How to use this calculator
- Enter or choose probability of A, P(A).
- Enter or choose probability of B, P(B).
- Enter or choose number of tries for A.
- The a and B both happen updates as you type. Use Copy result or Share to keep it.
Method
P(A and B) = P(A)·P(B); P(A or B) = P(A) + P(B) − P(A)·P(B); P(neither) = (1 − P(A))(1 − P(B)); P(A at least once in n) = 1 − (1 − P(A))ⁿ
- P(A), P(B)
- probabilities of the two independent events
- n
- number of independent tries
Worked example
- Two dice both showing six: 1/6 × 1/6 = 1/36 ≈ 2.78%.
- At least one six on two dice: 1/6 + 1/6 − 1/36 = 11/36 ≈ 30.56%.
- At least one six in 4 rolls: 1 − (5/6)⁴ = 1 − 625/1296 ≈ 51.77%.
Rolling at least one six in four throws is slightly better than even, at about 51.8%.
What the result means
Small chances add up over repeated tries: an event with a 1% chance each time has about a 63% chance of happening at least once in 100 independent tries.
Assumptions and limits
- A and B are independent: one happening does not change the chance of the other. Repeated tries are independent too.
- For dependent events (drawing cards without replacement, for example), P(A and B) = P(A) × P(B given A), which this calculator does not model.
Questions
Why do I subtract P(A and B) for “A or B”?
Adding P(A) and P(B) counts the cases where both happen twice, so one copy is removed.
Is this the famous de Méré problem?
Yes: the 17th-century gambler Chevalier de Méré noticed that “at least one six in four rolls” wins slightly more than half the time, which this example confirms at 51.77%.