Progressive systems
Martingale betting system: doubling the stake after a loss
The Martingale system doubles the stake after every loss, so the first win at odds of 2.00 recovers all losses and adds the starting stake. After 10 losses in a row 1,023 units are lost and the next stake is 1,024. Martingale does not change expected value: every stake has the same EV per unit, so the system only turns many small wins and rare large losses into the same average.
Contents
Key terms
- Martingale system
- A progression in which the stake doubles after every loss and returns to the starting stake after a win.
- Progression
- A rule that sets the stake from the results of earlier bets, not from an estimate of probability.
- Cycle
- A run of bets from the starting stake to the first win, or to the point where the money runs out.
- Bank
- The money set aside for betting. In a Martingale it decides how many losses in a row a cycle can survive.
How does the Martingale system work?
Odds of 2.00 and a starting stake of 1. The first bet loses, so the next stake is 2. That one loses too, so 4 follows. If the bet of 4 loses, 8 is staked, and if that wins, the return is 16.
The total staked is 1 + 2 + 4 + 8 = 15 and 16 comes back. The cycle ends at +1, the starting stake.
lost after n losses = 2^n − 1
next stake = 2^n| Losses in a row | Total lost | Next stake |
|---|---|---|
| 1 | 1 | 2 |
| 2 | 3 | 4 |
| 3 | 7 | 8 |
| 5 | 31 | 32 |
| 7 | 127 | 128 |
| 10 | 1,023 | 1,024 |
The stake grows faster than it looks: after 10 losses in a row you have to stake 1,024 to win one unit.
Why does Martingale not change expected value?
The expected value of a bet per unit staked is p × odds − 1. Every stake in the cycle is a new bet with the same p and the same odds, so it has the same EV per unit. The progression only changes how much is staked, not how much each staked unit returns on average.
Odds of 2.00 and p = 0.50, so EV is zero. A bank of 1,023 units survives 10 bets, because 1 + 2 + … + 512 = 1,023.
- the cycle ends at +1 with probability 1,023 / 1,024 = 99.90%;
- all 10 bets lose with probability 1 / 1,024 = 0.0977%, and then the loss is 1,023;
- the average cycle: (1,023 / 1,024) × 1 − (1 / 1,024) × 1,023 = 0.
Many small wins and one rare large loss give exactly the same average as a single bet with zero EV.
The same odds of 2.00, but the true probability is p = 0.48, so EV is −4% per unit.
- all 10 bets lose with probability 0.52^10 = 0.1446%, about once in 692 cycles;
- on average 12.01 units are staked per cycle, because more is staked after every loss;
- the average cycle is −0.48 units, exactly −4% of the average amount staked.
When EV per unit is negative, Martingale does not fix it. Because more is staked after losses, a cycle stakes more than flat betting would, so the average loss per cycle is larger as well.
How often do 10 losses in a row arrive?
A single cycle rarely fails, but there are many cycles. The probability that at least one of N cycles loses all 10 bets is 1 − (1 − q^10)^N, where q is the probability of a loss.
| Cycles | p = 0.50 | p = 0.48 |
|---|---|---|
| 100 | 9.31% | 13.47% |
| 500 | 38.65% | 51.49% |
| 1,000 | 62.36% | 76.46% |
| 5,000 | 99.24% | 99.93% |
After 1,000 cycles without a failure the profit is 1,000 units, and a single failed cycle takes 1,023. That is why a Martingale looks successful for a long time, until the streak arrives that on average had to arrive. How common such streaks are even with good bets is shown in the betting variance guide.
What if the odds are below 2.00?
For a cycle to end at +1, the stake after a loss has to cover all previous losses and one more unit. At odds of 1.50 the net win is 0.50 per unit, so the stake triples after every loss. The growth factor is odds / (odds − 1).
| Bet in the cycle | Odds 2.00 | Odds 1.50 |
|---|---|---|
| 1 | 1 | 2 |
| 2 | 2 | 6 |
| 3 | 4 | 18 |
| 4 | 8 | 54 |
| 5 | 16 | 162 |
| 6 | 32 | 486 |
At odds of 1.50, 242 is lost after 5 losses and the sixth stake is 486. Lower odds win more often, but every losing streak is far more expensive.
Stake limits and the reverse Martingale
- Bookmakers set a maximum stake. After a few losses the next stake can be above the limit, so the cycle cannot be completed.
- The reverse Martingale (Paroli) doubles the stake after a win. The pattern of wins and losses is different, but the EV per unit is the same.
- Stakes that depend on the estimate and the bankroll, not on earlier results, are covered in the Kelly criterion and flat vs Kelly guides.
Limitations
- The examples use odds of 2.00 and made-up probabilities. Real odds include a margin, so EV per unit is usually negative.
- The calculation assumes independent bets, with no tax on winnings and no stake limits.
- Martingale does not reduce risk, it moves it: instead of frequent small losses it gives rare large ones.
- No progression changes the expected value of a bet. Positive EV only comes from an estimate that is better than the odds.
Related guides
Sources
- Feller, W. (1968). An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd ed. New York: Wiley.