How this probability calculator works
Probability questions come in three everyday shapes, and each mode above handles one of them. A single event asks "out of all the equally likely ways this could go, how many go my way?" Two events ask "what is the chance both happen, or at least one does?" Repeated tries ask the question everyone actually types into a search bar at 1 a.m.: "if there is a 10% chance and I try 10 times, what are my real odds?" Whatever mode you pick, the result walks through the formula step by step with your numbers substituted in, so you can follow the math rather than trust a black box.
You can type probabilities however they live in your head: 25 means 25%, 0.25 is the decimal form, and 1/4 is the fraction. The calculator tells you how it read your input, so there are no silent surprises.
The formulas
Here P(A) and P(B) are the probabilities of each event, p is the chance of success on a single try, and n is the number of independent tries. The AND and OR formulas require the two events to be independent (more on that below).
AND vs OR, taught with two dice
Roll two dice. The chance the first die shows a six AND the second die shows a six is 1/6 × 1/6 = 1/36, about 2.8%. Both things must happen, so you multiply, and multiplying probabilities always makes the number smaller. Demanding more things shrinks your chances.
Now ask for a six on the first die OR the second die (at least one six). Tempting shortcut: 1/6 + 1/6 = 2/6. Wrong. That double-counts the roll where both dice show sixes. The correct move is to add, then subtract the overlap: 1/6 + 1/6 − 1/36 = 11/36, about 30.6%. You can verify by brute force: of the 36 equally likely rolls, exactly 11 contain at least one six. The subtraction is not a technicality; skip it and every OR answer you compute will be too big.
Worked examples
Single event: a bag holds 12 marbles and 3 are red. P(red) = 3 ÷ 12 = 0.25 = 25%. The complement (not red) is 75%, and the odds are 3 favorable to 9 unfavorable, which reduces to 1:3 in favor (3:1 against).
Two independent events: event A has a 50% chance, event B has 30%. P(A and B) = 0.5 × 0.3 = 15%. P(A or B) = 0.5 + 0.3 − 0.15 = 65%. P(neither) = 0.5 × 0.7 = 35%. P(exactly one) = 65% − 15% = 50%.
Repeated tries: a 10% chance per try, 10 tries. P(at least once) = 1 − 0.910 = 1 − 0.3487 = 65.13%. Not 100%. Not even close to a sure thing.
The "10 tries at 10% must be 100%" myth
This is the single most common probability mistake we see, so let's kill it properly. Ten tries at 10% each does not add up to certainty, because probabilities of independent events do not add. Each try can fail, and the tries do not remember each other. The right way to think about it is through the failures: each try fails 90% of the time, so all ten fail with probability 0.910 = 34.87%. That means you succeed at least once with probability 65.13%. More than a third of people who "should" have hit it by now have not, and the universe owes them nothing.
The practical version of this lesson: to be 95% confident of at least one success at 10% per try, you need about 29 tries, not 10. Chasing certainty gets expensive fast, because each extra try only shrinks the remaining failure probability by 10% of what is left. If you have ever wondered why a rare drop in a game takes so much longer than the drop rate suggests it should, this is why. This same instinct (adding OR probabilities without subtracting the overlap) is what the dice section above corrects.
When these formulas break: independence
Every multiplication on this page assumes the events are independent: the outcome of one tells you nothing about the other. Coin flips, dice rolls, and lottery drawings qualify. Drawing cards without replacement does not. Pull an ace from a deck and the chance the next card is an ace drops from 4/52 to 3/51, because the first draw changed the deck. Weather on consecutive days, free throws by a tiring player, and product failures with a shared cause are all everyday examples of dependence.
If your events are dependent, the honest answer is that you need conditional probability: P(A and B) = P(A) × P(B given A). For the card example, P(two aces in two draws) = 4/52 × 3/51 = 0.45%, noticeably less than the naive 4/52 × 4/52 = 0.59%. When in doubt, ask: if I learned the first result, would I change my bet on the second? If yes, the events are dependent and this calculator's two-event mode will be an approximation at best. For questions about distributions of many repeated events, our normal distribution calculator and confidence interval calculator pick up where single-event probability leaves off, and if you want to test whether an observed result beats chance, that is what a p-value is for. Want to feel the randomness instead of computing it? Roll a few with the dice roller.