The House Edge Explained in Double-Zero Roulette Games
Roulette is one of the most iconic casino games: simple to understand, visually appealing, and driven by straightforward odds. One key concept every player should understand before sitting at a double-zero (American) roulette wheel is the house edge. This article explains what the house edge is, how double-zero roulette creates a higher house advantage than single-zero variants, the math behind common bets, and what that means for your bankroll and play decisions.
What is the house edge?
The house edge is the long-run percentage of each bet that the casino expects to keep. It is a mathematical expectation, not a prediction about any single spin. If a game has a 5% house edge, the casino expects to retain about $5 for every $100 wagered over the long run. House edge is sometimes presented as “expected loss” or as RTP (return-to-player), which equals 1 − house edge. For example, a 5.26% house edge corresponds to an RTP of 94.74%.
Why double-zero roulette costs the player more
A double-zero roulette wheel (commonly called American roulette) has 38 pockets: numbers 1–36, plus 0 and 00. A single-zero (European) wheel has 37 pockets: 1–36 and 0. The extra 00 pocket increases the chance the casino wins on essentially every bet, which raises the house edge.
In most roulette bets, payouts are set as if there were fewer pockets (or no zeros). The difference between fair odds and the actual payout produces the house edge. In American roulette nearly every standard bet (straight-up, split, red/black, odd/even, dozens, columns, etc.) carries the same house edge of about 5.2632% because of the 2 zero pockets.
Basic math: how the 5.2632% is calculated
Take an even-money bet (red/black or odd/even). On an American wheel:
- Winning outcomes: 18 pockets (e.g., red)
- Losing outcomes: 20 pockets (18 black + 0 + 00)
- Total pockets: 38
Probability of winning = 18/38 ≈ 0.473684
Probability of losing = 20/38 ≈ 0.526316
If you bet $1, the expected value (EV) per spin is:
EV = (probability of win × payout for win) + (probability of loss × payout for loss)
For even-money, payout for win = +$1; for loss = −$1
EV = (18/38 × 1) + (20/38 × −1) = (18 − 20) / 38 = −2/38 ≈ −0.0526316
So the house edge is 2/38 = 0.0526316 = 5.26316%.
This same calculation applies to most other bets because the payouts are adjusted to the same (slightly unfair) structure. For example, a straight-up bet on a single number pays 35:1. The probability of hitting that number on American roulette is 1/38, so:
EV = (1/38 × 35) + (37/38 × −1) = (35 − 37)/38 = −2/38 = −5.2632%.
Exceptions and worse bets: the five-number bet
Most bets share the same house edge in double-zero roulette, but there is a notorious exception: the five-number bet (on 0, 00, 1, 2, 3), often available at American tables. It pays 6:1 but covers 5 numbers out of 38. The math:
Probability of win = 5/38
Payout = 6:1
EV = (5/38 × 6) + (33/38 × −1) = (30 − 33)/38 = −3/38 ≈ −0.078947
That is about a 7.8947% house edge — substantially worse than the usual 5.26%. Avoid this bet if you care about minimizing expected loss.
House edge versus volatility
House edge measures the average loss over the long run. Volatility (variance) measures how bumpy your results are in the short run. A single-number straight-up bet has the same house edge as an even-money bet in American roulette, but it is far more volatile: wins are rare but pay 35:1, losses are frequent. Even-money bets pay often but pay only 1:1. The long-term expectation (house edge) is the same, but your experience and bankroll swings will differ.
Practical examples: expected loss per hour
House edge lets you estimate expected losses. If you wager $10 per spin and play 50 spins per hour, your total wagers per hour are $500. With a 5.2632% house edge, expected loss per hour = $500 × 0.052632 ≈ $26.32. That is an average — actual results in any one session will vary widely.
Strategies and common misconceptions
- Betting systems cannot overcome the house edge. Progressive schemes like Martingale change bet sizing, but they do not change the underlying negative expectation; they only alter variance and increase the risk of catastrophic loss due to table limits or bankroll exhaustion.
- Choosing bets won’t change the house edge (except for the five-number bet). Whether you play inside or outside bets, the expectation per dollar wagered is generally the same on American roulette.
- The only ways to get a better expectation are: (a) find a European (single-zero) wheel or a table that offers la partage or en prison rules (these reduce house edge on even-money bets), or (b) find promotional offers, comps, or bonuses that offset part of the house edge.
- Bankroll management matters. Because roulette is high-variance, set bet sizes and loss limits consistent with what you can afford to lose. Decide on session lengths and stick to them.
Comparing single-zero vs double-zero
European roulette (single-zero) has 37 pockets, giving a house edge of 1/37 ≈ 2.7027% on most bets. That is roughly half the house edge of American roulette. If minimizing house edge is important to you, prefer single-zero games whenever possible.
Summary and takeaway
Double-zero (American) roulette carries a built-in house edge of 5.2632% on most standard bets because of the presence of both 0 and 00 among 38 pockets. This means for every $100 wagered you can expect, on average over the long run, to lose about $5.26. The five-number bet is worse (about 7.895% house edge) and should be avoided if you want to minimize expected loss. Remember that house edge is an average over many spins; short-term results can vary widely due to variance. No betting system changes the long-term math: the only practical ways to reduce expected loss are to choose single-zero wheels, prefer tables with favorable rules, or limit the total amount you wager.
