House edge (identical no matter how you bet)
House edge (identical no matter how you bet)-
Probability that streak happens within your rounds (-)-
Probability you reach your target profit before that streak ever happens-
Probability that streak happens before you reach your target-
Expected result per win-or-bust cycle-
Expected amount wagered per cycle-
House edge x amount wagered (same number)-
Your target profit is treated as - base-bet units, since a Martingale system, when it wins, nets exactly one base-bet unit of profit no matter how long the streak was.
This win probability is above 50% -- that's a player advantage, not a house edge. No real casino even-money bet offers this.
These two numbers are always equal. That's the whole point: expected loss is exactly the house edge times the amount wagered, regardless of how you size your bets. Martingale doesn't escape that -- it just trades a high chance of winning a little for a small chance of losing a lot.
How many times can a Martingale system double before it's stuck, and what's the real cost?
With a $1,000 bankroll and a $10 base bet, you can double 6 times in a row (bets of $10, $20, $40, $80, $160, $320) before the 7th bet, $640, would need more than what's left of the bankroll. Over 100 rounds of European roulette (18/37), that 6-in-a-row losing streak actually happens 59.73% of the time -- verified with the standard "no run of k losses" recurrence in tests/martingale-ref.py.
Steps to find your numbers
- Enter your bankroll, base bet, and table maximum (leave the table max blank if there isn't one).
- Pick a game preset (European or American roulette) or enter your own win probability.
- The calculator finds k -- the maximum number of times in a row your bankroll and table limit let you double -- by checking, bet by bet, whether the next doubled bet is still affordable.
- Enter how many rounds you plan to play; the calculator finds the probability that a losing streak of k actually happens within that many rounds.
- Enter a target profit; the calculator treats it as a number of base-bet units and finds the probability you reach it before that losing streak ever occurs.
The formulas behind the numbers
k = the largest integer such that bankroll >= unit x (2^(k+1) - 1) and unit x 2^k <= table max. Probability of busting in N rounds = 1 - u(N), where u(n) is found by the recurrence u(n) = p x sum_{j=0}^{k-1} q^j x u(n-1-j) for n >= k (with u(n)=1 for n<k), q=1-p.
- k = max consecutive losses your bankroll and table limit can absorb before the next doubled bet is unaffordable
- p = your win probability per round (edge = 1 - 2p when the bet pays even money)
- u(n) = probability that no run of k consecutive losses occurs in the first n rounds
- Cycle = one full attempt that ends either in a win (net +1 base-bet unit) or in a bust (a streak of k losses in a row)
Quick answers
Does the Martingale system work?
No -- expected loss is unchanged. It's exactly the house edge times the total amount wagered, regardless of bet sizing.
How many times can I double $10 with a $1,000 bankroll?
6 times in a row (verified: bets of $10 through $320 total $630; the 7th bet of $640 exceeds what's left).
What's the house edge on European roulette?
2.7027% (1/37) on an even-money bet, since only 18 of 37 pockets pay and the edge comes from the single zero.
What's the house edge on American roulette?
5.2632% (1/19) on an even-money bet -- the extra double-zero pocket roughly doubles the European edge.
What's the probability of a 6-loss streak in 100 rounds of European roulette?
59.73% -- verified with the standard runs recurrence and cross-checked against brute-force enumeration in tests/martingale-ref.py.
If my target is $100 profit (10 units) with k=6, what's my probability of getting there before busting?
83.11%, versus a 16.89% probability the losing streak happens first -- computed as (1 - q^k)^10 where q is the loss probability.
Does a bigger bankroll make Martingale safe?
It raises k and lowers your bust probability over a fixed number of rounds, but it never changes the house edge, and it raises the size of the eventual loss when the streak does happen.
Is there any betting system that beats the house edge?
No. Expected loss = house edge x total amount wagered for any previsible betting system (one that only uses past information) -- this is a mathematical certainty, not a strategy gap. Verified for Martingale and flat betting alike in tests/martingale-ref.py.
European roulette (18/37) example: bankroll $1,000, base bet $10
| Rounds played | Probability that a 6-loss streak happens |
|---|
| 10 | 5.40% |
| 20 | 13.97% |
| 50 | 35.28% |
| 100 | 59.73% |
| 200 | 84.41% |
Frequently asked questions
Does the Martingale system work?
No -- it does not change your expected loss. On any single bet, expected loss equals bet size times the house edge, and doubling after a loss doesn't change that per-bet math. What Martingale does change is the shape of your outcomes: a high probability of a small win, traded for a small probability of a loss large enough to wipe out every prior win many times over.
How many times can I double a $10 bet with a $1,000 bankroll?
6 times in a row (bets of $10, $20, $40, $80, $160, $320, totaling $630), assuming no table limit lower than $640. The 7th bet would need $640, which exceeds the $370 left in the bankroll.
What's the expected loss per Martingale cycle?
On European roulette (18/37) with a $10 base bet and k=6, the expected result per completed cycle (win or bust) is -$1.7352 -- and that's exactly the 2.7027% house edge times the $64.20 expected total wagered per cycle, to the fraction. Nothing about doubling changes that relationship.
Why does the probability of busting go up the longer I play?
Because every round is another independent chance to hit the k-losses-in-a-row streak that breaks the system. Over 10 rounds it's 5.40%; over 100 rounds it's 59.73%; over 200 rounds it's 84.41% -- the streak that ends the system becomes more and more likely the longer you keep playing, even though your odds on any single spin never change.
Can I raise my bankroll to fix this?
You can raise k (the number of doublings you can absorb), which lowers the bust probability for a fixed number of rounds -- but it never changes the house edge, and each extra doubling raises how much you stand to lose on the losing streak that finally does happen. It's a trade, not a fix.
This tool assumes a fixed, independent win probability every round (no memory between spins, matching real roulette) and a strict doubling Martingale that resets to the base bet after every win. It does not model table minimums that force a higher starting bet, partial/reduced final bets when a table limit is hit mid-sequence, or taxes/rake. Built for probability education -- not gambling advice.
Sources: Wizard of Odds -- Betting Systems, Wikipedia -- Gambler's ruin