How to use the streak probability calculator
- Enter the chance of success on each trial as a percent, for example 50 for a coin flip or 38 for a shooter who hits 38% of shots.
- Enter the run length you are asking about, such as 5 wins in a row, and the number of trials available.
- Read the headline chance that the run appears at least once. The stats add the chance of no run and the expected number of trials until the first run.
- Check the second table for how many trials it takes to reach a 50%, 90% or 99% chance of seeing the run.
Formula
state = length of the current run of successes, from 0 to k - 1
success: state + 1 (a run of k ends the search) failure: state returns to 0
expected trials until the first run of k = (1 - p^k) / ((1 - p) p^k)
The chance of a run is found by following how likely each run length is after every trial, which is exact. The expected wait has a closed form. A run longer than k also counts, because it contains a run of k.
Five heads in a row somewhere in 100 flips
With a fair coin (50%), a run length of 5 and 100 trials, the chance of at least one run of five heads is 81.011%, so a streak like that is more likely than not. The expected wait for the first one is 62 flips, and a 50% chance of seeing one is reached after 44 flips and 90% after 138.
A less likely run changes the picture. For a 30% chance of success, runs of 7 and 250 trials, the chance of at least one run is 3.677%, and the expected wait is 6,530.68 trials. With a 90% chance and a run of 10 in 50 trials the chance is 97.659%. For small cases such as 3 in a row within 10 fair flips, the exact answer of 50.78125% was confirmed by listing all 1,024 sequences.
Why streaks look surprising
People tend to expect a random sequence to alternate more than it does. Runs are common in long sequences: the longer the sequence, the more places a run could start. This page puts numbers on that, so you can tell whether a hot streak or a losing streak is unusual or only what luck produces.
Streaks of a given length can start at any trial, which is why the chance of at least one run in many trials is much larger than the chance that a particular set of trials is all successes.
Limits of the model
The model assumes every trial has the same chance of success and that trials do not influence each other. Skill changes, fatigue, opponents and momentum break that assumption in real games, so treat the result as the chance under pure repeated luck. For the number of successes in a fixed number of trials, use the Binomial Probability Calculator, and for waits for a single success the Geometric and Negative Binomial Calculator.
More probability and statistics tools
To convert scores, plan samples or study waiting times, try the Exponential Distribution Calculator, Geometric and Negative Binomial Calculator and Cohen's Kappa Calculator.
Assumptions and limits
- Each trial succeeds with the same chance, and trials are independent of each other.
- A run of at least k successes counts, so a longer run still counts as one run of k.
- Run length times the number of trials is limited so that the calculation finishes quickly in your browser.
- The searches for 50%, 90% and 99% stop after a fixed amount of work, so very rare runs show as beyond the limit.
- The example chances are an illustration. Replace them with your own.
Frequently asked questions
Is a run of 10 heads in 1,000 flips unusual?
It is less rare than most people think. With a fair coin the chance of at least one run of 10 in 1,000 flips is about 38.5%. Enter 50, 10 and 1000 to see the exact figure on this page.
What is the difference between this and the binomial?
The binomial counts how many successes there are in total, in any order. This page asks whether successes ever come in a row, which depends on the order, so it needs a different calculation.
Why is the expected wait so large when the chance is small?
Each extra success needed multiplies the difficulty by the chance of success, so the wait grows very quickly with the run length. A run of 7 at 30% takes about 6,530 trials on average.
Does a past streak make the next trial more likely to succeed?
Not if trials are independent. The chance on the next trial stays the same after any streak. Past results only matter if something real, such as skill, is changing.