Two events: the formulas
For independent events A and B with probabilities P(A) and P(B):
With P(A) = 0.5 and P(B) = 0.3: both happen 15% of the time, at least one 65%, exactly one 50% and neither 35%. Those last three, plus "both", always add up to 100%.
The "or" formula subtracts P(A and B) because simply adding the two chances counts the times both happen twice. That is why two 60% chances don't make 120%: at least one of them happens 1 − 0.4 × 0.4 = 84% of the time.
Repeated tries
The quickest question is the chance of at least one success: it is 1 minus the chance of failing every time.
In the 1600s the gambler Chevalier de Méré noticed that betting on at least one six in 4 rolls of a die won slightly more often than it lost: 1 − (5/6)⁴ = 51.77%. Betting on at least one double six in 24 rolls of two dice seemed equivalent, but lost money: 1 − (35/36)²⁴ = 49.14%. Pascal and Fermat's letters about his puzzle helped found probability theory.
For an exact number of successes, the calculator uses the binomial distribution:
C(n, k) counts the different orders the k successes can come in. In 10 tosses of a fair coin, exactly 5 heads has probability C(10, 5) × 0.5¹⁰ = 252/1024 = 24.61%, and at most 5 heads 62.3%. "At most" and "at least" add up the exact probabilities of every count in that range. The calculator also gives the expected number of successes, n × p, and its standard deviation, √(n × p × (1 − p)).
When these formulas don't apply
Everything here assumes independence and a chance that stays the same each time. Drawing cards without putting them back, picking lottery numbers that can't repeat, or quality failures that come in batches all break those assumptions, and the true answer can be quite different. Small-chance events that are linked, such as several homes in one street flooding, are especially easy to underestimate.
Questions people ask
How do I calculate the probability of two events both happening?
If the events are independent, multiply their probabilities: P(A and B) = P(A) × P(B). A 50% chance and a 30% chance happen together 15% of the time. If one event changes the chance of the other, such as drawing two cards without putting the first back, you need the conditional probability instead.
What is the probability of at least one success?
Work out the chance of failing every time and subtract it from 1: P(at least once) = 1 − (1 − p)^n. With a 1% chance per try, 100 tries give 1 − 0.99^100 = 63.4%, not 100%.
What does “independent” mean?
Two events are independent when one happening doesn’t change the chance of the other: coin tosses, dice rolls, or drawing a card and putting it back. Drawing cards without replacement, weather on consecutive days, or a machine wearing out are not independent, and the formulas here would be off.
If I’ve lost five times in a row, am I due a win?
No. With independent tries the chance is the same every time, however the last ones went. Believing otherwise is called the gambler’s fallacy. What does change with more tries is the chance of at least one win over the whole run.
How do I enter probabilities?
As a decimal (0.25), a percent (25%) or a fraction (1/4). For dice use fractions: rolling a six is 1/6, two sixes at once is 1/36.