The maximum run (or longest run) in n Bernoulli trials is the length of the longest consecutive block of successes. It is not the same as the total number of successes: two sequences can contain the same number of successes but have very different longest runs.
For independent trials with constant success probability p, the longest run typically grows on a logarithmic scale, roughly like log1/p(n) for large n. For a specific finite n and p, use an exact finite-state calculation when accuracy matters.
What “maximum run” means
Let X1, …, Xn be independent Bernoulli trials, with success probability p in every trial. A success might be heads in a coin-toss sequence. Define Ln as the largest number of consecutive successes anywhere in the sequence.
For example, the sequence S F S S F S S S contains six successes in total, but its maximum success run is three. The ordering matters as much as the count.
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This definition concerns success runs only. The longest block of either successes or failures is a different statistic and requires a different calculation.
Choose the probability question first
“The probability of a run” can refer to several different events. State the event, the number of trials, and the success probability before calculating.
- Longest run at most k: P(Ln ≤ k).
- Longest run at least k: P(Ln ≥ k) = 1 − P(Ln ≤ k − 1).
- Exactly k: P(Ln = k) = P(Ln ≤ k) − P(Ln ≤ k − 1).
These are not binomial probabilities. A binomial calculation gives the probability of a total number of successes, not their longest consecutive block.
Exact finite-sample calculation
To calculate P(Ln ≤ k), require that every block of k + 1 consecutive trials contain at least one failure. A finite-state recurrence tracks the current terminal success streak and discards any sequence that reaches k + 1.
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After each trial, keep probabilities for states 0, 1, …, k. State j means that the sequence currently ends with exactly j consecutive successes, while no run longer than k has occurred.
Transition rule
- A failure, with probability 1 − p, sends every state to state 0.
- A success, with probability p, sends state j to state j + 1 when j < k.
- A success from state k is excluded, because it creates a run of k + 1.
Recurrence
Let at,j be the probability that after t trials the current streak is j and no run exceeds k. Start with a0,0 = 1 and a0,j = 0 for j > 0. For each new trial:
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- at+1,0 = (1 − p) Σj=0k at,j
- at+1,j = pat,j−1 for 1 ≤ j ≤ k
Then P(Ln ≤ k) = Σj=0k an,j. This recurrence is exact for the stated iid model and can be implemented with an array of k + 1 numbers; it does not require enumerating all 2n sequences.
Practical algorithm
- Set an array
state[0..k]to zero and assignstate[0] = 1. - Repeat n times: compute the next array, putting
(1-p) * sum(state)in index 0 andp * state[j-1]in index j for each j from 1 through k. - Sum the final array to obtain P(Ln ≤ k).
- Use complements or differences for “at least” and “exactly” events.
When the total number of successes is fixed
Sometimes the question is conditional: exactly r of the n trials are successes, and the order is random. That is the event Sn = r, where Sn is the total success count.
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The relevant quantity is then P(Ln ≤ k | Sn = r). This is an arrangement problem over sequences containing r successes, not an unconditional iid calculation with success probability p. Philippou and Makri’s 1986 paper, Successes, runs and longest runs, gives a formula for this conditional distribution.
Do not replace a fixed-count question with a binomial model. The binomial model determines how likely each value of Sn is; once Sn is fixed, the calculation must account for the possible arrangements and their runs.
How the longest run grows for large n
For iid Bernoulli trials with constant p, the nominal large-sample scale is
Ln ≈ log1/p(n).
This is an asymptotic order-of-growth statement, not a guarantee that an observed sequence will contain exactly that run length. The discrete distribution can show integer effects and oscillations, and small samples can differ substantially from the logarithmic scale.
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The 2015 paper Laplace transform asymptotics and large deviation principles for longest success runs in Bernoulli trials develops this asymptotic behavior and gives an expansion for the mean containing a log1/p(n) term, a correction involving log1/p(1 − p), Euler’s constant γ ≈ 0.5772, and a residual described as small. That expansion is useful for asymptotic analysis; it should not be treated as an exact finite-sample mean.
Independent reader supportYour contribution helps us test, update, and keep practical guides available for everyone.A quick heuristic for the typical scale
A rule of thumb asks when the expected number of long runs is about one. A commonly cited approximation for the expected number of runs of length at least R is n(1 − p)pR. Setting this near one gives
R ≈ log1/p[n(1 − p)].
This estimate is a convenient intuition, not an exact distribution or a universal formula for the expected longest run. Its run-counting convention is approximate, and boundary effects, integer rounding, and dependence between overlapping candidate runs can matter.
Exact method or logarithmic estimate?
| Method | Best use | Strength | Limitation |
|---|---|---|---|
| Finite-state recurrence | Concrete n, p, and a specified event | Exact finite-sample probability under iid assumptions | Requires a separate calculation for each threshold or distribution value |
| Logarithmic scale | Large-n intuition and order-of-magnitude estimates | Shows how the longest run changes with n and p | Not an exact probability, and less reliable for small n or boundary values of p |
| Conditional arrangement calculation | Total successes fixed at r | Matches the information that the count is known | Not interchangeable with the unconditional Bernoulli model |
Assumptions and common pitfalls
- Independence: The recurrence and logarithmic results assume trials do not influence one another.
- Constant probability: Every trial uses the same p. If probabilities vary by trial, use a time-dependent model.
- Correct outcome: A longest-success-run result does not automatically describe the longest run of either outcome.
- Finite versus asymptotic: Logarithmic formulas describe large-sample behavior; exact recurrences are preferable when a decision depends on a particular finite probability.
- Extreme probabilities: Very small or very large p, and short sequences, make rough heuristics especially unreliable.
If outcomes are dependent, such as results generated by a changing physical process or a Markov chain, model that dependence directly rather than applying the iid Bernoulli formula by default.
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