Probability Calculator

For independent events, P(A and B) = P(A) × P(B), so heads on a coin (1/2) and a six on a die (1/6) happen together 1/12 of the time, or 8.33%. The chance of at least one six in 4 rolls is 1 − (5/6)^4 = 51.77%, and the calculator also finds exactly k successes in n tries.

By the Calculators With Steps team · Updated November 10, 2026 · How we check our math

Type a decimal (0.25), a percentage (25%) or a fraction (1/4).

P(A and B), both happen
0.083333
8.33%, about 1 in 12
P(A or B), at least one
0.58333358.33%
P(A xor B), exactly one
0.550%
P(neither A nor B)
0.41666741.67%
P(not A)
0.550%
P(not B)
0.83333383.33%
Step-by-step working
  1. Both: P(A and B) = P(A) × P(B) = 0.5 × 0.166667 = 0.083333
  2. At least one: P(A or B) = P(A) + P(B) − P(A and B) = 0.5 + 0.166667 − 0.083333 = 0.583333
  3. Exactly one: P(A xor B) = P(A) + P(B) − 2 × P(A and B) = 0.5 + 0.166667 − 0.166667 = 0.5
  4. Neither: (1 − P(A)) × (1 − P(B)) = 0.5 × 0.833333 = 0.416667
  5. Complements: P(not A) = 1 − 0.5 = 0.5, P(not B) = 1 − 0.166667 = 0.833333

These formulas assume A and B are independent: one happening doesn't change the chance of the other.

How to use the probability calculator

Choose one of three modes:

  • Two independent events: enter P(A) and P(B) to get the chance of both, at least one, exactly one or neither happening, plus each complement.
  • At least one success in n tries: enter the chance of success on each try and the number of tries. You also see how many tries you need for a 50%, 90% or 99% chance.
  • Exactly k successes in n tries: the binomial distribution, with cumulative probabilities (at most k, at least k) and a full table of outcomes for up to 60 tries.

Probabilities can be typed as 0.25, 25% or 1/4. A plain number between 1 and 100, such as 30, is read as a percentage, and the result tells you when that happens.

Probability formulas for two independent events

Two events are independent when one happening does not change the chance of the other, such as flipping a coin and rolling a die. With P(A) = 0.5 (heads) and P(B) = 1/6 (rolling a six):

What you want Formula Heads and a six
Both A and B P(A) × P(B) 0.083333 (1 in 12)
A or B (at least one) P(A) + P(B) − P(A) × P(B) 0.583333
Exactly one (A xor B) P(A) + P(B) − 2 × P(A) × P(B) 0.5
Neither (1 − P(A)) × (1 − P(B)) 0.416667
Not A 1 − P(A) 0.5
Not B 1 − P(B) 0.833333

The “or” formula subtracts the overlap because adding P(A) and P(B) counts the case where both happen twice. Note that “or” plus “neither” always makes 1: 0.583333 + 0.416667 = 1.

Worked example: at least one six in four rolls

The French gambler known as the Chevalier de Méré asked why betting on at least one six in four rolls of a die seemed to win, while betting on at least one double six in 24 rolls of two dice seemed to lose. His puzzle is one of those discussed in the 1654 letters between Blaise Pascal and Pierre de Fermat that laid the foundations of probability theory.

The trick is to look at the chance of failing every time:

  1. The chance of not rolling a six on one roll is 5/6 = 0.833333.
  2. The rolls are independent, so the chance of no six in four rolls is 0.833333⁴ = 0.482253.
  3. The chance of at least one six is 1 − 0.482253 = 0.517747, just over 51.77%. A small edge for the bettor.

For a double six in 24 rolls, the chance per roll is 1/36, and 1 − (35/36)²⁴ = 0.491404, just under 50%. De Méré’s instinct was right, even though the two bets look alike.

Exactly k successes: the binomial formula

P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ

  • n = number of independent tries
  • k = number of successes you want
  • p = chance of success on each try
  • C(n, k) = n! ÷ (k! × (n − k)!), the number of ways to choose which tries succeed

Exactly 2 sixes in 10 rolls of a die: C(10, 2) = 45, (1/6)² = 0.027778 and (5/6)⁸ = 0.232568, so P = 45 × 0.027778 × 0.232568 = 0.29071, about 29%. The chance of at most 2 sixes is 0.775227, and of at least 2 sixes is 0.515483. The mean number of sixes is n × p = 1.6667.

How many tries until success is likely?

Solving 1 − (1 − p)ⁿ ≥ target for n gives n = ln(1 − target) ÷ ln(1 − p), rounded up. For an item with a 1% drop rate in a game, or any 1-in-100 chance:

Target chance Tries needed at p = 1% Tries needed at p = 1/6
50% 69 4
90% 230 13
99% 459 26

Even after 100 tries at 1%, there is still a 36.6% chance of no success at all. Bad luck streaks are more common than intuition suggests.

Common probability mistakes

  • Adding chances across tries. Six rolls at 1/6 each is not a 100% chance of a six. It is 1 − (5/6)⁶ = 66.51%.
  • The gambler’s fallacy. After five tails in a row, a fair coin is still 50% heads on the next flip. Independent tries have no memory.
  • Treating dependent events as independent. Drawing two aces from a deck without replacement changes the odds on the second draw, so the simple product rule does not apply.
  • Mixing percentages and decimals. 30% is 0.3, not 30. Multiplying 30 × 50 instead of 0.3 × 0.5 gives nonsense.

To draw fair random numbers for a game or experiment, use the random number generator. For probabilities under a bell curve, such as test scores or measurements, use the z-score calculator. The fraction calculator is handy for keeping dice and card probabilities exact.

Probability Calculator: frequently asked questions

How do you calculate the probability of A and B?

For independent events, multiply the probabilities: P(A and B) = P(A) × P(B). The chance of flipping heads (0.5) and rolling a six (1/6) is 0.5 × 1/6 = 1/12, about 0.083333 or 8.33%.

How do you calculate the probability of A or B?

Add the two probabilities and subtract the overlap so it isn't counted twice: P(A or B) = P(A) + P(B) − P(A and B). For heads or a six, that is 0.5 + 0.166667 − 0.083333 = 0.583333, or 58.33%.

What is the probability of at least one success in several tries?

Work out the chance of failing every time and subtract it from 1: P = 1 − (1 − p)ⁿ. With a 1% chance per try and 100 tries, P = 1 − 0.99¹⁰⁰ = 0.633968, about 63.4%, not 100%.

What are the odds of rolling at least one six in six rolls?

1 − (5/6)⁶ = 0.665102, or about 66.5%. Many people guess 100% because six rolls times a 1/6 chance looks like 1, but probabilities of repeated tries do not simply add.

What is the difference between independent and mutually exclusive events?

Independent events don't affect each other, like two separate coin flips, so P(A and B) = P(A) × P(B). Mutually exclusive events can't both happen, like rolling a 2 and a 5 on one die, so P(A and B) = 0 and P(A or B) = P(A) + P(B). This calculator's two-event mode assumes independence.

How do I calculate the probability of exactly k successes?

Use the binomial formula P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ. The chance of exactly 5 heads in 10 fair coin flips is 252 × 0.5⁵ × 0.5⁵ = 252/1024 = 0.246094, about 24.6%.

Can a probability be greater than 1?

No. Every probability is between 0 (impossible) and 1 (certain), or 0% to 100%. If you get a number above 1, you have probably added probabilities that overlap or that belong to repeated tries.

How do I convert odds into a probability?

Odds of 1 to 3 in favor (often said as 3 to 1 against) mean 1 success for every 3 failures, so the probability is 1 ÷ (1 + 3) = 1/4 = 0.25, or 25%. Type 1/4 straight into the calculator.

How we checked this page: every figure in the examples and FAQs above is produced by this calculator and verified by automated tests against known answers, and the rules and rates come from the sources below. Last updated . Spotted a problem? Tell us.

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