Coincidence Cluster (Streak Paradox)

why long streaks of coin flips are far more common than you think

Starting values — Coincidence Cluster (Streak Paradox)

Flips per sequence (N)	100
Streak length (L)	6
Simulation trials ℹ️	1000

Change any control above and the results recompute from these.

Loading interactive simulation...

The wheel was not the strange part 🖖

On 18 August 1913 the roulette wheel at Monte Carlo landed on black twenty-six times running, and the gamblers around it lost fortunes betting heavier and heavier on red, which was surely due. Red was not due. Every spin still paid 18 in 37, and the run itself was merely rare: the chance that any given spin starts twenty-six blacks is about one in 137 million. The name stuck — the gambler’s fallacy is still called the Monte Carlo fallacy. Set this tool to 100 flips and a run of six shows up 80.7% of the time; what no setting changes is the next flip.

Why long runs are almost guaranteed 🖖

Our gut expects heads and tails to alternate neatly, but randomness clumps. The longest run of identical outcomes grows with the number of flips — roughly log₂(N). Flip a fair coin 100 times and a streak of six-in-a-row appears more often than not. This tool lets you watch that intuition break: raise N and the streak curve climbs steeply.

Spotting a faked coin-flip log 🖖

Ask students to secretly fabricate 200 coin flips instead of tossing a real coin, and a statistician can usually pick the fakes at a glance. People inventing "random" sequences almost never write a run of six or seven identical results, yet genuine tosses nearly always contain one. This classroom demonstration (popularized by mathematician Theodore Hill) reveals that humans smooth out the very clumps real randomness insists on.

Problem solved in full

6 identical results in a row from 100 flips of a fair coin 5 steps
100 flips of a fair coin, and the question is whether 6 identical results in a row turn up somewhere in them. Count the sequences that escape — but 2¹⁰⁰ sequences will not be counted one at a time.
1. Cut a sequence into its maximal runs. Escaping a streak of 6 means every run is 5 long or shorter, so what you are counting is the ways to write 100 as an ordered sum of parts from 1 to 5, doubled because the first flip can be either face. Call that count R(i): a leading run of length j leaves R(i − j) behind it, and the five allowed lengths give five terms.
  
  $R (i) = R (i - 1) + R (i - 2) + R (i - 3) + R (i - 4) + R (i - 5), R (0) = 1$
2. Divide by 2^N to work in probabilities instead of counts, and each part of length j brings a factor 2^−j with it. Anything under 6 flips is safe outright, so q₁ through q₅ are 1, and the first real value is checkable by hand: exactly 2 of the 64 six-flip sequences are all-same, so q₆ = 31/32.
  
  $q_{N} = \frac{2 R ( N )}{2 ^{N}} = j = 1 \sum 5 2^{- j} q_{N - j}, q_{6} = \frac{31}{32}$
3. Up to 11 flips there is a shortcut, because two separate runs of 6 need 12 flips to fit side by side. Below that no sequence can contain two, so the probability of a streak is exactly the expected number of them — 2/64 for a run starting at flip 1, and 1/64 for each later start, which also needs the flip before it to differ. At N = 10 that is 6/64, so q₁₀ = 29/32 = 0.90625 with nothing approximated.
  
  $1 - q_{N} = \frac{2}{64} + \frac{N - 6}{64} = \frac{N - 4}{64} (6 \leq N \leq 11)$
4. Past 11 flips the events overlap and the counting shortcut dies, but the recurrence settles into steady geometric decay. Substituting q_N = Ax^N leaves five powers of 2x summing to 1, and the geometric-series formula folds those five terms into one: y = 2 − y⁻⁵, with y = 2x. Iterate from y = 2 and it converges in four steps — 1.968750, 1.966190, 1.965969, 1.965950 — to y = 1.965948, so x = 0.982974.
  
  $1 = j = 1 \sum 5 (2 x)^{- j} ⟹ y = 2 - y^{- 5}, y = 2 x = 1.965948$
5. Ninety flips separate the exact anchor at N = 10 from the target at N = 100. Geometric decay has barely started at N = 10, so this is an approximation, but the correction dies fast enough that it reproduces both decimals the tool prints from its exact recurrence.
  
  $q_{100} \approx q_{10} x^{90} = 0.90625 \times 0.98297 4^{90} = 0.1932 ⟹ P = 0.8068$
Answer

19.32% of 100-flip sequences escape, so the table's 80.68% says a run of 6 is the ordinary outcome, not the eerie one. The number that carries the lesson is 0.982974: each extra flip cuts your chance of escaping by 1.7%, halving it every 40.4 flips. Run that backwards and even odds arrive at 45 flips; run it forwards and 180 flips leave under 5% of sequences clean. The simulated column beside the exact one is a different species of number. A 1000-trial estimate of an 80.68% event carries a standard deviation of √(0.8068×0.1932/1000) = 1.25 percentage points, so it moves by more than a whole point between runs, and even at this tool's ceiling of 10,000 trials it is still ±0.39 points. The exact column's second decimal means something and the simulated one's cannot — which is precisely what the z row underneath is measuring.

$\frac{ln 2}{- ln 0.982974} = 40.4, σ = \frac{0.8068 \times 0.1932}{1000} = 0.0125$

References (2)

Insight block 3 — why a fabricated coin-flip log gives itself away: T. P. Hill, "The Difficulty of Faking Data." CHANCE 12(3), 27–31, 1999.
And the general result behind the tool: P. Diaconis and F. Mosteller, "Methods for Studying Coincidences." Journal of the American Statistical Association 84(408), 853–861, 1989 — why surprising runs are expected rather than remarkable.

By the 42Math team · Updated 2026-07-29

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

100 flips, streak of 6 - With 100 coin flips and looking for a streak of 6 heads or tails, the exact probability is 80.7%. Most people think 6 in a row is highly unusual, but it's more likely than not to happen in 100 tosses!
50 flips, streak of 5 - With 50 flips, a streak of 5 consecutive heads or tails is highly expected, with a 82.1% probability of occurring naturally.
150 flips, streak of 7 - With 150 flips, a streak of 7 consecutive heads or tails has a 69.7% probability. This shows how clustering makes runs of identical outcomes inevitable in longer sequences.