Red ten times in a row on roulette: the probability explained
Before the first spin, ten reds in a row has a probability of (18/37)^10 on a single-zero wheel, about 0.074%, or 1 in 1,347. On a double-zero wheel it is (18/38)^10, about 0.057%, or 1 in 1,758. After nine reds, the next spin is still 18/37 or 18/38.
The short answer
A single-zero wheel has 18 red pockets out of 37, so one spin lands red with probability 18/37. Ten independent spins all landing red has probability (18/37)^10, about 0.074%. That is roughly one chance in 1,347, which is rare but far from impossible.
The same arithmetic underlies the gambler's fallacy, the belief that a long run makes the opposite colour due. This guide shows the calculation so you can check every figure yourself. Gambling is addictive: play responsibly, and only if you are 18 or over.
Step 1: the rules of the wheel
The probability follows from the layout of the wheel, so we start there. A European (single-zero) wheel has 37 pockets: 18 red, 18 black and 1 green (the 0). An American (double-zero) wheel has 38 pockets: 18 red, 18 black and 2 green (0 and 00). The differences between the two are set out in the guide to European and American roulette.
We assume a fair wheel, meaning every pocket is equally likely and each spin is unaffected by earlier ones. A licensed operator in Kenya must display the rules, odds and house edge of its casino games (regulation 20(2) of the 2026 Conduct of Gambling Operations Regulations), so the pocket count should be visible before any stake.
Step 2: the chance of red on one spin
With equally likely pockets, the chance of an outcome is the number of pockets that match divided by the total number of pockets.
European: P(red) = 18/37 ≈ 0.4865 (48.65%)
American: P(red) = 18/38 = 9/19 ≈ 0.4737 (47.37%)
Red is below one half on both wheels because the green pockets belong to neither colour. This is also why a red bet pays even money yet carries a house edge.
Step 3: multiply for independent spins
When spins are independent, the chance that all of them happen is the product of the individual chances. Two reds in a row is (18/37) × (18/37) = 324/1,369, about 23.67%. Ten reds multiplies ten copies of 18/37.
European: (18/37)^10 = 18^10 / 37^10 = 3,570,467,226,624 / 4,808,584,372,417,849 ≈ 0.000743
American: (9/19)^10 = 9^10 / 19^10 = 3,486,784,401 / 6,131,066,257,801 ≈ 0.000569
To state a probability as "1 in N", divide 1 by it. That gives N ≈ 1,347 for the single-zero wheel (0.0743%) and N ≈ 1,758 for the double-zero wheel (0.0569%).
Run lengths from 1 to 10
The same product works for every run length k: the chance of k reds in a row, starting from the next spin, is (18/37)^k or (18/38)^k. The table lists each value, rounded.
| Reds in a row (k) | Single zero (18/37)^k | Single zero, 1 in | Double zero (18/38)^k | Double zero, 1 in |
|---|---|---|---|---|
| 1 | 48.65% | 2.1 | 47.37% | 2.1 |
| 2 | 23.67% | 4.2 | 22.44% | 4.5 |
| 3 | 11.51% | 8.7 | 10.63% | 9.4 |
| 4 | 5.60% | 17.9 | 5.03% | 19.9 |
| 5 | 2.72% | 36.7 | 2.38% | 41.9 |
| 6 | 1.33% | 75.4 | 1.13% | 88.5 |
| 7 | 0.645% | 155 | 0.535% | 187 |
| 8 | 0.314% | 319 | 0.253% | 395 |
| 9 | 0.153% | 655 | 0.120% | 833 |
| 10 | 0.0743% | 1,347 | 0.0569% | 1,758 |
Each extra red multiplies the chance by about 0.49 on the single-zero wheel and 0.47 on the double-zero wheel. The figures fall quickly, but a run of five or six is common enough to be unremarkable. The same point appears in the guide to theoretical return against short-session results.
Two different questions
People often mix these up, and the mix-up is the gambler's fallacy.
- Before any spin: what is the chance that the next ten spins are all red? Answer: (18/37)^10, about 0.074%.
- After nine reds: what is the chance that the tenth spin is red? Answer: 18/37, about 48.65%.
The second answer comes from the first. The chance of ten reds is the chance of nine reds times the chance of one more, so dividing out the nine reds that have already happened leaves (18/37)^10 ÷ (18/37)^9 = 18/37. The nine reds are now history and cost nothing in probability. The wheel has no memory, and neither does a tested random number generator that stands in for one.
Why ten reds is not special
Take any specific sequence of ten spins that contains only reds and blacks, such as red, black, red, black and so on. Each position has probability 18/37, so the whole sequence has probability (18/37)^10 as well. Ten reds feels unusual only because we notice patterns and ignore messy sequences.
Sequences that include a green pocket have a different, smaller probability each, because green has only 1 or 2 pockets. Nothing about a run of one colour sets it apart mathematically from its mixed neighbours.
How often does a run of ten appear in a long series?
The chance that at least one run of ten reds appears somewhere in a long series needs a step-by-step recurrence, and this guide does not give a figure for it. What can be stated exactly is the expected number of windows. In a series of n spins there are n − 9 windows of ten consecutive spins, and each is all red with probability (18/37)^10.
In 1,000 spins there are 991 windows of ten. On the single-zero wheel the expected number of all-red windows is 991 × 0.000743 ≈ 0.74. On the double-zero wheel it is 991 × 0.000569 ≈ 0.56.
This is an average over many series, not a probability. Overlapping windows are not independent, so a run of eleven reds counts as two windows. In any single series the number can be zero, one or more.
What a streak does not change
A streak does not alter the house edge. The edge on a red bet is 1/37, about 2.70%, on the single-zero wheel, and 2/38 = 1/19, about 5.26%, on the double-zero wheel. The derivation is in the guide to the roulette house edge.
Ten reds have just landed. A stake of KES 100 on red or black for the next spin has an expected loss of 100 × 1/37 ≈ KES 2.70 on the single-zero wheel, or 100 × 1/19 ≈ KES 5.26 on the double-zero wheel. This is the same figure as before the streak began.
A change of colour does not become more likely, and a continuing run does not become more likely either. If a run makes a stake feel urgent or "due", that feeling is the signal to step back. Setting a time and loss limit before a session starts is covered in the guide to the gambler's fallacy and session limits; the responsible gambling page lists the regulator's toll-free line and the self-exclusion process.
Checking the numbers yourself
Anyone can repeat the calculation with a calculator that has a power key. Enter 18 ÷ 37, raise it to the tenth power, and compare it with 0.000743. For the double-zero wheel, enter 18 ÷ 38. For the "1 in N" form, take 1 divided by the result.
Questions and answers
What is the probability of red ten times in a row on a European wheel?
It is (18/37)^10 = 3,570,467,226,624 / 4,808,584,372,417,849, which is about 0.0743%, or roughly 1 in 1,347. The wheel has 37 pockets and 18 of them are red.
What is it on an American wheel?
It is (18/38)^10 = (9/19)^10 = 3,486,784,401 / 6,131,066,257,801, about 0.0569%, or roughly 1 in 1,758. The second zero means 38 pockets, still with 18 red.
After nine reds, is black more likely on the tenth spin?
No. The chance of red on the next spin is 18/37 (about 48.65%) on a single-zero wheel and 18/38 (about 47.37%) on a double-zero wheel, whatever happened before. Each spin is independent of the earlier ones.
Is ten reds in a row less likely than a mixed sequence such as red, black, red, black?
No. Any particular sequence of ten spins that contains only reds and blacks has the same probability, (18/37)^10 on the single-zero wheel. A run of ten reds only looks special because it is easy to describe.
Does a run of ten reds mean the wheel or the game is faulty?
Not by itself. A run of ten has a small but real chance on every ten spins, and over many spins such runs are expected to appear now and then. A fault is judged by testing a very large number of results, not by one streak.
Does a long run change the house edge?
No. The edge on a red or black bet is 1/37 (about 2.70%) on the single-zero wheel and 2/38 = 1/19 (about 5.26%) on the double-zero wheel, before and after any streak.