Roulette Martingale & Betting System Simulator

Test Martingale, Fibonacci, D'Alembert, and Labouchère across 10,000 simulated roulette sessions in your browser. Set your starting bankroll, base unit bet, table maximum, and wheel geometry (Single-Zero 2.70%, Double-Zero 5.26%, or Triple-Zero 7.69%) to see the exact bust rate, average net loss, and sample bankroll trajectories.

Progression system Martingale (Double After Every Loss)Fibonacci (Step Up 1, 1, 2, 3, 5, 8, 13...)D'Alembert (+1 Unit on Loss, −1 Unit on Win)Labouchère / Cancellation ([1, 2, 3, 4] Line) Wheel geometry European Single-Zero (0) · 2.70% edge American Double-Zero (0, 00) · 5.26% edge Triple-Zero (0, 00, 000) · 7.69% edge Starting bankroll ($) Base unit bet ($) Table max bet ($) Spins per session 50 spins (~1 hour) 100 spins (~2 hours) 200 spins (~3.5 hours) 500 spins (weekend trip)

Run 10,000 Sessions

10,000-Session Monte Carlo Results

Bust rate (hit limit or broke)

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Sessions finishing ahead

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Average net result / session

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Theoretical house-edge loss

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Realized house edge

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Average total wagered

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Median session outcome

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Worst session loss

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Avg. spins survived

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Closed-Form Martingale Ceiling for Your Settings

Max consecutive losses

6 straight losses

Peak survivable bet

$320

Loss when streak hits

$630

Bust prob. per cycle (q^L)

2.126%

Bust prob. in 100 cycles

88.3%

Bust prob. in 250 cycles

99.5%

Five sample session bankroll trajectories from the current 10,000-session run. Notice the characteristic Martingale sawtooth: steady upward steps of +1 base unit punctuated by vertical drops when a losing streak exceeds the bankroll or table cap.

The Exact Mathematics of the Martingale Progression

Under the classic Martingale on an even-money bet (Red, Black, Even, Odd, 1–18, or 19–36), you start with a base wager b₀ (say $10) and double your bet after every loss: b_k = b₀ · 2^k after k consecutive losses. The first time you win at step k, your winning bet of b₀ · 2^k recovers all prior losses in the cycle:

Cumulative prior losses = b₀(1 + 2 + 4 + … + 2^(k−1)) = b₀(2^k − 1)

and leaves a net profit of exactly +b₀ ($10). If you had an infinite bankroll, no table limit, and infinite time, every cycle would eventually terminate with +$10. In a real casino, two hard ceilings stop the doubling sequence at a finite step L: your remaining bankroll and the table maximum bet.

Why a Single-Cycle Expected Value Is Always Negative

Suppose a player brings $630 to an American double-zero table ($10 base bet, $500 table max) so they can survive up to L = 6 consecutive losses ($10, $20, $40, $80, $160, $320). On each spin, the probability of losing an even-money bet is q = 20/38 ≈ 0.526316.

• Probability of busting the 6-step cycle: q⁶ = (20/38)⁶ ≈ 2.1256% (about 1 in 47 cycles), losing −$630.
• Probability of winning the cycle: 1 − q⁶ ≈ 97.8744%, winning +$10.
• Expected value of one cycle: (0.978744 × +$10) + (0.021256 × −$630) = +$9.7874 − $13.3911 = −$3.6037 per cycle.

Notice that −$3.6037 divided by the average total amount wagered per cycle ($68.47) is −5.2632% (−2/38)—the exact house edge of the American wheel, down to the last decimal place. Changing the bet sizing from flat betting to Martingale, Fibonacci, D'Alembert, or Labouchère never alters the ratio of expected loss to total dollars wagered; it only increases the average amount you wager per spin and concentrates your losses into tail events.

Comparing the Four Classic Progression Systems

Frequently asked questions

Why does the Martingale system fail in roulette?

Because doubling your bet after every loss grows exponentially (1, 2, 4, 8, 16, 32, 64, 128...) while your profit on a completed cycle is only 1 base unit. At a $10 minimum / $1,000 maximum table with a $1,000 bankroll, you can only survive 6 consecutive losses ($10 + $20 + $40 + $80 + $160 + $320 = $630; step 7 requires $640, exceeding your remaining $370 bankroll). On an American double-zero wheel, 6 losses in a row happens with probability (20/38)^6 = 2.13% per cycle, wiping out 63 winning cycles in a single loss.

Does changing the progression to Fibonacci, D’Alembert, or Labouchère beat the house edge?

No. By linearity of expectation (and Doob’s Optional Stopping Theorem), a sum of negative-expectation bets always has negative expectation equal to the house edge times the average total amount wagered: E[Net] = −edge × E[Total Wagered]. On an American double-zero wheel, every progression system loses 5.26% of the total dollars put across the felt; on European single-zero, 2.70%; and on a triple-zero (000) wheel, 7.69%.

Why do so many Martingale sessions finish with a small profit?

Martingale trades a high probability of a small gain for a small probability of a catastrophic loss. In a 200-spin session, roughly 70% to 85% of sessions survive without hitting the fatal losing streak and finish up $50 to $150, while the 15% to 30% of sessions that hit the streak lose their entire bankroll. When you average the many small winners and the few total wipeouts across 10,000 sessions, the average net result is always a loss equal to the house edge times total action.

Why do casinos put maximum bet limits on roulette tables?

Table limits protect the casino from extreme single-spin variance, not from positive-EV progressions. Even without a table limit, every player has a finite bankroll, and a finite bankroll acts as an exact mathematical cap on how many times a wager can be doubled.

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