The Martingale System, Explained

After reading this you will be able to compute the chance a doubling-after-loss system busts before you plan to stop, and see why the expected loss does not change no matter how you stake.

The Martingale is the oldest betting system that sounds like a free lunch. You bet on an even-money outcome. If you win, you pocket one base unit and start over. If you lose, you double the next bet, so a single win recovers every prior loss plus one unit of profit. On paper each sequence ends in a small win. In practice the doubling runs into a wall: the table limit, your bankroll, or both.

Here is the hook. A base bet of 10 on European roulette, with a table that lets you double seven times, produces a profit on roughly 97 out of 100 sequences. That feels like a machine that prints money. It is not. The 3 sequences that fail cost so much that they wipe out all the small wins and then some. This article shows the exact arithmetic.

What the system actually does

A single Martingale sequence is a run of doubled bets that stops the moment you win, or stops in disaster when you hit the doubling limit. If your base bet is b and you are allowed n doublings, your bet sizes in one sequence go:

b,\ 2b,\ 4b,\ \dots,\ 2^{n}b

That is n+1 possible bets in the sequence. The first one is the base bet, and each later one is twice the previous. If you win on any of them, the doubling structure guarantees a net profit of exactly b for the whole sequence. The reason is simple: the sum of all previous losses is always one unit less than the current bet.

b + 2b + 4b + \dots + 2^{k-1}b = (2^{k} - 1)\,b

When you win the bet of size 2^{k}b, you gain that amount and have lost (2^{k}-1)b before it. The difference is b. So every sequence that ends in a win nets exactly one base unit, regardless of how many doublings it took.

The catastrophe is the other branch. If you lose all n+1 bets, you lose the whole geometric sum:

L = (2^{n+1} - 1)\,b

With b = 10 and n = 7, that is (2^{8} - 1) \times 10 = 2550. One bad sequence costs 2550 units to win the 10-unit prize.

The two numbers that decide everything

Let p be your probability of winning a single round. On European roulette an even-money bet (red, black, odd, even) wins when the ball lands in 18 of the 37 pockets, so p = 18/37 \approx 0.4865. The probability of losing one round is q = 1 - p \approx 0.5135.

A sequence fails only if you lose every one of the n+1 bets. Those rounds are independent, so the bust probability of a single sequence is:

q_{\text{seq}} = q^{\,n+1}

With q \approx 0.5135 and n+1 = 8, that is 0.5135^{8} \approx 0.01907, about 1.9%. So a single sequence succeeds about 98.1% of the time and fails about 1.9% of the time.

The win chance per sequence looks reassuring precisely because the failure is rare and large. A rare, large loss is the hardest kind of risk for intuition to price. The math below prices it for you.

Expected value of one sequence

Each sequence has two outcomes. With probability 1 - q^{\,n+1} you win b. With probability q^{\,n+1} you lose (2^{n+1}-1)b. The expected value of one sequence is:

E_{\text{seq}} = b\,(1 - q^{\,n+1}) - (2^{n+1} - 1)\,b\,q^{\,n+1}

This simplifies to a clean result. Factor out b and collect terms:

E_{\text{seq}} = b\,\big(1 - (2q)^{\,n+1}\big)

Everything hinges on 2q. If 2q = 1, meaning p = 0.5 and a fair game, then E_{\text{seq}} = 0 for any number of doublings. The Martingale neither helps nor hurts in a fair game. If q \gt 0.5, as it is in any casino game, then 2q \gt 1 and (2q)^{n+1} grows, so E_{\text{seq}} is negative and gets more negative with more doublings.

For the roulette numbers: 2q \approx 1.027, and 1.027^{8} \approx 1.2306. So E_{\text{seq}} \approx 10 \times (1 - 1.2306) = -2.31 units per sequence.

Worked example: the demo numbers

Base 10, win 48.65%, 7 doublings, 100 sequences

Loss probability per round: q = 1 - 0.4865 = 0.5135.
Rounds per sequence: n+1 = 7+1 = 8.
Bust probability per sequence: q^{8} = 0.5135^{8} \approx 0.01907, so 1.907%.
Cost of a bust: (2^{8}-1)\times 10 = 2550 units.
EV per sequence: 10 \times (1 - (2\times 0.5135)^{8}) \approx -2.31 units.
Over 100 sequences the expected total is 100 \times (-2.31) \approx -231 units.
Chance of surviving all 100 sequences without a single bust: (1 - 0.01907)^{100} \approx 0.1447, about 14.5%.

Read that last line again. Playing 100 sequences, you have only a 14.5% chance of never hitting the wall. The 85.5% chance of at least one bust is where the money goes. The expected result is a loss of about 231 units, which is close to the 240 units of house edge you would expect from flat-betting the same total action.

The chance of hitting the table limit at least once climbs past 50% by around 37 sequences and reaches about 84% by 100 sequences.

Interpreting the results

The calculator gives you four things worth reading together.

Bust probability per sequence: The chance one sequence loses every bet, q^{\,n+1}. Small, but not negligible.
Overall bust probability: The chance of at least one bust across all the sequences you plan, 1 - (1 - q^{\,n+1})^{r} for r sequences. This grows fast.
Maximum single-sequence loss: (2^{n+1}-1)b. The amount you must have on hand to complete one full sequence.
Expected value: The total you expect to lose, r \times b \times (1 - (2q)^{\,n+1}). Always negative in a casino game.

The key comparison is the maximum loss against the target profit. In the demo you risk 2550 units per sequence to win 10. That ratio, 255 to 1, is the honest price of the "almost always wins" feeling.

Move the doublings and watch the risk

Common mistakes

The most expensive mistake is believing that a long losing streak "must" end soon. Roulette spins are independent. After six losses in a row your chance of losing the seventh is still 51.35%. The wheel has no memory and no obligation to balance out.

Three more errors show up constantly:

Ignoring the table limit. Casinos cap the maximum bet exactly to stop unlimited doubling. A table with a 500 unit max and a 10 unit base allows only about 5 doublings, not the 7 you assumed. Fewer doublings means a higher bust chance per sequence.
Confusing "usually wins" with "wins money". A 98% win rate per sequence still loses money because the 2% loss is 255 times larger than the win. Frequency of winning is not the same as expected profit.
Underfunding the bankroll. You need (2^{n+1}-1)b liquid just to finish one sequence. Running out of money mid-sequence turns a recoverable position into a locked-in loss.

When to use it, and when not

The Martingale changes the shape of your results, not the average. It trades many small wins for rare large losses, converting a steady drip of house edge into a lumpy one. If your goal is a short entertaining session with a high chance of a small win, and you accept the small chance of a large loss, the system does that. It does not beat the house, and it cannot turn a negative-edge game positive.

If you have a genuine edge, do not double after losses. Size your bets by the Kelly Criterion Calculator instead, which grows a bankroll fastest when the expectation is in your favor. To estimate the chance of losing a fixed bankroll before reaching a goal under any staking plan, use the Risk of Ruin Calculator. If you want to understand how likely those scary losing streaks are in the first place, see the Streak Probability tool. And if you are chasing a casino bonus rather than a system, the Bonus Wagering EV calculator shows what the offer is worth after the rollover.

Frequently asked questions

Does the Martingale work on a fair coin?

Even at p = 0.5 the expected value of each sequence is exactly zero, because 2q = 1 and (2q)^{n+1} = 1. You break even on average, but you still risk large losses for small wins. It does not create an edge; it only reshapes variance.

Why is the house edge unchanged if I usually win each sequence?

Because the rare loss is huge. In the demo you win 10 units about 98% of the time and lose 2550 units about 1.9% of the time. Multiply out: 0.981 \times 10 - 0.019 \times 2550 \approx -2.5 units. The frequent small wins cannot cover the occasional wipeout.

What bankroll do I need to run the demo settings?

At least 2550 units to complete a single worst-case sequence with base 10 and 7 doublings. To play 100 sequences with a real chance of surviving a bust, you would need many multiples of that, and you would still expect to end down about 231 units.

Would a higher table limit make the system safe?

No. More doublings lower the per-sequence bust chance but raise both the maximum loss and the expected loss. At 10 doublings the bust chance drops to 0.259%, yet the maximum loss climbs to 20470 units and the expected loss per sequence grows to about 3.60. There is no setting where the expectation turns positive.

Is a reverse Martingale (doubling after wins) better?

It has the same zero-edge-in-a-fair-game, negative-edge-in-a-casino expectation. It flips the shape: many small losses and rare large wins. The average result is still the house edge times your total action.

Try it yourself in the tool