Probability Calculator

When every outcome is equally likely, probability is favorable ÷ possible. Rolling a 6 on a fair die is 1 favorable outcome out of 6, so 1/6 = 0.167 = 16.7%. The value always lands between 0 (impossible) and 1 (certain), and anything outside that range means the counting went wrong. The phrase "equally likely" is doing more work than it appears to. It is the assumption the whole formula rests on, and it fails more often than people notice. The six faces of a die qualify. The sums of two dice do not: there are 36 equally likely pairs, but only one produces a total of 2 while six produce a total of 7. Counting sums as if they were equally likely — 11 possible totals, so 1/11 each — gives the wrong answer for every one of them. So the real skill is not the division. It is choosing a sample space whose outcomes genuinely are interchangeable, and then counting both parts of the fraction within that same space.

A fair coin has landed heads six times in a row. What is the chance the next flip is tails? Still 50%. The coin has no mechanism for remembering the previous six flips, no ledger to balance, no tendency to correct. The belief that it does is common enough to have a name — the gambler's fallacy — and it is the single most expensive misunderstanding in this entire subject. What confuses the intuition is a true statement sitting next to a false one. It is true that six heads in a row is unlikely: (1/2)⁶, about 1.6%. But that figure applies before any flips have happened, to the whole sequence. Once six heads have already landed, they are facts, not probabilities, and the seventh flip starts from nothing. The genuine pattern behind "things even out" is duller than the fallacy suggests. A long run does not get cancelled by an opposite run; it gets diluted by the sheer volume of later flips. After ten thousand more tosses, the initial excess of six heads is a rounding error in the overall proportion. Nothing corrected it — it was simply outweighed. One caution: this reasoning holds only for genuinely independent trials. Drawing cards without replacement is a different situation entirely, because each draw really does change what remains in the deck.

Three rules cover almost everything. AND — for two independent events to both happen, multiply. Two heads in a row is 1/2 × 1/2 = 1/4. Rolling two sixes is 1/6 × 1/6 = 1/36. Multiplying always shrinks the number, which matches the intuition that demanding more conditions makes success rarer. OR — for mutually exclusive events, where no outcome satisfies both, add. The chance of rolling a 1 or a 2 is 1/6 + 1/6 = 1/3. When the events can overlap, plain addition double-counts the overlap and you must subtract it back: the chance a drawn card is a heart or a king is 13/52 + 4/52 − 1/52 = 16/52, not 17/52, because the king of hearts was counted twice. NOT — the chance an event does not happen is 1 − p. This looks trivial and is the most useful of the three. It shines on "at least one" questions, which are painful head-on and easy in reverse. The chance of getting at least one six in four rolls of a die means adding up the cases for exactly one, two, three and four sixes. Or you flip it: the chance of no six on a single roll is 5/6, so across four independent rolls it is (5/6)⁴ ≈ 0.482, and the answer is 1 − 0.482 = 51.8%. Whenever a question contains the words "at least one", compute the complement.

Two events are independent when knowing one happened tells you nothing about the other. Coin flips are independent. Two draws from a deck with replacement are independent. Two draws without replacement are not: removing a card changes both parts of the fraction for the next draw. When events are linked, you need conditional probability, written P(A | B) and read "the probability of A given B". Its definition is P(A and B) ÷ P(B) — you shrink the sample space down to the cases where B happened, then ask how many of those also contain A. The practical consequence is that the multiplication rule needs a correction. The chance of drawing two aces from a full deck without replacement is not (4/52)². It is 4/52 × 3/51, because after the first ace there are only three left among fifty-one cards. One warning that reaches beyond card games: P(A | B) and P(B | A) are different numbers, and treating them as interchangeable is a well-documented source of bad reasoning in medical and legal contexts. The probability that a test is positive given a condition is not the probability of the condition given a positive test — and when the condition is rare, the two can differ enormously.

How many people must be in a room before there is a better-than-even chance that two of them share a birthday? The usual guess is somewhere near 180. The answer is 23. With 23 people the probability is about 50.7%. At 50 people it is roughly 97%, and by 70 it is over 99.9% — practically a certainty in a group that most people would call small. The result stops feeling impossible once you notice what is actually being counted. The question is not "does someone share my birthday", which really would need a large crowd. It is whether any pair matches, and pairs multiply far faster than people. Twenty-three people form 23 × 22 ÷ 2 = 253 distinct pairs, each with its own chance to collide. The calculation itself uses the complement rule from earlier. Work out the probability that all birthdays differ — 365/365 × 364/365 × 363/365 … — and subtract from 1. Each new person shaves the product a little further, and it crosses one half at the twenty-third. The lesson generalises well beyond birthdays: whenever you are looking for a collision anywhere in a set rather than a match to one specific target, the odds are much higher than intuition reports. Hash collisions, duplicate IDs and coincidences in data all follow the same arithmetic.