Calculators
Probability Calculator
Work out the probability of a single event, combine two events, or find the chance of at least one success in repeated trials — every result shows the formula behind it.
1. Single event
Probability of one event from favorable and total outcomes.
Probability
16.67%
P = 1 / 6 = 0.1667
Odds for
1 : 5
Odds against
5 : 1
2. Two events
Combine two events. Enter probabilities as decimals between 0 and 1.
P(A and B)
0.15 (15%)
P(A) × P(B) = 0.5 × 0.3 = 0.15
P(A or B)
0.65 (65%)
P(A) + P(B) − P(A)P(B) = 0.5 + 0.3 − 0.15 = 0.65
P(not A)
0.5 (50%)
1 − P(A) = 1 − 0.5 = 0.5
3. At least once in n trials
Chance of succeeding at least once across repeated independent trials.
Probability of at least one success
65.13%
1 − (1 − 0.1)10 = 1 − 0.3487 = 0.6513
Probability of no success at all: 34.87%
Formulas assume equally likely outcomes and independent trials where stated. Calculated on your device; nothing is sent anywhere.
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Frequently asked questions
How do I calculate the probability of a single event?
Divide the number of favorable outcomes by the total number of equally likely outcomes. Rolling a six on a fair die is 1 favorable outcome out of 6, so the probability is 1/6 ≈ 0.1667, or 16.67%. Drawing a heart from a full deck is 13/52 = 1/4 = 25%. This only works when every outcome is equally likely — fair dice, coins and well-shuffled cards, not “it either rains or it doesn’t”.
What is the difference between probability and odds?
Probability compares favorable outcomes to all outcomes; odds compare favorable outcomes to unfavorable ones. Rolling a six has probability 1/6 but odds of 1 : 5 for, or 5 : 1 against. To convert odds of x : y for into a probability, divide x by x + y — odds of 3 : 2 mean 3/5 = 60%.
When do I multiply probabilities and when do I add them?
Multiply for “and” when the events are independent: two coin flips both landing heads is 0.5 × 0.5 = 0.25. Add for “or”, but subtract the overlap: P(A or B) = P(A) + P(B) − P(A and B). If the events are mutually exclusive they cannot happen together, the overlap is zero, and you simply add — rolling a 1 or a 2 on one die is 1/6 + 1/6 = 1/3.
How does the “at least once” formula work?
The chance of at least one success in n independent tries is 1 minus the chance of failing every time: 1 − (1 − p)ⁿ. With a 10% chance per try and 10 tries, that is 1 − 0.9¹⁰ ≈ 65.1% — not 100%, a very common mistake. To reach a 90% chance of at least one success at 10% per try you need 22 tries, because 0.9²² ≈ 0.098.
Can it calculate the probability of exactly k successes?
Not directly — this page covers single events, two combined events and “at least once”. For exactly k successes in n independent trials use the binomial formula C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ. Exactly two sixes in five rolls is 10 × (1/6)² × (5/6)³ ≈ 16.1%. You can use the two-event section to check the pieces: (1/6)² is P(A and B) with both probabilities set to 0.1667.
What about events that are not independent, like drawing cards without replacement?
Then the second probability changes after the first event, and you multiply by the conditional probability instead: P(A and B) = P(A) × P(B given A). Drawing two aces in a row from a 52-card deck is 4/52 × 3/51 = 1/221 ≈ 0.45%, not (4/52)² ≈ 0.59%. Enter the already-conditioned values — 0.0769 and 0.0588 — as independent to get the product.
About this tool
The page has three calculators for the three questions people ask most. The single-event section turns “favorable out of total” into a fraction reduced to lowest terms, a decimal, a percentage, and odds for and against in simplified form. The two-event section combines P(A) and P(B) into P(A and B), P(A or B) and P(not A), using the correct rule for independent or mutually exclusive events and refusing mutually exclusive inputs that add up to more than 1. The repeated-trials section answers “what is the chance this happens at least once if I try n times?” for up to 100,000 trials.
Every result prints its formula with your numbers substituted, so you can follow each step — useful for checking homework, and for spotting when you have chosen the wrong rule. Results update as you type. Probabilities in the second and third sections are entered as decimals between 0 and 1, so 35% is 0.35 and 1 in 8 is 0.125.
Worked examples. Dice: the chance of rolling at least one six in four rolls is 1 − (5/6)⁴ ≈ 51.8%, the bet that famously made money for gamblers in 17th-century France. Games: a 2% drop chance over 50 attempts gives 1 − 0.98⁵⁰ ≈ 63.6%, and over 100 attempts ≈ 86.7% — never a guarantee. Reliability: if each of three independent backup systems fails 5% of the time, all three fail together with probability 0.05³ = 0.0125%, while at least one fails with probability 1 − 0.95³ ≈ 14.3%.
The most common errors are treating dependent events as independent (cards drawn without replacement, people chosen from a small group), adding probabilities of events that can happen together, and expecting “1 in n” odds to succeed within n tries. A 1-in-100 event over 100 tries succeeds at least once only about 63% of the time; you need about 230 tries to reach 90%. All calculations run in your browser.