Watch a “sure thing” fall apart
Betting on Heads
Generate 1,000 simulated runs with your current min, max, and bankroll to estimate how long before the system goes bankrupt or hits the max bet wall and cannot proceed.
The Martingale is a betting system for even-money wagers (like red/black in roulette or heads/tails on a coin). You start with a small bet. Every time you lose, you double your next bet. As soon as you win, you go back to your original stake. The idea is simple: one win wipes out all previous losses and leaves you ahead by your original bet. It’s been around since the 18th century and sounds foolproof, but it isn’t.
On paper each bet is fair, but Martingale quietly trades lots of tiny wins for a single huge loss. Long losing streaks are rare—but not impossible. And with a table limit and a finite bankroll, one streak is all it takes.
Each time you lose, your bet doubles. That means your total losses after a streak of losses grow exponentially.
Start with minimum bet b. After n consecutive losses you’ve lost:
b + 2b + 4b + … + 2n−1b = b(2n − 1)
Your next bet would be b·2n. With a table limit (max bet M), you can only double until b·2n ≤ M, so you can withstand at most n = log2(M/b) losses in a row. Past that, you can’t double. You’ve hit the ceiling and the next loss wipes you out.
On a fair 50/50 bet, the chance of k losses in a row is (½)k. So the chance of eventually hitting a streak long enough to break you is positive—and as the number of bets grows, that probability approaches 1. You will eventually hit that streak.
Each round is fair (expected value zero). Martingale doesn’t change that, it just packs many small wins into one big eventual loss. With a table limit and finite bankroll, you inevitably go broke.
There’s no simple closed-form formula for the exact expected number of flips given min bet b, max bet M, and bankroll B. Failure can happen two ways (bankrupt or losing at max bet), and the bet size depends on the current streak, so the process is a state-dependent random walk with two absorbing boundaries.
What we can say: the number of consecutive losses needed to reach max bet is n = ⌈log2(M/b)⌉. On a fair coin, the expected number of flips until you get n consecutive losses (ignoring bankroll) is:
2n+1 − 2
With an infinite bankroll, failure would typically occur on the order of 2n+1 flips, when that streak finally appears. With a finite bankroll you can go broke earlier, so the real expected flips is lower and depends on B, b, and M. The practical way to estimate it is to simulate many runs and average the flip count at failure. Use the “Go” button in the Bulk Simulate section above with your current settings to see a Monte Carlo estimate.
The only way to guarantee a win with this system is to have no table limit and an unlimited bankroll. Then you could keep doubling until you win. In practice, no casino offers no limit, and no one has infinite money.
In the real world, with a finite bankroll and enough time, the system always loses—and the house always wins.