Maths Probability

Probability calculator

Mode
Favourable outcomes
Total outcomes
Probability 16.66667 %
1 out of 6
Probability against 83.33333 %
Odds for 1 : 5
One in 6
Decimal odds 6
Multiply only for independent events

The two-event and repeated-trial tabs both assume independence. The panel cannot check that assumption and will multiply whatever you give it.

Probability is favourable outcomes over total outcomes. For independent events, both happening is the product; at least one is 1 minus the product of the complements. Over n tries at probability p, at least once is 1 − (1−p)ⁿ, so ten tries at 10% gives 65.1% rather than certainty.

How to calculate probability

1 Pick the tab that matches the situation: one event, two events, or one chance repeated.
2 Enter probabilities as percentages, or as favourable and total outcomes on the first tab.
3 Read the row you need. Every tab shows the complements as well as the headline figure.
4 Before trusting the two-event tab, satisfy yourself that the events really are independent.

Three tabs, three questions. One event turns a count into a probability and into odds. Two events combines a pair of probabilities four ways. Repeated trials answers the question people most often get wrong, which is what happens to a small chance given many attempts. All three are short formulas, and the difficulty in probability has never been the arithmetic.

Independence is a claim about the world

Multiplying two probabilities is only correct when one event happening leaves the other unchanged, and nothing on this page can verify that. Two dice qualify. Two cards drawn without replacement do not, because the first draw alters what is left in the deck. Rain on consecutive days does not. Two components in the same rack sharing a power supply, a batch and a room temperature do not, and reliability estimates built on the assumption that they do are the reason redundant systems fail together more often than the arithmetic predicts.

Where events are dependent the multiplication rule needs a conditional probability: P(A and B) = P(A) × P(B given A). The simple product then overstates the answer when the events reinforce each other and understates it when one crowds the other out. The "at least one" row inherits the same problem in the opposite direction, since it is built from the same product.

The base rate does most of the work

Take a test that is 99 per cent accurate in both directions, looking for something that affects one person in a thousand. Screen ten thousand people. Ten of them have the condition and about ten of those test positive. The other 9,990 do not have it, and one per cent of them, about a hundred people, test positive anyway. That is roughly a hundred and ten positive results of which ten are real, so a positive result carries about a nine per cent chance of being right.

Nothing in that calculation is subtle, and almost nobody guesses it. The intuition wants to read "99 per cent accurate" as "99 per cent chance the answer is correct", and the false positives drawn from the vast healthy majority swamp the true ones drawn from the tiny affected minority. Raise the rate to one in a hundred and the same test gives a positive result a 50 per cent chance of being right. The test did not change. The population did.

Small chances and many tries

The repeated-trials tab exists because "one in a thousand" reads as safe and stops being safe under repetition. A one-in-a-thousand failure per attempt is 63 per cent likely to happen at least once across a thousand attempts, and 99.995 per cent likely across ten thousand. The complement is the easy way to see it: never failing is 0.999 multiplied by itself a thousand times, and repeated multiplication by something slightly under one goes to zero faster than intuition expects.

Read the expected-successes row alongside the headline. At ten tries of a ten per cent chance it shows 1, and one expected success sits beside a 34.9 per cent chance of no successes at all. An average is not a forecast of any particular run.

What people use it for

  • Working out dice, card or lottery odds
  • Estimating the chance of at least one failure over many attempts
  • Combining two independent risks
  • Converting between probability, odds and one-in-n
  • Probability homework

Questions

For independent events, multiply for both happening. For at least one, use 1 − (1−a)(1−b), which is not the same as adding them.

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