Placing opposite bets results in loss
john grochowski firstname.lastname@example.org May 23, 2012 4:04PM
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Updated: July 3, 2012 8:49AM
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: Not quite. The red/black bets are outside bets, and satisfy the $5 minimum. However, the bet on the zeroes is an inside bet, and inside bets are a separate $5 minimum. If you bet $1 on the zeroes, you still have to 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 percent it is as 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. Same deal on each of the 18 red numbers.
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.
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 percent of his wagers, the normal house edge at double-zero roulette.
If Eric wants the zeroes as his “real” bet, the house edge is the same if he just bets them on their own.
John Grochowski is a local freelance writer. Look for him on Facebook (http://tinyurl.com/7lzdt44); Twitter (@GrochowskiJ) and at casinoanswerman.com.