Martingale calculator
- Bets you can cover
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- Loss if a run busts
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- Chance a run busts
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- Expected value per run
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- Chance of at least one bust
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Key facts
- A Martingale run wins one base bet or loses every bet in the sequence: with room for n bets, the loss is base x (2 to the power n, minus 1).
- In our calculation, a $5 base and a $2,000 bankroll covers 8 doubling bets. A run busts 0.48% of the time (about 1 in 207) and costs $1,275.
- Over 100 runs at single zero, the chance of at least one bust is 38.4%. One bust wipes out 255 winning runs.
- Every run loses 2.70% of the money it puts at risk on a single zero wheel, the same as flat betting. Doubling cannot change the house edge.
How to use the Martingale calculator
Enter your base bet, then limit each run either by your bankroll or by the number of bets in a row you are prepared to make. Pick the chance of winning each bet: an even-money bet on single zero roulette (18 in 37), on double zero (18 in 38), a fair coin for comparison, or a custom chance. Add how many runs you plan to play. The calculator lists every bet in the doubling sequence and shows the loss if a run busts, the chance that happens, the expected value per run and the chance of at least one bust over all your runs.
A run is one cycle of the system: it ends at the first win, or when you cannot cover the next doubled bet. When you limit by bankroll, the calculator works out the most bets your bankroll covers. A table maximum works the same way, so if the table limit stops you before your bankroll does, enter the number of bets the limit allows instead.
The Martingale formulas
With a base bet B, room for n bets and a chance p of winning each bet (so q = 1 - p of losing):
- Bet number k = B x 2 to the power (k - 1)
- Loss if the run busts = B x (2 to the power n, minus 1)
- Chance a run busts = q to the power n
- Expected value per run = B x (1 - q to the power n) - loss x q to the power n
- Bets you can cover with bankroll W = the largest n where B x (2 to the power n, minus 1) is no more than W
- Chance of at least one bust in R runs = 1 - (1 - q to the power n) to the power R
These are the same lines the calculator runs, and every figure on this page was checked against them in python.
Worked example: $5 base, $2,000 bankroll, single zero
Your bankroll of $2,000 covers 8 bets, because the ninth would take the total to $2,555. In our calculation:
| Bet | Stake | Total lost if this bet loses | Chance you reach this bet |
|---|---|---|---|
| 1 | $5 | $5 | 100% |
| 2 | $10 | $15 | 51.35% |
| 3 | $20 | $35 | 26.37% |
| 4 | $40 | $75 | 13.54% |
| 5 | $80 | $155 | 6.95% |
| 6 | $160 | $315 | 3.57% |
| 7 | $320 | $635 | 1.83% |
| 8 | $640 | $1,275 | 0.94% |
- The run wins $5 with a chance of 99.52% and loses $1,275 with a chance of 0.48%, about 1 in 207.
- Expected value per run: 0.9952 x $5 - 0.0048 x $1,275 = -$1.19.
- Average amount put at risk per run: $44.00. And $1.19 is 2.70% of $44.00, exactly the single zero house edge.
- Over 100 runs, the chance of at least one bust is 38.4%. Making it through all 100 runs would net $500, and one bust costs $1,275, so a single bad run wipes out 255 good ones.
These are the calculator's default inputs. Switch the chance to 18 in 38 and the same 8 bets bust about 1 in 170 runs, the expected loss rises to $2.54 per run, and the chance of a bust in 100 runs climbs to 44.6%.
A bigger bankroll changes the timing, not the result
The usual answer to a bust is a bigger bankroll. Here is what that buys with a $5 base at single zero, in our calculation:
| Bankroll | Bets covered | Loss on a bust | Chance a run busts | EV per run | Chance of a bust in 100 runs |
|---|---|---|---|---|---|
| $100 | 4 | $75 | 6.95% (1 in 14) | -$0.56 | 99.9% |
| $500 | 6 | $315 | 1.83% (1 in 55) | -$0.87 | 84.3% |
| $1,000 | 7 | $635 | 0.94% (1 in 106) | -$1.03 | 61.2% |
| $2,000 | 8 | $1,275 | 0.48% (1 in 207) | -$1.19 | 38.4% |
| $5,000 | 9 | $2,555 | 0.25% (1 in 403) | -$1.36 | 22.0% |
| $10,000 | 10 | $5,115 | 0.13% (1 in 784) | -$1.53 | 12.0% |
| $20,000 | 11 | $10,235 | 0.07% (1 in 1,527) | -$1.70 | 6.3% |
Each doubling of the bankroll adds one bet and roughly halves the bust risk, but the bust itself doubles in size and the expected loss per run goes up, because a deeper sequence puts more money at risk on average. A $20,000 bankroll makes busts rare; it does not make them small. You are still risking $10,235 to win $5 at a time.
Pundit tip: run the calculator with the fair coin option. At 50% the EV is exactly zero, which shows the system adds nothing: all of the loss you see on a real wheel is the house edge.
Why the average never moves
The Martingale changes the shape of your results, not their average. With the fair coin option the calculator shows an EV of $0.00 for any bankroll; on a real wheel the loss per run is always 2.70% (single zero) or 5.26% (double zero) of the money put at risk. That is what every system that chases losses does, as our guides to the Martingale strategy and roulette strategy explain, and the belief that a win is "due" after a run of losses is the gambler's fallacy. The expected value calculator shows the same edge on a single bet, and roulette odds lists the chance behind every bet on the layout.
If you play roulette, the safest version of this calculator is the one that tells you the bankroll is money you can afford to lose in full. Set that limit before you start; our guide to bankroll management covers how.
Roulette and the Martingale in Australia
Roulette is legal in person at licensed Australian casinos, where state regulators approve the game rules; in Victoria the VGCCC approves the rules of table games at the Melbourne casino. Online roulette is a different matter: under the Interactive Gambling Act 2001, online casino games cannot legally be offered to people in Australia, although it is not an offence for you to play, and ACMA blocks illegal sites. If chasing losses has become a habit, free and confidential help is available around the clock.
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Frequently asked questions
How does the Martingale strategy work?
You bet on an even-money chance such as red, double your stake after every loss and go back to your base bet after a win. The first win in a run recovers all the losses in that run plus one base bet.
How much bankroll do I need for a Martingale?
To cover n bets you need base x (2 to the power n, minus 1). With a $5 base: $75 covers 4 bets, $635 covers 7, $1,275 covers 8 and $10,235 covers 11. Each extra bet doubles the bankroll needed but only halves the bust risk, roughly.
What are the odds of losing a Martingale run?
On an even-money bet at single zero roulette, the chance of losing n bets in a row is (19/37) to the power n. In our calculation that is 6.95% for 4 bets, 0.94% for 7, 0.48% for 8 and 0.13% for 10.
Can the Martingale beat roulette?
No. It wins small amounts often and loses a large amount rarely, and the average comes out at the normal house edge on everything bet: 2.70% on a single zero wheel and 5.26% on double zero, in our calculation.
Does the Martingale work on sports betting?
No. Every bet carries the bookmaker's margin, so at $1.90 on a true 50/50 each run loses 5% of what it puts at risk. Worse, at odds under $2.00 doubling stops covering your losses: at $1.90, a win on the fifth bet or later leaves the run behind, in our calculation.
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