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

Probability Calculator

Free probability calculator: compute single events, AND/OR for independent events, complements, percent chance, and odds — with Kolmogorov axioms.

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 of the event you selected, as a decimal in [0, 1]. For single mode this is P(A); for independent-and it is P(A)·P(B); for independent-or it is P(A) + P(B) − P(A)·P(B); for complement it is 1 − P(A).
Probability as a percentage
50.00%
Complement P(not event) = 1 − P
0.5
Odds (P / (1 − P))
1

Background.

The Quanta probability calculator handles the four operations that make up the core of every introductory probability course: a single event P(A), the joint probability of two independent events P(A and B), the probability that at least one of two independent events occurs P(A or B), and the complement P(not A) = 1 − P(A). Pick the mode, enter your probabilities as decimals between 0 and 1, and the calculator returns the raw probability, the same value expressed as a percent chance, the complement, and the odds — all four representations you might need to communicate the answer to a different audience.

Probability is, at its foundation, simpler than most people remember from school. Andrey Kolmogorov's 1933 monograph Grundbegriffe der Wahrscheinlichkeitsrechnung distilled the entire subject into three axioms: every event has a probability that is at least zero, the probability of the certain event (the whole sample space) is exactly one, and the probability of a union of disjoint events is the sum of their individual probabilities. Every rule the calculator above implements — multiplication for independent events, inclusion-exclusion for unions, the complement rule — follows from those three axioms by a few lines of algebra.

The thing that goes wrong, over and over, when people try to compute probabilities in real life is not the arithmetic. It is one of two confusions about what the inputs actually mean. The first confusion is between independent events and mutually exclusive events. Two events are independent when knowing one occurred tells you nothing about whether the other did — a coin flip in Nairobi and a die roll in Mombasa, or two successive spins of a roulette wheel. Two events are mutually exclusive when they cannot both happen at the same time — rolling a 1 and rolling a 6 on a single die, or the same horse finishing both first and second. Independent AND is multiplication: P(A and B) = P(A) · P(B). Mutually exclusive OR is plain addition: P(A or B) = P(A) + P(B). General OR for events that are neither necessarily independent nor mutually exclusive needs inclusion-exclusion: P(A or B) = P(A) + P(B) − P(A and B). Confusing these is how people end up claiming a 'fifty-fifty' probability for the wrong question.

The second confusion is the gambler's fallacy — the belief that a streak of one outcome makes the opposite outcome 'due'. If a fair coin lands heads ten times in a row, the eleventh flip is still 50/50. Independence is a permanent property of the random mechanism, not a tendency that fixes itself over time. The misconception arises because in the long run heads and tails do balance out to roughly 50/50 by the law of large numbers, but the balancing happens by overwhelming the streak with a much larger volume of subsequent flips, not by the coin remembering its past behaviour.

A related confusion is base rate neglect: if a medical test for a rare disease is 99% accurate and you test positive, your actual probability of having the disease can still be under 10% if the disease is rare enough — because the false positives from the healthy 99.9% of the population outnumber the true positives from the sick 0.1%. Bayes' rule formalises the correction: posterior odds = prior odds × likelihood ratio. The probability calculator above does not implement full Bayesian updating, but it does give you the odds representation that makes Bayesian reasoning natural — odds multiply, probabilities do not, which is why every Bayesian statistician quietly works in odds-space and only converts back to probabilities at the end for human consumption.

The remaining gotcha is the difference between probability and percentage. A probability of 0.25 and a percent chance of 25% are mathematically identical, but the percent form invites a category error: weather forecasts that say '30% chance of rain' are reporting a probability, not a fraction of the sky that will be wet or a fraction of the day during which it will rain. The calculator returns both representations so you can pick whichever one your audience understands without ambiguity. Worked examples below cover the canonical classroom problems — two fair coins both landing heads (probability 0.25, or 25%), at least one head out of two coins (probability 0.75, or 75%), rolling a six on a fair die (1/6 ≈ 16.67%, odds 1:5 against) — alongside notes on when to convert to odds and when to leave the answer as a probability, the difference between independent and mutually exclusive events spelled out with examples, a short note on conditional probability and Bayes' rule for the curious, and the reasons why even an apparently simple probability question can have the wrong answer if you mis-identify the underlying experiment.

What is probability calculator?

Probability is the formal mathematical measure of how likely an event is to occur, expressed as a number between 0 (impossible) and 1 (certain). The modern definition is due to Andrey Kolmogorov, whose 1933 monograph Grundbegriffe der Wahrscheinlichkeitsrechnung set out the three axioms that define a probability measure: non-negativity (every event has probability ≥ 0), normalisation (the whole sample space has probability 1), and countable additivity (the probability of a union of pairwise disjoint events is the sum of their probabilities). Every other rule in elementary probability is a theorem derivable from those three axioms. The complement rule, P(not A) = 1 − P(A), comes from applying additivity to A and its complement (which together cover the entire sample space). The general addition rule for any two events, P(A or B) = P(A) + P(B) − P(A and B), comes from inclusion-exclusion — subtract the intersection because it has been counted twice. Independence is a separate definition layered on top of the axioms: two events A and B are independent if and only if P(A and B) = P(A)·P(B). When that condition holds the addition rule simplifies to P(A or B) = P(A) + P(B) − P(A)·P(B), which is the formula the calculator uses for independent-or mode. Conditional probability is defined as P(A | B) = P(A and B) / P(B), and Bayes' theorem rearranges that into P(A | B) = P(B | A)·P(A) / P(B), giving the canonical recipe for updating a prior belief in light of new evidence. Odds are an equivalent representation: the odds in favour of A are P(A) / (1 − P(A)), and odds against are the reciprocal. Bookmakers, geneticists, epidemiologists, and Bayesian statisticians all prefer odds because Bayes' rule on odds is multiplicative: posterior odds = prior odds × likelihood ratio, no division required. Convert between probability and odds with p = odds / (1 + odds), and odds = p / (1 − p). The two representations carry exactly the same information; which to use is a question of audience, not mathematics.

How to use this calculator.

  1. Pick the mode that matches your question. 'Single event' just echoes P(A) back with its complement, percent, and odds — useful when you want the percent or odds form of a probability you already know. 'Both events occur (independent)' multiplies P(A) and P(B). 'At least one occurs (independent)' applies the addition rule with the independence simplification. 'Complement' returns 1 − P(A).
  2. Enter probabilities as decimals between 0 and 1, not percentages. A 25% chance is 0.25, not 25. The calculator rejects values outside [0, 1] because they cannot be valid probabilities — a 'probability' of 1.5 or −0.1 is a sign that you converted units incorrectly or are using the wrong rule.
  3. Confirm your events really are independent before using the AND or OR modes. Two coin flips, two die rolls, two roulette spins — yes. The two halves of a tennis match, the two legs of a parlay where the same team plays both, two stock returns on the same trading day — no. If the events influence each other you need conditional probability instead, which means computing P(A) and P(B | A) separately and multiplying, not P(A) · P(B).
  4. Read the percent output when you are presenting the result to a non-technical audience and the raw probability when you are feeding it into another formula. 0.25 and 25% are the same number; using the wrong representation in a calculation (multiplying by 100 twice, or forgetting to divide) is one of the most common arithmetic mistakes in applied probability.
  5. Use the odds output when the question is naturally about betting, risk ratios, or Bayesian updating. Odds of 4 (often written 4:1 on) mean the event is four times more likely to happen than not; odds of 0.25 (1:4 against) mean it is four times more likely not to happen than to happen. Conversion: odds = p / (1 − p); p = odds / (1 + odds).
  6. Apply the complement output as a sanity check. Probability + complement must equal exactly 1 — if you can compute the complement more easily than the event itself, do that and subtract. The classic example is 'at least one head in ten flips of a fair coin': the complement is 'no heads at all' = (1/2)^10 ≈ 0.000977, so the answer is 1 − 0.000977 ≈ 0.999, which is much easier than summing ten binomial terms.

The formula.

P(A∩B) = P(A) × P(B)

Four formulas cover every mode of the calculator, all derivable from Kolmogorov's three axioms. Single event: result = P(A). Nothing to compute beyond unit conversions — this mode exists so you can quickly translate a known probability into the percent, complement, and odds forms. AND for independent events (multiplication rule): result = P(A and B) = P(A) · P(B). The rule follows directly from the definition of independence: two events A and B are independent if and only if P(A ∩ B) = P(A)·P(B). It does not apply when A and B influence each other — for dependent events you need the general multiplication rule P(A and B) = P(A) · P(B | A) instead, which requires the conditional probability and is not what this mode computes. OR for any two events (inclusion-exclusion): P(A or B) = P(A) + P(B) − P(A and B). The intersection is subtracted because it would otherwise be counted once in P(A) and again in P(B). When A and B are independent this becomes result = P(A) + P(B) − P(A)·P(B), which is the formula used by the calculator's independent-or mode. When A and B are mutually exclusive the intersection is empty and the formula collapses to P(A or B) = P(A) + P(B), but this calculator does not assume mutual exclusivity — if your events are mutually exclusive, set their joint probability to zero by setting one of them to zero and adding manually, or use single mode on each and add. Complement rule: result = P(not A) = 1 − P(A). This is an immediate consequence of normalisation: A and its complement together make up the whole sample space, which has probability 1. Once result is known the calculator derives three secondary outputs: resultPercent = result × 100 (the same number in percent form), complement = 1 − result (the probability the event does not occur), and odds = result / (1 − result) (the ratio form used by bookmakers and Bayesians). The odds output is capped at 1×10^308 when result = 1, to keep the JSON serialisable; treat that sentinel as 'infinite odds' meaning certainty. All arithmetic runs through Decimal.js to avoid IEEE-754 rounding artefacts that would otherwise make 0.1 + 0.2 read as 0.30000000000000004 in the result panel.

A worked example.

Example

Worked example: flip two fair coins. What is the probability that both land heads? Each coin lands heads independently with probability 0.5, so P(A and B) = 0.5 × 0.5 = 0.25. The calculator returns result = 0.25, resultPercent = 25%, complement = 0.75 (the probability that at least one of the two flips is tails), and odds = 0.333 (one in three, or 1:3 against). Switch the mode to 'At least one occurs (independent)' with the same inputs and you get a different question: what is the probability of at least one head out of two flips? Inclusion-exclusion gives P(A or B) = 0.5 + 0.5 − 0.5·0.5 = 0.75 (75%, complement 0.25, odds 3:1 on). Sanity check: 'at least one head' is the complement of 'no heads at all' = 0.5·0.5 = 0.25, so 1 − 0.25 = 0.75 confirms the answer. Try a single-die example: probability of rolling a six is 1/6 ≈ 0.1667. Put 0.1667 into single mode and the calculator returns result ≈ 0.1667, resultPercent ≈ 16.67%, complement ≈ 0.8333 (the probability of rolling something other than six), and odds ≈ 0.2 — usually quoted as 1:5 against, which matches the intuition that there are five non-six outcomes for every six outcome on a fair die.

modeindependent-and
probability B0.5
probability A0.5

Frequently asked questions.

What is the difference between independent and mutually exclusive events?
Two events are independent if knowing that one happened tells you nothing about whether the other happened — formally, P(A and B) = P(A)·P(B). Two coin flips, two die rolls, two roulette spins are all independent. Two events are mutually exclusive if they cannot both happen at the same time — formally, P(A and B) = 0. Rolling a 1 and rolling a 6 on a single die are mutually exclusive. The two concepts are sometimes confused because both come up in 'AND' and 'OR' questions, but they are very different: independent AND uses multiplication, P(A)·P(B); mutually exclusive OR uses plain addition, P(A) + P(B). Independence is almost never compatible with mutual exclusivity — if two events both have positive probability and cannot both happen, they cannot be independent, because knowing one happened forces the other not to have happened.
How do I calculate the probability of A and B when they are not independent?
Use the general multiplication rule: P(A and B) = P(A) · P(B | A), where P(B | A) is the conditional probability of B given that A has occurred. The Quanta calculator's independent-and mode assumes independence, so do not use it when A influences B. Example: drawing two aces in a row from a 52-card deck without replacement. P(first ace) = 4/52. P(second ace | first was an ace) = 3/51 (one ace removed, one card removed). So P(both aces) = (4/52)·(3/51) ≈ 0.0045. If you mistakenly multiplied (4/52)·(4/52) you would get 0.0059 — a 30% overestimate, because you would have treated the two draws as independent when in fact the first draw makes the second less likely.
What is Bayes' theorem and when do I need it?
Bayes' theorem is the rule for updating a probability in light of new evidence: P(A | B) = P(B | A)·P(A) / P(B). It is the right tool whenever you have a prior belief about A, observe some related event B, and want to revise your belief. The most famous example is medical testing: if a disease has prevalence 0.1% and a test has 99% sensitivity and 99% specificity, then a positive test does not mean a 99% probability of disease — it means about 9%, because the false positives from the 99.9% healthy population swamp the true positives from the 0.1% sick population. The probability calculator on this page does not implement full Bayes (it expects raw probabilities as input), but the odds output is the natural starting point: Bayes on odds is multiplicative, posterior odds = prior odds × likelihood ratio, which is much cleaner than the probability form.
When should I convert a probability to odds, and which way around is 1:5 versus 5:1?
Convert to odds when the audience is bettors, geneticists, epidemiologists, or Bayesians — domains where the natural quantity is a ratio rather than a fraction. Odds in favour of an event are P / (1 − P); odds against are (1 − P) / P. A fair coin has odds 1 (or 1:1, often called 'evens'). A 25% probability has odds 0.333, usually quoted as 1:3 against, meaning the event is three times more likely not to happen than to happen. A 6 on a fair die has odds 0.2, or 1:5 against — five non-six outcomes for every six. The notation is the source of most confusion: '1:5' and '5:1' are not interchangeable. By convention 1:5 against (or 'odds of 5 to 1 against') means the event is unlikely; 5:1 on (or 'odds of 5 to 1 on') means it is likely. When in doubt convert back to probability with p = odds / (1 + odds) and check the result is on the side of 0.5 you expected.
Is probability the same thing as percent chance?
Mathematically yes — multiplying a probability by 100 produces the same number in percentage form, and 0.25 and 25% carry identical information. In communication, no. Saying '25% chance of rain' is a probability statement about whether it will rain at all; saying '25% rain' invites the misreading that 25% of the sky will be wet or that it will rain for 25% of the day. Probabilities are scalar measures of uncertainty about whole events, not fractions of the event itself. The calculator returns both representations on every run so you can pick the one your audience reads correctly. For statistical calculations always use the raw probability — feeding a percentage into a formula that expects a probability is one of the most common arithmetic errors in applied work.
What is the gambler's fallacy and why is it wrong?
The gambler's fallacy is the belief that a streak of one outcome makes the opposite outcome 'due'. If a fair coin lands heads ten times in a row, the eleventh flip is still 50/50 — the coin has no memory, and the probability of heads on any individual flip is independent of every other flip. The misconception arises because the law of large numbers does guarantee that the long-run proportion of heads converges to 0.5, but it does so by drowning the streak in a much larger volume of future flips, not by the coin somehow correcting itself. After 10 heads in a row, the next 1,000 flips will probably contain about 500 heads — bringing the overall proportion from 10/10 = 1.0 closer to 510/1,010 ≈ 0.505, then 5,010/10,010 ≈ 0.5005, and so on. The streak is never erased; it is diluted by sheer volume. Treating each flip as independent (which it is) and computing P(eleventh head) = 0.5 is the correct answer.
What is base rate neglect and how does it affect probability calculations?
Base rate neglect is the cognitive bias of ignoring the prior probability of an event when interpreting new evidence. The classic example: a test for a disease has 99% sensitivity (true positive rate) and 99% specificity (true negative rate); you test positive; what is the probability you have the disease? Most people answer 99%. The correct answer depends on the disease's base rate — if 1 in 1,000 people have the disease, then in a population of 100,000, about 100 have it and 99,900 do not. The test will correctly catch 99 of the 100 sick and incorrectly flag 1% of the 99,900 healthy ≈ 999 false positives. So out of about 1,098 positive tests, only 99 are real — a posterior probability of about 9%, not 99%. Bayes' theorem formalises this; the lesson is that any conditional probability statement is meaningless without the base rate.
Can a probability be greater than 1 or less than 0?
No, never. Kolmogorov's first axiom requires every event to have probability ≥ 0, and the normalisation axiom forces the certain event (the whole sample space) to have probability exactly 1; from these two it follows that every probability lies in the closed interval [0, 1]. If your computation produces a 'probability' of 1.2 or −0.1, you have either made an arithmetic error, double-counted an intersection (a common mistake in OR problems — use inclusion-exclusion), or confused probabilities with odds or with raw frequencies. The calculator rejects out-of-range inputs for exactly this reason: rather than silently propagate a meaningless 'probability' it stops and flags the problem so you can find where the units went wrong.
How do I compute the probability that at least one of several independent events occurs?
Use the complement: the easiest path is to compute the probability that none of the events occur, then subtract from 1. If A, B, C are independent with probabilities pA, pB, pC, then P(none) = (1 − pA)·(1 − pB)·(1 − pC), and P(at least one) = 1 − (1 − pA)·(1 − pB)·(1 − pC). For two independent events this is identical to the inclusion-exclusion formula P(A or B) = pA + pB − pA·pB that the calculator's independent-or mode uses, but the complement form generalises much more cleanly to three or more events. Example: probability of at least one six in four rolls of a fair die = 1 − (5/6)^4 ≈ 0.518, just over even money. This was the original problem solved by Chevalier de Méré and Blaise Pascal in 1654, which kicked off the formal theory of probability.
Why do I get strange decimal artefacts like 0.30000000000000004 in some online probability calculators?
IEEE-754 binary floating-point arithmetic cannot represent 0.1, 0.2, 0.3 exactly — only the nearest binary approximation — so adding 0.1 + 0.2 in JavaScript yields 0.30000000000000004. The Quanta calculator routes all probability arithmetic through Decimal.js, which uses arbitrary-precision decimal arithmetic, so 0.1 + 0.2 + 0.3 + … sums to clean decimal answers and the percent and odds outputs are exact to the precision shown. This matters most in iterated computations — multiplying many small probabilities together in a chain — where each binary rounding compounds. For one-off calculations the visible artefact is usually only the last digit and can be ignored, but if you are auditing a calculator's accuracy, the cleanest test is to feed it 0.1 and 0.2 in single mode and check that the percent output reads exactly 10% and 20%, not 10.000000000000002%.

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