You are offered three doors. Behind one is a car; behind the other two are goats. You pick door 1. The host, who knows where the car is, opens door 3 to reveal a goat. Now he asks: stay with door 1, or switch to door 2? Most people stay — it feels like a 50/50 choice. It is not. Switching wins two times out of three. Probability is full of results like that: obvious until you count properly, and the counting is simpler than most students expect.

The one formula: favourable over total

Probability is a ratio. P(event) = favourable outcomes ÷ total outcomes — as long as every outcome is equally likely. Roll one fair die: there are 6 outcomes, and only one of them is a 6, so P(rolling a 6) = 1/6 ≈ 16.67%. P(rolling an even number) = 3/6 = 1/2 = 50%, because 2, 4 and 6 are three of the six faces. P(not rolling a 6) = 5/6 ≈ 83.33% — the complement rule, P(not A) = 1 − P(A). The same ratio works on a deck of cards: 4 aces in 52 cards means P(drawing an ace) = 4/52 = 1/13 ≈ 7.69%. The method never changes: count the outcomes you want, count all the outcomes, divide.

AND and OR: multiply, and add (carefully)

Independent events are the next step — events where the first result does not change the second. Flip a fair coin twice: P(two heads in a row) = 1/2 × 1/2 = 1/4 = 25%. Three heads in a row is (1/2)³ = 1/8 = 12.5%. For a coin and a die, P(heads then a 6) = 1/2 × 1/6 = 1/12 ≈ 8.33%.

OR is where the mistake hides. P(heads or a 6) is not 1/2 + 1/6 — that double-counts the outcome where both happen. The full rule is P(A or B) = P(A) + P(B) − P(A and B), so 1/2 + 1/6 − 1/12 = 7/12 ≈ 58.33%.

Two dice: why 7 is the most likely sum

Roll two dice and there are 36 equally likely outcomes (6 × 6). A sum of 7 can happen six ways — 1+6, 2+5, 3+4, 4+3, 5+2, 6+1 — so P(rolling a 7) = 6/36 = 1/6 ≈ 16.67%, the most likely sum of all. A sum of 2 has only one way (1+1), and a sum of 12 only one way (6+6), so each is 1/36 ≈ 2.78%. Two 6s together is also 1/36 ≈ 2.78%. The middle sums are common because they have the most combinations, not because the dice "like" them.

The complement trick: "at least one"

"At least one" questions look hard until you flip them around. P(at least one 6 in 4 rolls) = 1 − P(no 6 in 4 rolls). Each roll misses the 6 with probability 5/6, so four misses in a row is (5/6)⁴ = 625/1296 ≈ 48.23%. Subtract from 1: 671/1296 ≈ 51.77% — about 51.8%. So in four rolls you are more likely than not to see at least one 6, even though a single roll only gives you 1/6. The same trick works on coins: P(at least one head in 3 flips) = 1 − (1/2)³ = 7/8 = 87.5%. And exactly 2 heads in 3 flips is 3/8 = 37.5% — the three sequences HHT, HTH and THH.

Conditional probability: when the first result changes the second

Some events are not independent. Draw two cards without replacement: P(second card is an ace | first was an ace) = 3/51 ≈ 5.88%, because only 3 aces are left out of 51 cards. P(ace then king, no replacement) = (4/52) × (4/51) = 0.60%. And a die example that surprises people: P(rolling a 6 | the roll is even) = (1/6) ÷ (1/2) = 1/3 ≈ 33.33% — knowing the roll is even narrows the world to {2, 4, 6}, and the 6 is one of three.

The Monty Hall switch: why 2/3 is real

Back to the doors. Your first pick has a 1/3 chance of hiding the car, which means the other two doors together hold a 2/3 chance. The host opens one of those two doors and shows a goat — he does not choose randomly, he always reveals a goat, and that information does not change the 2/3. The whole 2/3 now sits behind the one remaining unopened door. Switching wins 2/3 = 66.67% of the time; staying wins 1/3 = 33.33%. Still suspicious? With 100 doors, switching wins 99/100 = 99% — and the 3-door game is the same game, just smaller.

The gambler's fallacy: the coin has no memory

Five heads in a row? The next flip is still 1/2 = 50% heads. P(6 heads in a row) is (1/2)⁶ = 1/64 ≈ 1.56% — but that is the chance before the run starts. Once five heads have already happened, the next flip starts fresh. A die that has not rolled a 6 in ten rolls is not "due" — every roll is still 1/6. Independence means the past does not push the future around.

Want the steps shown instead of just the answers? A free probability calculator works through all five modes — single events, AND/OR of independent events, dice and coin outcomes, conditional probability, and the full Monty Hall explainer. If the doors still bother you, the monty hall problem walks through the 3-door game and the 100-door version step by step, and the guide to how to calculate probability covers the favourable-over-total rule with worked examples. Check your own numbers on a free probability calculator and the traps stop working on you.