Audited 27 Jul 2026·Last updated 28 Oct 2026·5 citations·Tier 1·0 uses

Probability Calculator

Combine probabilities for independent events — both happen, at least one happens — or take a complement, and see the result as a decimal, percent and odds.

Probability Calculator

What do you want to compute?
Enter as a decimal between 0 and 1. A fair coin landing heads is 0.5; rolling a six on a fair die is about 0.1667 (1/6); drawing an ace from a full deck is about 0.0769 (4/52).
Only used by the AND and OR modes. Leave at default for a single-event or complement calculation. Must be in [0, 1].
Probability
0.5
The probability the chosen rule produces, as a decimal from 0 to 1: P(A), P(A and B), P(A or B) or P(not A).
Probability as a percentage
50.00%
Complement P(not event) = 1 − P
0.5
Odds (P / (1 − P))
1

Background.

If there is a 50% chance of rain on Saturday and a 50% chance on Sunday, is a wet weekend certain? Adding the two says 100%, and that is wrong: the sum counts twice every weekend on which it rains both days. Take that overlap out and, if the two days were independent, the chance of at least one wet day is 0.5 + 0.5 − 0.25 = 75%. Weather isn't independent from one day to the next — a wet Saturday makes a wet Sunday more likely — so the real figure is usually lower still. Two small corrections, and between them they cover most of what goes wrong when people combine chances by instinct.

Four rules are on offer. A single probability can be restated as a percentage, a complement and odds. Two independent probabilities can be combined as both happening, P(A) × P(B), or at least one happening, P(A) + P(B) − P(A) × P(B). And the complement turns the chance of something into the chance of it not happening, 1 − P(A), which is often the quickest route to an awkward answer: the chance of at least one six in four rolls of a die is 1 − (5/6)⁴ = 0.5177.

Probabilities go in as decimals from 0 to 1, so a 30% chance is 0.3. The two-event rules are right only when the events are independent, meaning that learning one happened leaves the chance of the other unchanged: separate coins, separate dice, draws that are put back. Cards dealt from one deck, two symptoms of the same illness and the weather on neighbouring days are not independent, and multiplying their probabilities can give an answer that is badly wrong.

What is probability calculator?

A probability is a number from 0 to 1 measuring how likely an event is, with 0 meaning impossible and 1 certain. When all the outcomes of an experiment are equally likely, it is the number of favourable outcomes over the total, so a six on a fair die is 1/6. The equally-likely condition is easy to break without noticing. The totals of two dice run from 2 to 12, but they are not equally likely: a 7 can be made six ways out of 36, so its chance is 6/36, while a 2 needs double ones, 1/36. More generally, the rules come from three axioms Andrey Kolmogorov set out in 1933: no probability is negative, the probability that one of all the possible outcomes occurs is 1, and the probabilities of events that cannot happen together simply add. Every rule used here follows from those three plus one definition. Two events are independent when P(A and B) = P(A) × P(B), the formal way of saying that knowing one tells you nothing about the other. Mutually exclusive is a different idea: the events can't both happen, so P(A and B) = 0. Odds carry the same information as a probability in ratio form: the chance an event happens divided by the chance it doesn't, P ÷ (1 − P).

How to use this calculator.

  1. Pick the rule in What do you want to compute?: Single event to restate one probability, Both events occur for P(A and B), At least one occurs for P(A or B), Complement for P(not A).
  2. Enter Probability of A as a decimal: divide a percentage by 100, or work out a fraction first — 1/6 is 0.1667 and 4/52 is 0.0769. Values below 0 or above 1 are rejected. Rounding a fraction to four places costs almost nothing: 0.1667 × 0.1667 gives 0.0278, the same as 1/36 to four decimals.
  3. For the two-event rules, enter Probability of B as well; Single event and Complement ignore it. Before trusting the result, check that one event does not make the other more or less likely.
  4. Read the result as a decimal and as a percentage. The complement line is 1 minus the result, so in Complement mode it hands back P(A); the odds line is the result divided by its complement.
  5. For three or more independent events, chain the steps: combine A and B, then enter that result as A and the next event as B. For at least one success over many identical tries, 1 − (1 − p)ⁿ is quicker.

The formula.

P(A∩B) = P(A) × P(B)

Single event: result = P(A). Both, for independent events: P(A and B) = P(A) × P(B). At least one, for independent events: P(A or B) = P(A) + P(B) − P(A) × P(B). Complement: P(not A) = 1 − P(A). Whatever result R the rule produces, the percentage is 100 × R, the complement is 1 − R and the odds in favour are R ÷ (1 − R).

The OR rule is the general addition rule, P(A) + P(B) − P(A and B), with independence supplying the overlap as P(A) × P(B). Without the subtraction the overlap is counted twice, which is how two 50% chances turn into a false certainty. For events that cannot happen together the overlap is zero and plain addition is correct, but this rule always subtracts P(A) × P(B), so it is the wrong tool for them. Rolling a 1 or a 6 with one die is 1/6 + 1/6 = 0.3333. Putting 1/6 and 1/6 through the OR rule gives 11/36 = 0.3056, which answers a different question: the chance of at least one six when two separate dice are rolled.

Dependent events need a conditional probability in the multiplication rule: P(A and B) = P(A) × P(B given A). Drawing two aces in a row from one deck is 4/52 × 3/51 = 0.0045, or 1 in 221, because the first ace leaves three in a smaller deck. Treating the draws as independent, 4/52 × 4/52, gives 0.0059, about 31% too high.

Odds and probability convert with P = odds ÷ (1 + odds) and odds = P ÷ (1 − P). At P = 1 the odds are infinite, and the result displays 1.000 × 10³⁰⁸ as a stand-in.

A worked example.

Example

Two fair coins are tossed, and the question is the chance that both land heads. Each toss gives heads with probability 0.5 and neither coin can influence the other, so the Both events occur rule applies: P(A and B) = 0.5 × 0.5 = 0.25, or 25%. The complement, 1 − 0.25 = 0.75, is the chance that at least one coin shows tails. The odds in favour are 0.25 ÷ 0.75 = 0.3333, one to three, which a bookmaker would call 3 to 1 against. Listing the outcomes confirms it. HH, HT, TH and TT are equally likely, and only HH qualifies: one out of four. The same list answers the at-least-one-head question as well, three outcomes out of four, matching 0.5 + 0.5 − 0.25 = 0.75 from the OR rule.

modeindependent-and
probability B0.5
probability A0.5

Frequently asked questions.

Can two events be both independent and mutually exclusive?
Only if one of them has probability zero. Mutually exclusive events cannot happen together, so P(A and B) = 0, while independence needs P(A and B) to equal P(A) × P(B), which is positive whenever both chances are. Rolling a 1 and rolling a 6 on one die are mutually exclusive and strongly dependent: seeing a 1 tells you for certain the roll was not a 6.
What if my events affect each other?
Multiply the first probability by the probability of the second given the first. A box of 10 bulbs with 2 faulty: picking two without putting the first back, the chance both are faulty is 2/10 × 1/9 = 0.0222. The independent rule would give 0.2 × 0.2 = 0.04, nearly double, because it ignores that the first faulty bulb leaves fewer behind. When the pool is enormous next to the sample, as with a poll of 1,000 voters drawn from millions, removing each pick barely changes the rest, and the independent rule is a close approximation.
A coin has landed heads ten times running. Is tails due?
No. The coin has no memory, so the eleventh toss is still 0.5 either way. Eleven heads in a row was unlikely from the start, 0.5¹¹ = 4.883 × 10⁻⁴, or 1 in 2,048, but once ten have happened only the last toss is uncertain. Long-run proportions settle toward one half because later tosses swamp an early streak, not because the coin compensates for it.
How do I turn bookmaker odds into a probability?
For fractional odds of a to b against, the implied probability is b ÷ (a + b), so 5/1 is 1/6, about 0.1667. For decimal odds, divide 1 by the price: 2.50 implies 0.4. Add up the implied probabilities for every outcome of an event and the total comes to more than 1; the excess is the bookmaker's margin, which is why those figures overstate each outcome's true chance.
Why does a positive result on a 99%-accurate test often turn out to be wrong?
Because the condition may be rare. Suppose 1 person in 1,000 has it and the test catches 99% of cases while clearing 99% of healthy people. Among 100,000 people, 99 of the 100 who have it test positive, but so do 999 of the 99,900 who don't, so only about 9.02% of positives are real. Combining a prior with test evidence is a job for Bayes' theorem.
What are the chances of at least one hit over many tries?
Take the chance of missing every time and subtract it from 1: 1 − (1 − p)ⁿ. A 10% chance per attempt reaches 52.17% over seven attempts. The same arithmetic explains the so-called 100-year flood, which has a 1% chance in any year: over 30 years the chance of at least one is 1 − 0.99³⁰ = 26.03%, and over 100 years it is still only 63.40%. To find how many tries push the chance past one half, divide ln 0.5 by ln(1 − p): at 10% per try that is 6.58, so the seventh attempt crosses it.
Why did the Chevalier de Méré lose money on double sixes?
He reasoned in proportion. Betting on at least one six in four rolls of a die wins with probability 1 − (5/6)⁴ = 0.5177, so the bet paid over time. Double sixes are six times rarer, so he expected 24 rolls of two dice to be just as good, but 1 − (35/36)²⁴ = 0.4914, a slow loser. The puzzle, put to Blaise Pascal in 1654, helped start the mathematical study of chance.
How do I combine three or more independent events?
Apply the rules repeatedly. For all of them, multiply: three independent backup systems that each fail 5% of the time all fail together with probability 0.05³ = 1.25 × 10⁻⁴. For at least one, subtract the product of the misses from 1. Check the independence first, though; backups sharing a power supply or a software bug can fail together far more often than the product suggests.
What does a probability of 0.7 mean for something that happens only once?
It describes how often events judged that way occur, not a fraction of the event itself. A forecaster is well calibrated if, across all the occasions they say 0.7, the event happens about 70% of the time. A single outcome can't confirm or refute a 0.7 forecast; a long record of forecasts can, which is how weather services and prediction models are scored.
Why do the odds read 1.000 × 10³⁰⁸ when the probability is 1?
Odds divide the chance of an event by the chance of it not happening, and for a certain event the second figure is zero, so the odds are infinite. The display uses the largest practical number as a stand-in rather than an error. At the other end, a probability of 0 gives odds of 0, an event with no chance at all.

How this page was produced

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Quanta Calculator
Primary sources
5 cited below
Method
P(A∩B) = P(A) × P(B)
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