5 reds in a row: the odds — both of them

Two different questions hide in this query. A specific window: the next 5 spins all red = (18/37)⁵ = 2.725% (1 in 37). Your actual session: the chance of seeing a run of 5+ reds at least once is 77.0% in 100 spins, 95.0% in 200, and it’s practically certain — computed exactly with a Markov chain (method), a figure most sites never touch.⚙ computed

Will you see it? Session odds

Session length	P(streak ≥ 5 at least once)	Expected count of such runs
50 spins	50.87%	0.66
100 spins	77.04%	1.36
200 spins	94.99%	2.76
500 spins	99.95%	6.95
1000 spins	100.00%	13.95

Five is where players start narrating — yet it still averages roughly one appearance per couple hundred spins: a scheduled guest, not an omen.

Why this streak matters: it’s a Martingale killer

Betting black through 5 reds on a $5 Martingale means your next bet is $160 and climbing. Our million-session Martingale run shows exactly how often these walls arrive.

The fallacy checkpoint

After 5 reds, the next spin is still red with probability 18/37 = 48.65%. The wheel has no memory — streaks are what independence looks like, not evidence against it. The 1913 Monte Carlo table famously ran 26 blacks while the room bet ever-heavier on red; the episode named the gambler’s fallacy — its place in the game’s full timeline here.

Other streak lengths

3 in a row 4 in a row 6 in a row 7 in a row 8 in a row 9 in a row odds hub →

Frequently asked

What are the odds of 5 reds in a row in roulette?

For a fixed window: (18/37)^5 ≈ 2.725% on a European wheel, about 1 in 37. On American wheels it’s rarer: (18/38)^5 ≈ 2.385%.

Will I ever see 5 reds in a row?

More often than intuition says: at least once in 95% of 200-spin sessions by exact Markov computation. Long streaks are a normal feature of independent spins.

After 5 reds, is black due?

No — each spin is independent, so black is still 48.65% (EU). Betting bigger on “due” colors is the gambler’s fallacy, and it’s exactly how streaks eat Martingale bankrolls.