Martingale “bust risk” is the chance you hit a losing streak long enough that you can’t place the next required double-up bet; you can estimate the bankroll you need by (1) defining your base bet and limits, (2) converting your bankroll into the maximum number of consecutive losses you can survive, and (3) translating that losing-streak threshold into a probability over the number of spins you plan to play. The key insight: even when your bet wins almost half the time (like roulette even-money bets), the probability of eventually encountering a long losing streak rises quickly with time, so “safe” bankrolls are usually much larger than intuition suggests.
Why Martingale fails: bust comes from streaks, not “bad luck”
Martingale on an even-money wager (red/black, odd/even) works like this: you start with a base bet b. After each loss, you double: b, 2b, 4b, 8b… The first win recovers all previous losses and nets +b profit (ignoring table rules and special outcomes).
The strategy doesn’t lose because each spin is “rigged” against you in an obvious way; it loses because:
- Streaks are inevitable over enough spins, even with near-50/50 outcomes.
- Your required next bet grows exponentially, but your bankroll and table limits do not.
- In roulette specifically, the house edge and the zero (and sometimes double zero) slightly reduce the win probability on even-money bets, making streaks marginally more likely than a pure coin flip.
So the real question is not “Will Martingale win a few cycles?” It often will. The real question is: How likely am I to encounter a streak longer than my bankroll can handle within N spins? The three steps below answer that.
Step 1: Define your inputs (b, bankroll, and the hard cap that actually stops you)
You need three practical inputs before any calculation is meaningful.
1) Base bet (b)
Pick the smallest unit you’ll actually use (for example, $5). Smaller b increases how many doubles you can afford, reducing bust risk.
2) Effective bankroll (B)
This is the amount you are truly willing to lose on this session, not your total available funds. Martingale concentrates losses into rare “blowups,” so using a realistic stop-loss matters.
3) Your real stopping point: bankroll cap and table cap
You bust when you cannot place the next required bet. That can happen due to:
- Bankroll limit: you don’t have enough money to make the next bet.
- Table limit: the casino cap prevents your next double even if you have the money.
Use the stricter of the two. If the table max is M, the biggest bet you can place is M, so the most doubles you can execute is limited even with a huge bankroll.
4) Win probability per spin on your bet (p)
For roulette even-money bets:
- European wheel: p is 18/37 (you win on 18 numbers, lose on 19 including zero)
- American wheel: p is 18/38 (lose on 20 including 0 and 00)
If you’re unsure what wheel/rules you’re on, you can often verify the RTP and rules disclosure for the specific game. For example, you can find RTP info on rouletteuk.co.uk/online-roulette that demonstrates how roulette variants can differ in expected return depending on wheel and rules—information that directly affects p and therefore your streak risk.
Step 2: Convert bankroll (and table limit) into “max losses in a row you can survive”
This step turns your budget into a single number: L, the maximum number of consecutive losses you can withstand before you bust.
Bankroll-based maximum losses (L_bankroll)
After k consecutive losses, your total amount lost is:
b + 2b + 4b + … + 2^(k-1)b = b(2^k – 1)
To be able to keep playing through k losses, you must be able to cover that cumulative loss and still place the next bet. A simple conservative rule is:
You can survive k consecutive losses if B is at least b(2^(k+1) – 1)
Why this slightly stronger condition? Because after k losses, the next required bet is 2^k b, so you need bankroll to both have absorbed previous losses and still stake the next bet.
Practical shortcut:
- Compute the largest k such that b(2^(k+1) – 1) <= B
- Then L_bankroll = k
Table-limit maximum losses (L_table)
If the table maximum is M, the largest bet you can place is M. Starting from base bet b, the biggest doubling step you can reach is where 2^k b <= M. So:
- L_table = largest k with 2^k b <= M
Your true survivable streak length (L)
Set:
- L = minimum of Lbankroll and Ltable
Interpretation: if you can survive L losses in a row, you bust on the (L+1)th consecutive loss, because you can’t place the next required bet.
Example (turn money into L)
Assume:
- Base bet b = 10
- Session bankroll B = 1,270
- Table max high enough to ignore for now
Check k:
- Need B >= b(2^(k+1) – 1)
Try k = 6: 10(2^7 – 1) = 10(128 – 1) = 1,270 ✓
Try k = 7: 10(2^8 – 1) = 2,550 ✗
So L = 6. You can survive 6 consecutive losses; 7 losses in a row busts you.
This is the most important translation in the whole method: bankroll doesn’t buy “more time,” it buys “more losses in a row.”
Step 3: Translate L into bust probability over N spins (a usable risk number)
Now you estimate: “What is the chance I see a streak of (L+1) losses somewhere in my next N spins?”
Let:
- q = probability of losing a spin = 1 – p
- S = L + 1 (the bust streak length)
- N = number of spins you plan to play
A practical approximation you can compute quickly
A common, usable approximation is:
Approx bust risk ≈ 1 – (1 – q^S)^(N – S + 1)
Explanation in plain terms:
- There are (N – S + 1) possible starting positions for a run of S losses.
- q^S is the probability that a specific block of S spins is all losses.
- This treats blocks as mostly independent (they’re not fully independent because runs overlap), so it’s an approximation—usually good enough for planning and typically optimistic (it can understate risk a bit).
Worked example using the L from above
From Step 2: L = 6, so S = 7.
Assume European roulette even-money:
- p = 18/37 ≈ 0.4865
- q = 19/37 ≈ 0.5135
Compute q^7:
- 0.5135^2 ≈ 0.2637
- ^4 ≈ 0.0695
- ^7 = ^4 ^2 ^1 ≈ 0.0695 0.2637 0.5135 ≈ 0.0094 (about 0.94%)
Now pick N = 200 spins:
- Number of starting positions: 200 – 7 + 1 = 194
- Approx bust risk ≈ 1 – (1 – 0.0094)^194
Estimate (1 – 0.0094)^194:
- Using a quick log approximation: ≈ exp(-0.0094*194) = exp(-1.82) ≈ 0.16
So bust risk ≈ 1 – 0.16 = 0.84, or about 84% over 200 spins.
This surprises many people: a bankroll that “feels large” relative to a 10-unit base bet still has a high chance of eventual blowup if you keep spinning.
Turning the process around: bankroll needed for a target risk
Often you want: “What bankroll keeps bust risk under R over N spins?”
You can solve it in a practical loop:
- Choose b, N, and a maximum acceptable bust risk R (for example, 5%).
- Guess L (start with 6, 7, 8…).
- Compute S = L + 1, then approximate bust risk using the formula above.
- Increase L until the risk is under R.
- Convert that L back to a bankroll requirement using Step 2:
– Required bankroll B_needed ≈ b(2^(L+1) – 1)
– Then apply table limit constraints: also ensure 2^L b is not above the table max.
A quick sense of scale: each extra survivable loss roughly doubles the bankroll needed, because the geometric series is dominated by the last term.
Important adjustments most calculators miss (and why they matter)
1) House edge changes p, which changes streak odds
Even small differences in p materially change q^S when S is large, because you’re exponentiating q. That’s why verifying wheel type/rules (Step 1) is not a formality; it changes the bust curve.
2) “Profit target per cycle” does not reduce bust probability
Some players stop after winning 1 base bet, thinking it “locks in profit.” It doesn’t change the probability that a long losing streak happens before you stop, unless you also strictly cap the number of spins or cycles. Bust risk is driven by how long you keep sampling spins.
3) Overlap makes the approximation slightly optimistic
Runs overlap (a sequence of 10 losses contains multiple 7-loss windows). The approximation in Step 3 tends to undercount that overlap effect a bit. If you need a safety margin, treat the computed risk as a lower bound and require a lower target (for example, aim for 2% if you want “about 5%”).
4) Table limits often dominate before bankroll does
Even if you can afford deeper doubling, the table maximum can force a bust earlier. Always compute L_table; otherwise you may overestimate survivability.
Quick Summary
To calculate Martingale bankroll needs, first define base bet, session bankroll, table max, and win probability; then convert bankroll and limits into the maximum losing streak you can survive; finally estimate the chance that a longer streak appears within your planned number of spins. The takeaway is structural: because required bets grow exponentially, bust risk rises quickly with time, so “comfortable” bankrolls often translate into surprisingly high blowup probabilities over realistic session lengths.