Martingale levels vs bankroll: levels, not luck
To keep a doubling ladder ahead for longer, give it more steps with a smaller base stake; a bigger bankroll on the same steps does not keep it ahead any longer.
Chapter 3 of the spinner roulette research (chapters 0 to 9): Levels, not luck.
Each extra step makes losing the whole ladder about half as likely, so it roughly doubles how long the typical session stays ahead. With a $5 base and a bankroll of exactly one ladder, 5 steps hold to spin 39, 6 to 70, 7 to 150 and 8 to 290; 9 and 10 steps last past spin 500. Ten steps at $5 from $5,000 end with a typical session of +$1,185.
Doubling the bankroll to $10,000 with the same 8 steps at $25 leaves the turn at spin 290. The extra money lets sessions survive one lost ladder, but the worst 5 % now lose $10,000 and the average gets worse.
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The numbers
| Levels | Predicted half-life (spins) | Typical session ahead until | Typical at the end | Average | Worst 5 % |
|---|---|---|---|---|---|
| 5 | 40 | spin 39 | -$155 | -$63.25 | -$155 |
| 6 | 78 | spin 70 | -$315 | -$99.34 | -$315 |
| 7 | 151 | spin 150 | -$635 | -$151.36 | -$635 |
| 8 | 295 | spin 290 | -$180 | -$209.14 | -$1,275 |
| 9 | 574 | still ahead at spin 500 | +$1,130 | -$290.70 | -$2,555 |
| 10 | 1,117 | still ahead at spin 500 | +$1,185 | -$327.56 | -$5,115 |
| Setup | Typical session ahead until | Typical at the end | Average | Worst 5 % | Ended broke |
|---|---|---|---|---|---|
| Martingale on red, $25 base, $5,000 | spin 290 | -$5,000 | -$898.23 | -$5,000 | 51.2% |
| Martingale on red, $25 base, $10,000 | spin 290 | -$375 | -$1,265.78 | -$10,000 | 22.2% |
| Martingale on red, $5 base, 10 levels, $5,000 | still ahead at spin 500 | +$1,185 | -$324.32 | -$5,000 | 15.6% |
Takeaways
- Count your bankroll in base stakes: that number, not the dollars, sets how long the typical session stays ahead.
- A smaller base stake under the same table maximum buys more steps.
- The average loss grows with every step: −$63.25 at 5 steps, −$327.56 at 10.
The math
Each level multiplies q by 1 − p = 19/37 ≈ 0.51, so n½ = ln 2 / (q·p) roughly doubles per level. With a $5 unit and bankroll 2^L − 1 units: L5 q 0.0357, n½ 40; L6 0.0183, 78; L7 0.0094, 151; L8 0.0048, 295; L9 0.0025, 574; L10 0.0013, 1,117. Measured medianAheadUntil: 39, 70, 150, 290, 500 and 500 (censored at the session end), each inside its 10-spin checkpoint.
The bankroll enters only through L, the number of levels that fit under the table maximum and the bankroll. A $10,000 bankroll at $25 still fits 8 doublings under the $5,000 maximum bet, so q and n½ are unchanged (measured 290), while a bust costs twice as much: median −$375, mean −$1,265.78, CVaR 5 % −$10,000, bust 22.2 %. Mean loss grows with L because turnover does: −$63.25 (L5) to −$327.56 (L10).
q = (19/37)^L
n½ = ln 2 / (q × 18/37)
L = largest integer with u × (2^L − 1) ≤ bankroll and u × 2^(L−1) ≤ table max