Probability of Repeated Trials (n in a Row, At Least Once)

Enter the probability for one trial and the number of trials. Events A and B are calculated at the same time, and the probabilities of their combinations are shown too.

Probability of A in one trial

Number of trials for A

Probability of B in one trial

Number of trials for B

Enter probabilities as numbers from 0 to 1 (for example, 25% is 0.25) and numbers of trials as whole numbers from 1 to 1000. If you only need one event, still put placeholder values for event B (for example, 0.5 and 1).

Result

Enter the probabilities and the numbers of trials in the fields on the left and press "Calculate". The result will appear here.

[ CLOSE ]

What you can do on this page
What is this calculation used for?
How to Use
Formula
Symbols and terms
- Symbols
- Terms
Good to know before you start
How to calculate it in Excel
How to calculate it in Google Sheets
How to calculate it in Python
How to write it in LaTeX and other math languages (copy and paste)
How to have ChatGPT do the calculation
DataChef Features
Related Features
NumberChef Calculators List

What you can do on this page

Enter the probability for one trial and the number of trials, and you get the probability that it happens \(n\) times in a row on the spot
It also calculates the probability that it happens at least once and the probability that it never happens, all at once
It works with two events, A and B, at the same time, so you can also get combinations such as "A happens at least once and B never happens"
A plain-language explanation of the formulas and copy-and-paste formulas for Excel, Google Sheets and Python are all on this page

This works only when the probability is the same every time and the results do not affect each other (they are independent). Dice, coins, and loot boxes with published odds are fine. It does not work when the probability changes from one try to the next, as in "drawing tickets without putting them back".

What is this calculation used for?

Estimating your odds with loot boxes and prize draws in games

"If I open a loot box with a 1% drop rate 100 times, I should get the item at least once." In fact, the chance of at least one drop is \(1 - 0.99^{100} \approx 0.634\) (63.4%), so about 1 in 3 players still get nothing after 100 tries.
Knowing the "at least once" formula lets you plan your spending and number of tries with numbers instead of gut feeling.

Not panicking over a "false positive" in a screening test (health)

A test can come back "positive" even when you do not have the disease (a false positive), and this happens a few percent of the time on each test. If it is 5% per test, the chance of having at least one false positive over 10 years of yearly screening is as high as \(1 - 0.95^{10} \approx 0.4\) (40%).
Knowing that a flagged result is not the same as having the disease, and that almost anyone may see one if they repeat a test often enough, helps you stay calm when asked to come back for a retest. It is one of the most useful ways this formula helps sort out a big worry with numbers.

What "a 100-year flood" really means (disaster preparedness)

A "100-year flood" is a flood with a 1% chance of happening in any given year. Over a 30-year mortgage, the chance of at least one such flood is \(1 - 0.99^{30} \approx 0.26\) (26%), more than 1 in 4. (Real flood risk assessments use more detailed models, but the idea is the same.)
Once you understand this formula, you can see that "it did not happen this year" does not make you safe, and that the probability steadily builds up as the period gets longer. This helps with decisions such as buying flood insurance or keeping emergency supplies.

Quality control (the probability that every part is good)

In a product with 10 parts, each with a 0.5% defect rate, the probability that not a single part is defective is \((1 - 0.005)^{10} \approx 0.951\) (95.1%).
The more parts there are, the lower the chance that "every part is good". Manufacturers use this formula to estimate the pass rate of the whole product and work backward to find the defect rate allowed for each part.

How many times to enter a sweepstakes or lottery to have a realistic chance

If you enter a drawing with a 10% chance of winning 7 times, the chance of winning at least once is \(1 - 0.9^{7} \approx 0.52\) (52%), just over even odds.
Being able to calculate "how many tries give me about a 50% chance" makes it easier to decide where to put your time and effort.

Formula

Probability that it happens \(n\) times in a row

Standard notation (the usual math form)

In words (symbols replaced with words)

\(P\) \(=\) \(p\) \(n\)

③ \(P\): probability of \(n\) in a row \(=\) ① \(p\): probability in one trial ② \(n\): number of trials

The formula in words

① Take the \(p\): probability in one trial

② multiply it by itself as many times as the \(n\): number of trials

③ and you get the \(P\): probability of \(n\) in a row

Quick example

The probability of getting heads on a coin (probability 0.5) 3 times in a row is

\(P\): 3 in a row \(=\) one trial (0.5) trials (3)

\(0.5 \times 0.5 \times 0.5 = 0.125\ \ (12.5\%)\)

Key idea

This formula works only when the probability is the same every time and does not depend on earlier results (the trials are independent). Dice, coins, and loot boxes with published odds are fine. It does not work when the probability changes from one try to the next, as in "drawing tickets without putting them back".

Probability that it never happens in \(n\) trials

Standard notation (the usual math form)

\(P\) \(=\) \((1-p)\) \(n\)

In words (symbols replaced with words)

③ \(P\): probability it never happens \(=\) ① \((1-p)\): probability it does not happen in one trial ② \(n\): number of trials

The formula in words

① Take the \((1-p)\): probability it does not happen in one trial

② multiply it by itself as many times as the \(n\): number of trials

③ and you get the \(P\): probability it never happens

Quick example

If you open a loot box with a 0.25 chance of a rare item 4 times, the probability of missing all 4 times is

\(P\): missing all 4 times \(=\) missing once (0.75) trials (4)

\(0.75 \times 0.75 \times 0.75 \times 0.75 \approx 0.316\ \ (31.6\%)\)

Key idea

The probability that it does not happen in one trial comes from the complement: 1 − (the probability that it happens in one trial). If the chance of a rare item is 0.25, the chance of missing is \(1 - 0.25 = 0.75\).

Probability that it happens at least once in \(n\) trials

Standard notation (the usual math form)

\(P\) \(=\) \(1\) \(-\) \((1-p)\) \(n\)

In words (symbols replaced with words)

④ \(P\): probability of at least once \(=\) ③ \(1\): total probability \(-\) ① \((1-p)\): probability it does not happen in one trial ② \(n\): number of trials

The formula in words

① Take the \((1-p)\): probability it does not happen in one trial

② multiply it by itself as many times as the \(n\): number of trials to get the probability that it never happens

③ subtract that from the \(1\): total probability

④ and you get the \(P\): probability of at least once

Quick example

If you open a loot box with a 0.25 chance of a rare item 4 times, the probability of getting at least one rare item is

\(P\): at least one rare item \(=\) \(1\): total probability \(-\) missing once (0.75) trials (4)

\(0.75 \times 0.75 \times 0.75 \times 0.75 \approx 0.316\)

\(1 - 0.316 = 0.684\ \ (68.4\%)\)

Key idea

If you try to count "at least once" directly, you have to add up "exactly once", "exactly twice", and so on, which is a lot of work. The opposite side (the complement), "never", takes just one formula. So you find that and subtract it from the total probability, 1. This is the key idea of the formula. Also, adding more trials only brings the probability closer to 1 (100%); it never reaches 100%. "Open a loot box with a 1% chance 100 times, and the chance of at least one rare item is only about 63%."

For repeated trials, the basic rule is: multiply the probability for one trial by itself as many times as there are trials. For "at least once", the trick is to find the complement, "never happens", and subtract it from the total, 1.

Symbols and terms

Symbols

\(p\)	p	The probability that the event happens in one trial. (Example - a loot box with a 0.25 chance of a rare item)
\(n\)	n	The number of trials. (Example - for 4 tries, \(n = 4\))
\(p^n\)	p to the n-th power	\(p\) multiplied by itself \(n\) times. The small \(n\) at the upper right is the exponent, meaning "multiply \(n\) times". (Example - \(0.5^3 = 0.5 \times 0.5 \times 0.5 = 0.125\))
\(1-p\)	one minus p	The probability that the event does not happen in one trial (the probability of the complement).
\(P\)	capital P	The probability you want to find. On this page it stands for things like the probability of \(n\) in a row or the probability of at least once.

Terms

event	Something that either happens or does not, such as "rolling a 6" or "winning a raffle". In probability, such outcomes are called events.
trial	One single run of an action with a chance outcome, such as one roll of a die or one draw of a ticket. The formulas on this page are for repeating the same trial \(n\) times.
independent	When an earlier result does not change how likely the next one is. Dice and coins have the same probability on every throw, however many times you throw them, so the throws are independent.
repeated trials	Doing the same trial under the same conditions many times. In statistics courses, independent repeated trials with two outcomes (happens or not) are called Bernoulli trials, and they are the basis of the binomial distribution.
complement	The opposite event, that the event does not happen. The complement of "happens at least once" is "never happens", and this switch is the key to the third formula on this page.
exponent (power)	Multiplying the same number by itself several times. In \(0.5^3\), the small number at the upper right (the exponent) tells how many times to multiply.

Good to know before you start

Here is what helps you use the calculation on this page with real understanding, not just by pressing the button.

Basic probability (Grade 7)	Knowing that a probability is a number from 0 to 1, and that the probability something happens and the probability it does not add up to 1 Knowing that the probability of independent events happening one after another is found by multiplying
Exponents (Grade 6)	Knowing that "to the \(n\)-th power" means multiplying the same number \(n\) times, as in \(0.5^3 = 0.5 \times 0.5 \times 0.5\)
The complement (Grade 7 to high school)	Being able to see that the opposite of "happens at least once" is "never happens" Knowing that when a probability is hard to count directly, you can calculate the opposite side and subtract it from 1
Multiplying decimals (Grades 5–6)	Being able to multiply decimals, as in \(0.75 \times 0.75\) Having a feel for how the answer keeps getting smaller each time you multiply by a number less than 1

How to calculate it in Excel

Copy the whole table below and paste it into cell A1 in Excel. It works as is.

Table to find the probability of n in a row

Probability in one trial p	0.25
Number of trials n	4
Probability of n in a row	=B1^B2

Table to find the probability of never in n trials

Probability in one trial p	0.25
Number of trials n	4
Probability of never	=(1-B1)^B2

Table to find the probability of at least once

Probability in one trial p	0.25
Number of trials n	4
Probability of at least once	=1-(1-B1)^B2

After pasting, B1 and B2 are your inputs and B3 is calculated automatically.
"^" is the symbol for a power (how many times to multiply). "=B1^B2" means "multiply the value in B1 by itself B2 times".
In the third table, for example, B3 shows about 0.684 (about 68%). Just replace B1 and B2 with your own numbers.

How to calculate it in Google Sheets

Copy the whole table below and paste it into cell A1 in Google Sheets. It works as is.

Table to find the probability of n in a row

Probability in one trial p	0.25
Number of trials n	4
Probability of n in a row	=B1^B2

Table to find the probability of never in n trials

Probability in one trial p	0.25
Number of trials n	4
Probability of never	=(1-B1)^B2

Table to find the probability of at least once

Probability in one trial p	0.25
Number of trials n	4
Probability of at least once	=1-(1-B1)^B2

The same formulas as in Excel work as is. Copy the whole table, paste it into cell A1, and replace B1 and B2 with your own numbers.

How to calculate it in Python

probability_per_trial = 0.25  # probability in one trial
number_of_trials = 4          # number of trials

p_every_time = probability_per_trial ** number_of_trials        # probability of n in a row
p_never = (1 - probability_per_trial) ** number_of_trials       # probability of never
p_at_least_once = 1 - p_never                                   # probability of at least once

print(f"Probability of {number_of_trials} in a row: {p_every_time}")
print(f"Probability of never: {p_never}")
print(f"Probability of at least once: {p_at_least_once}")

Runs with the standard library only. "**" is the symbol for a power (to the n-th power). Change the probability and the number of trials at the top, and run it.

How to write it in LaTeX and other math languages (copy and paste)

Probability that it happens \(n\) times in a row

P = pⁿ

P = p^{n}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>P</mi>
    <mo>=</mo>
    <msup><mi>p</mi><mi>n</mi></msup>
  </mrow>
</math>

P = p^n

p^n

P := p^n;

P = p^n;

P = p^n

Probability that it never happens in \(n\) trials

P = (1 − p)ⁿ

P = (1-p)^{n}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>P</mi>
    <mo>=</mo>
    <msup>
      <mrow><mo>(</mo><mn>1</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow>
      <mi>n</mi>
    </msup>
  </mrow>
</math>

P = (1 - p)^n

(1 - p)^n

P := (1 - p)^n;

P = (1 - p)^n;

P = (1 - p)^n

Probability that it happens at least once in \(n\) trials

P = 1 − (1 − p)ⁿ

P = 1 - (1-p)^{n}

<math xmlns="http://www.w3.org/1998/Math/MathML" display="block">
  <mrow>
    <mi>P</mi>
    <mo>=</mo>
    <mn>1</mn>
    <mo>&#x2212;</mo>
    <msup>
      <mrow><mo>(</mo><mn>1</mn><mo>&#x2212;</mo><mi>p</mi><mo>)</mo></mrow>
      <mi>n</mi>
    </msup>
  </mrow>
</math>

P = 1 - (1 - p)^n

1 - (1 - p)^n

P := 1 - (1 - p)^n;

P = 1 - (1 - p)^n;

P = 1 - (1 - p)^n

How to have ChatGPT do the calculation

You are a probability calculation assistant. Do the following calculation by actually running Python code, and base your answer only on the numbers from the execution result (do not answer by mental math or guessing).

I open a loot box 4 times. Each time, the chance of a rare item is 0.25. The results are independent.
Find each of the following:
1. The probability of getting a rare item all 4 times
2. The probability of missing all 4 times
3. The probability of getting at least one rare item

Show the formulas you used and the numbers from the execution result.

How to Use

1

Enter your numbers

Type the numbers you want to calculate with into the input fields
2

Calculate

Press the "Calculate" button
3

Check the result

The result appears on the spot. The same page also explains the idea behind the calculation and the formula