Odds & strategy math · Free tool
Chance Calculator: Probability of a Win, a Miss or a Losing Streak
Quick answer
The chance of at least one success in n independent tries is 1 − (1 − p)^n, where p is the chance per try. Expected successes are n × p. For red on a single-zero roulette wheel (48.65%), the chance of at least one run of 8 or more losses somewhere in 100 spins is about 20%, which this calculator works out exactly.
| At least one success | 1 − (1 − p)^n |
|---|---|
| Zero successes | (1 − p)^n |
| Expected successes | n × p |
| 8-loss streak in 100 red bets (single zero) | About 20.2% (exact calculation) |
| Each try | Independent: past results don't change p |
This chance calculator works out the probability of at least one success, zero successes and a losing streak of any length across a number of independent tries. Enter the chance per try as a percentage or as "1 in N", how many tries you'll make, and the streak length you want to check. The results update as you type.
The default is betting red on a single-zero roulette wheel: 18 winning pockets out of 37, or 48.65% per spin. Over 100 spins you expect about 48.65 wins, and there's a 20.2% chance that at some point you lose 8 or more spins in a row.
How this is calculated
Each try is treated as independent: the result of one spin or roll doesn't change the next. That lets us use the multiplication rule for independent events, where the chance of several things all happening is the product of their chances.
P(zero successes) = (1 − p)^n P(at least one) = 1 − (1 − p)^n expected successes = n × pp is the chance per try as a decimal (48.65% = 0.4865); n is the number of tries.
The losing-streak figure needs more work, because streaks can start anywhere and overlap. We track the probability that no run of L misses has happened yet after each try. Call that a(i), and let q = 1 − p be the chance of a miss:
a(i) = 1 for i < L a(L) = 1 − q^L a(i) = a(i−1) − p × q^L × a(i−L−1)Chance of at least one run of L misses in n tries = 1 − a(n). The last term counts the sequences where the first run of L misses ends exactly at try i: no run up to try i−L−1, then one success, then L misses.
This is exact, not a simulation. The calculator runs it in your browser for up to 1,000,000 tries, and the default result above was computed the same way when this page was built.
Worked example: a 1-in-37 chance (one roulette number), 37 tries.
- p = 1 ÷ 37 = 0.02703, so 1 − p = 0.97297
- Zero successes = 0.97297^37 ≈ 0.363, or 36.3%
- At least one success = 1 − 0.363 = 63.7%
- Expected successes = 37 × 0.02703 = 1
Picking one number for 37 spins gives you one hit on average, yet more than a third of the time you get none at all.
Losing streak odds for even-money roulette bets
Losing runs are where most bankrolls get hurt, so here are the exact figures for 100 bets on red with each wheel's win chance from our roulette data:
| Wheel | Win chance | 5+ losses in a row | 6+ in a row | 8+ in a row | 10+ in a row |
|---|---|---|---|---|---|
| European (single zero) | 48.65% | 84.6% | 59.7% | 20.2% | 5.6% |
| American (double zero) | 47.37% | 87.6% | 64.5% | 23.6% | 6.9% |
| Triple zero | 46.15% | 90.0% | 68.9% | 27.1% | 8.4% |
A run of 6 losses is more likely than not over 100 spins on any wheel. That matters for doubling systems: our martingale strategy calculator shows how many losses in a row a bankroll can cover, and this table shows how often that many arrive.
Checking the formula against a simulation
Formulas are only useful if they match what actually happens. Our plinko simulation dropped 1,000,000 balls per row setting. On a 16-row board, a ball lands in one of the 8 outer slots 2.13% of the time. Plug that into the zero-successes formula for 100 balls:
(1 − 0.02127)^100 = 11.65%Our simulation found no outer-slot hit in 100 balls 11.65% of the time.
Reading the results without the gambler's fallacy
Two things trip people up. First, "1 in N" doesn't mean it happens within N tries: the at-least-once chance for N tries is only about 63% for any large N. Second, a streak doesn't make the opposite result more likely. Believing a win is due after a run of losses is the gambler's fallacy, which wrongly treats independent plays as connected.
The streak figure answers "how often do runs like this show up?", which is useful for sizing bets and setting limits. It says nothing about the next spin. In the long run, results settle towards the true probability, which is the law of large numbers explained in our guide to odds and probability.
Using it for casino games
- Roulette: use 18 of 37 (48.65%) for red on a single-zero wheel, or 1 in 37 for a single number. Our roulette strategy test shows what streaks do to betting systems.
- Crypto dice: your win chance is the target divided by 100 at roll-under. The dice odds calculator gives the matching payout.
- Crash: the chance of reaching a 2x cash-out at a 1% edge is 49.5%, per our crash gambling guide.
None of these numbers change the house edge. They show how bumpy the ride is, not how to win it. More probability guides are in our odds and house edge hub.
Embed this calculator
Frequently asked questions
How do you calculate the chance of something happening at least once?
Work out the chance it never happens, then subtract from 1. If each try succeeds with probability p, the chance of no successes in n independent tries is (1 − p)^n, so at least one success is 1 − (1 − p)^n. A 1-in-37 chance tried 37 times gives 1 − (36/37)^37 ≈ 63.7%, not 100%.
If something is 1 in 100, will it happen in 100 tries?
Not necessarily. The chance of at least one hit in 100 tries at 1 in 100 is 1 − 0.99^100 ≈ 63.4%. About a third of the time it won't happen at all. You expect one success on average, but averages hide a lot of variation in short runs.
How likely is a long losing streak in roulette?
More likely than most people think. Betting red on a single-zero wheel loses 19 spins in 37. Our exact calculation puts the chance of at least one run of 8 or more losses in 100 spins at about 20%, and a run of 6 or more at about 60%. Use the calculator to try your own numbers.
Does a losing streak make a win more likely next time?
No. Roulette spins, dice rolls and crash rounds are independent, so the chance on the next try is the same as on every other try. Believing a win is 'due' is the gambler's fallacy. The streak figures here describe how often runs occur, not what happens next.
What is the difference between expected successes and the chance of success?
Expected successes (n × p) is the long-run average count. The chance of at least one success is a probability between 0% and 100%. With p = 1% and n = 300 you expect 3 successes on average, and the chance of getting at least one is about 95%.
Sources
- Two Basic Rules of Probability — OpenStax, Rice University. Multiplication rule; independent events
- Terminology — OpenStax, Rice University. Probability as long-run relative frequency; law of large numbers
- Gambling: Chances, probabilities and odds — Encyclopaedia Britannica. Probability = favourable/total; odds = unfavourable to favourable
- Doctrine of the maturity of the chances — Encyclopaedia Britannica. Gambler's fallacy: falsely assumes plays are dependent
- Roulette Basics — Wizard of Odds. House edge: double-zero 5.26%, five-number bet 7.89%; single-zero 2.70%; triple-zero 7.69%; la partage about 1.35% on even-money bets