Down The Road

Playing Opposites in Roulette

By on Monday, June 06, 2011

On a busy night in a crowded casino, I sat at a coffee shop counter and overheard a man and a woman planning their roulette attack. One would bet red, the other black. Wins would balance losses. They wouldn’t lose any money, but they’d get their comps for free. Or so they thought. They’d neglected to take into account that whenever the spin was 0 or 00, they’d lose BOTH bets. Let’s say the two of them were each betting $5 on red, for $10 at risk on each spin of the wheel. In a perfect sequence of 38 spins in which each number — 1 through 36, along with 0 and 00 — occurred once, they would risk a total of $380. On the 18 red numbers, they’d collect $180 in winnings plus keep their $180 in wagers. So at the end of the sequence, they’d have $360, and the house would have have $20 of the original $380. Now let’s say one wagers $5 on red and the other wagers $5 on black. On 36 of the 38 numbers, the red and black wagers would cancel out. One bet would win, the other would lose.  That leaves two more numbers, 0 and 00. On those, both red and black lose, meaning the couple loses $10 on 0 and $10 more on 00. That’s a total of $20 in losses. At the end of our 38-spin sample, the couple has $360 of its original $380 in wagers, and the house has $20, the same as if they’d bet opposite sides. Those $20 in losses from $380 in wagers comes to 5.26 percent in the casino coffers. Yep, the same house edge we’ve been talking all along.  Sorry, no free comps here.

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