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Betting on the zeroes isn't backing house side

Betting on the zeroes isn't backing house side

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Among the seemingly countless myths against gambling, some die harder than others.

Take the myth of roulette players betting with the house. There's no such thing, but the myth has used up far more than nine lives and still has believers.

Let's let a reader describe the myth. Calling himself "Mr. Zero," he emailed to ask, "Everybody says the house gets its edge from the zeroes. No zeroes, no edge, right? So why not play on the side of the house? Shouldn't everybody bet the zeroes, or would the volume of zero bets put casinos out of business?"

The zeroes are important components of the house edge, but if you bet on them, you're not betting with the house. You face the same edge as on any other number.

The real source of the house edge is that roulette pays winning bets less than true odds. That's how the house gets its edge on every game. It collects full fare on losing bets, but pays less than true odds on winners. As long as the game plays according to normal odds and approaches the mathematical expectation of wins and losses, that will lead to a profit for the house.

I've written about this before, but the myth is so hardy readers keep coming back to it. I usually demonstrate by showing house edge calculations for single-number bets and for 18-number bets such as red or black. The house edges are the same: 2.7 percent on a single-zero wheel, 5.26 percent with both zero and double-zero, and 7.69 percent on the few wheels that have added a triple-zero.

Just to be different, let's try it with a two-number split bet such as 17-18, 23-26 or any two numbers that are adjacent either horizontally or vertically on the table felt. Heck, on a double-zero wheel, you could do it with 0-00.

On any split, if the ball lands on either of your numbers, you're paid at 17-1 odds

If there were only 36 numbers on the wheel, 17-1 would be true odds and there would be no house edge. If you bet $1 on a 17-18 split on each of 36 spins in which each number came up once, you'd win twice. On each win, you'd keep the $1 bet and get $17 in winnings. At the end of the trial, you'd have your fill $36 and the house would have nothing. House edge: Zero.

Try that again while adding a single zero. Now there are 37 numbers, so you risk $37. Your two wins still leave $36 on your side of the table, but this time the house keeps $1. That's its edge. When you divide $1 to the house by $37 wagered, then multiply by 100 to convert to percent, you get 2.7 percent.

Now try it with 0 and 00. That brings the wheel to 38 numbers, and your risk on 38 spins is $38. The two wins once again give you $36. The house keeps $2 and the calculation shows a 5.26 percent spin.

With 0, 00, and 000, there are 39 numbers. A sequence in which each shows up once will cost $39 in $1 wagers, and the two wins leave you with $36. The house keep $3, or 7.69 percent.

The house grows with each zero added, but you don't put yourself on the house side by betting on zeroes or including zeroes in your splits. The house still pays your winners less than the true odds of winning, and the house maintains its edge.

Look for John Grochowski on Facebook and Twitter (@GrochowskiJ).

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