Getting the Roulette Edge

2 July 2017

By John Grochowski

Questions readers ask most frequently go in cycles.

For the first couple of years I wrote this column, when legalized gaming was new to much of the country, “How do I get a casino job?” was top of the list. Soon that was replaced by “How do I beat slots?” and then by strategy questions on video poker and blackjack.

One that’s always been part of the mix has been, “How does the casino get its edge?” That’s the question that’s top of the charts right now. In the last month, I’ve received more than a dozen emails asking that question for a variety of games.

So let’s spend the next couple of weeks answer that question. For some veteran players, this is treading old ground, and I apologize for that. But for those who’ve emailed and those who’ve wondered without asking, let’s start with roulette.

Regardless of whether you’re betting single numbers, 18 numbers at a time (red/black, odd/even, first 18/last 18) or almost anything in between, roulette pays at odds that would yield an even game if there were 36 numbers on the wheel.

But there aren’t. There are 38 numbers on a double-zero game and 37 numbers on games with only one 0.

If you win a single-number bet, it pays at 35-1 odds. But on the double-zero wheels that most U.S casinos use, the true odds are 37-1, so the house gets an edge by paying less than true odds.
Pretend you were at a wheel with just numbers 1 through 36 and no zeroes, and you bet $1 on No. 17 on each spin. If you did that for 36 spins and each number came up once. You’d risk a total of $36. On the one winning spin you’d be paid at 35-1 odds, so you’d get $35 in winnings and keep your $1 wager.

So at the end of the run you’d have wagered $36 and you’d still have $36. Winners are paid at true odds and there’s no house edge.

We can try that run again for an 18-number bet such as red. On a 36-number wheel, there would be 18 red numbers and 18 black, making the true odds 18-18, or 1-1. An even-money payout would yield no house edge. If you bet $1 on red for 36 spins, you’d have 18 winners. On each win, you’d keep the $1 wager and get $1 in winnings, so at the end of the run you’d have $36, the same amount you risked.

But with 0 and 00, there are 38 numbers. Now, to have each number occur once, our imaginary run has to be 38 spins and you have to risk $38 instead of $36.

On a winning single-number bet, you’re still paid 35-1, and you still have $36 on your side of the table when the trial ends. But this time, the house has kept $2 of the $38 in wagers. Divide the $2 that goes to the house by the $38 you risked, then multiply by 100 to convert to percent, and you have a 5.26 percent house edge.

Same deal with 38 bets on red. This time there are 18 red numbers, 18 blacks and two greens – the 0 and 00. True odds against winning no longer are 1-1, they’re 20-18, or 1.11-1. At the end of the trial, you still have $36, but that’s $2 less than the $38 you’ve risked – the same 5.26 percent edge.

We could walk through this for any bet on the layout, but it comes down to the same thing. The house pays less than true odds.

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