AM (890) in Chicago. Look for John Grochowski on Facebook and Twitter @GrochowskiJ.