# Probability Calculator

> Find the probability of A and B, A or B, exactly one, neither, and of A happening at least once in several tries, for independent events.

Category: Statistics & probability · Applies to: Worldwide · Last updated: 2026-10-29
Interactive calculator: https://calculators.geekmidst.com/calculators/probability-calculator

Enter the chances of two independent events as decimals, fractions or percentages to combine them, and see how likely A is over repeated tries.

Also called: Independent events calculator.

## Inputs
- Probability of A, P(A) (default 1/6)
- Probability of B, P(B) (default 1/6)
- Number of tries for A (default 4)

## Results
- A and B both happen
- A or B (at least one)
- Exactly one of A and B
- Neither happens
- A does not happen
- B does not happen
- A at least once in the tries
- A on every try

## Formula

```
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
```

Where:
- P(A), P(B): probabilities of the two independent events
- n: number of independent tries

## Worked example
1. Two dice both showing six: 1/6 × 1/6 = 1/36 ≈ 2.78%.
2. At least one six on two dice: 1/6 + 1/6 − 1/36 = 11/36 ≈ 30.56%.
3. At least one six in 4 rolls: 1 − (5/6)⁴ = 1 − 625/1296 ≈ 51.77%.

**Result:** 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%.

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Source: GM¹X CalcVerse — https://calculators.geekmidst.com/calculators/probability-calculator