Bankroll Simulator
See how your bankroll is expected to grow over many bets — and how likely you are to go bust at a given stake size.
Your own estimate of the chance of winning each bet — not the odds' implied probability.
Enter your bankroll, stake, odds, win probability and number of bets to see your simulated bankroll trajectory.
This simulation is a mathematical approximation based on your own estimate of the win probability and the assumption of unchanged, flat staking. It is not a guarantee of real-world results and does not replace responsible gambling.
A bankroll simulator shows how a bankroll is statistically expected to develop under flat staking over many bets — and how likely you are to lose the entire bankroll (go bust) along the way. The tool is built on basic probability theory and gives you a realistic picture of the swings behind a given betting plan, not just the average outcome.
How the simulation works
Each bet is treated as an independent random outcome: you win with probability p and lose with probability (1 − p). Given your stake s and decimal odds o, the expected profit and variance per bet are:
μ = s × (p × (o − 1) − (1 − p))
σ² = s² × (p × (o − 1)² + (1 − p) − (p × (o − 1) − (1 − p))²)
Over N independent bets at the same stake, the expected bankroll becomes B + N×μ, with a standard deviation of √(N×σ²) around that value. The risk of ruin (losing the entire bankroll) is estimated with a classic approximation for a random walk with drift:
Risk of ruin ≈ e^(−2 × μ × B ÷ σ²)
The formula is an approximation that assumes you keep staking the same amount indefinitely, not just for the N bets entered above. If the bet has no positive expected value (μ ≤ 0), the risk of eventually losing the entire bankroll is mathematically 100%, no matter how large it is.
Worked example
Say you have a bankroll of 10,000 and stake 100 per bet at odds 2.00, with an estimated win probability of 55%. The expected profit per bet is μ = 100 × (0.55 × 1 − 0.45) = 10, with a variance of σ² = 100² × (0.55 × 1 + 0.45 − 0.10²) = 9,900 (a standard deviation of about 99.50 per bet). After 100 bets, the expected bankroll is 10,000 + 100 × 10 = 11,000, with a standard deviation of √(100 × 9,900) ≈ 995. The risk of ruin is estimated at e^(−2 × 10 × 10,000 ÷ 9,900) ≈ almost 0% — a large bankroll relative to the stake makes this setup safe despite the swings.
Frequently asked questions
What does 'risk of ruin' mean?
Risk of ruin is the estimated probability that your bankroll eventually reaches zero and you have no money left to bet with. It depends on both your edge (how much you're expected to win on average) and your variance (how much results swing) — a bigger edge and a larger bankroll relative to your stake both lower the risk.
Why is the risk of ruin 100% if I have no edge?
Without a positive expected value, your expected result is zero or negative over time, while variance keeps swinging the bankroll up and down. Mathematically (classic gambler's ruin theory), this means the bankroll eventually hits zero if you keep betting indefinitely, no matter how large it started out.
Is the risk of ruin exact, or just an estimate?
It's an approximation based on a continuous random-walk model (Brownian motion), which is standard in bankroll-management theory for advantage play. It assumes you keep staking the same amount indefinitely — not just for the N bets you entered — so it's a guide to the long-run risk, not an exact probability for your specific N bets.
