To calculate probability, count the favourable outcomes, count all the possible outcomes, and divide: P(event) = favourable outcomes / total outcomes. Rolling a 6 is 1 favourable outcome out of 6, so P(6) = 1/6 = 16.67%. For two events, multiply for AND and add (minus the overlap) for OR; for “at least one”, use 1 minus the chance of none.
Probability is just counting: two numbers plus three rules.
Check any value instantly with a free probability calculator.
Step 1: The Basic Formula — Favourable Over Total
P(event) = favourable outcomes / total outcomes. A fair die has 6 outcomes (1–6); a fair coin has 2 (heads, tails); a standard deck has 52 cards.
One die: P(6) = 1/6 = 16.67%; P(even) = 3/6 = 50%; P(6 or 5) = 2/6 = 33.33%; P(not 6) = 5/6 = 83.33%.
One card: P(ace) = 4/52 = 7.69%; P(red) = 26/52 = 50%; P(king of hearts) = 1/52 = 1.92%.
Count favourable first, total second, and reduce the fraction before converting — 6/36 is correct but unreadable; 1/6 = 16.67% is the same number, clearer.
Step 2: AND Means Multiply, OR Means Add (Minus the Overlap)
AND — both happen: multiply. P(A and B) = P(A) × P(B), for independent events. P(two heads in a row) = 1/2 × 1/2 = 1/4 = 25%. P(three heads) = (1/2)^3 = 1/8 = 12.5%. P(heads then a 6) = 1/2 × 1/6 = 1/12 = 8.33%. Each event cuts the chances down. If your AND answer beats either single probability, you added instead of multiplying.
OR — at least one happens: add, then subtract the overlap. P(A or B) = P(A) + P(B) − P(A and B). P(heads or a 6) = 1/2 + 1/6 − 1/12 = 7/12 = 58.33% — “heads and a 6” (1/12) is counted twice if you only add. P(rain or traffic) = 0.3 + 0.4 − 0.12 = 58%. But P(rolling a 6 or a 5) = 1/6 + 1/6 = 1/3 = 33.33% — one die cannot show both, so there is no overlap. Quick check: P(A or B) sits between the bigger single probability and 1.
Step 3: Dice — The Two-Dice Sums Table and “At Least One 6”
Two dice have 36 outcomes (6 × 6), but the sums are not equally likely — 7 is most common, 2 and 12 rarest:
| Sum | Pairs | Probability |
|---|---|---|
| 2 | 1 | 1/36 = 2.78% |
| 3 | 2 | 2/36 = 5.56% |
| 4 | 3 | 3/36 = 8.33% |
| 5 | 4 | 4/36 = 11.11% |
| 6 | 5 | 5/36 = 13.89% |
| 7 | 6 | 6/36 = 1/6 = 16.67% |
| 8 | 5 | 5/36 = 13.89% |
| 9 | 4 | 4/36 = 11.11% |
| 10 | 3 | 3/36 = 8.33% |
| 11 | 2 | 2/36 = 5.56% |
| 12 | 1 | 1/36 = 2.78% |
Counts add to 36, probabilities to 36/36 = 1 — the table checks itself. So P(rolling a 7) = 6/36 = 16.67%, P(two 6s) = 1/36 = 2.78%.
“At least one 6” — the complement trick. Counting one, two, three and four 6s is four cases — flip it: count “no 6 at all” and subtract from 1. P(no 6 in 4 rolls) = (5/6)^4 = 48.23%, so P(at least one 6) = 1 − 48.23% = 51.77% (about 51.8%). It scales: 2 rolls, 11/36 = 30.56%; 6 rolls, 66.51%.
Step 4: Coins — Exactly k Heads in n Flips
The formula is the binomial count: P(exactly k heads in n flips) = C(n,k) / 2^n, where C(n,k) counts the ways to choose which flips are heads.
Three flips (8 outcomes): exactly 0 heads 1/8 = 12.5%; exactly 1 head 3/8 = 37.5%; exactly 2 heads 3/8 = 37.5%; exactly 3 heads 1/8 = 12.5%; at least one head 7/8 = 87.5%.
Four flips (16 outcomes): exactly 1 head 4/16 = 25%; exactly 2 heads 6/16 = 37.5% (C(4,2) = 6); exactly 3 heads 4/16 = 25%; exactly 4 heads 1/16 = 6.25%. Counts add to the total (1+3+3+1 = 8; 1+4+6+4+1 = 16) — if not, a case was missed.
Step 5: Conditional Probability — When Something Already Happened
P(A|B) = P(A and B) / P(B) — shrink the world to the outcomes where B happened, then count A inside it. P(6 | the roll is even) = 1/3 = 33.33%: the even outcomes are {2, 4, 6} and one is a 6. Knowing the roll is even doubled the chance from 16.67%.
Cards without replacement — the deck changes. P(second ace | first ace) = 3/51 = 5.88%: 3 aces remain out of 51 cards. P(ace then king) = (4/52) × (4/51) = 16/2652 = 0.60%. Coins have no memory — after five heads the next flip is still 50% — but a deck without replacement does. Direction matters: P(6 | even) = 1/3, but P(even | 6) = 1. The Monty Hall version is in Monty Hall Problem Explained.
Common Mistakes When Calculating Probability
Adding for AND. P(two heads) is 1/2 × 1/2 = 25%, not 1/2 + 1/2 = 100%. Forgetting the OR overlap. P(heads or a 6) is 7/12 = 58.33%, not 2/3. Counting “at least one” case by case. Use the complement: 1 − (5/6)^4 = 51.77%. Treating dependent events as independent. Two aces without replacement is (4/52) × (3/51) = 0.45%. The gambler’s fallacy. After five heads, the next flip is still 50%. Forgetting to reduce the fraction. Only 1/6 — not 6/36 — shows that rolling a 7 is as likely as rolling a single 6.
Frequently Asked Questions
How do you calculate probability?
Divide favourable by total outcomes: P = favourable/total. Rolling a 6 is 1/6 = 16.67%. Multiply for AND, add-minus-overlap for OR; for “at least one”, use 1 minus the chance of none.
What is the probability of rolling a 7 with two dice?
6/36 = 1/6 = 16.67%. Six pairs make 7 — (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Seven is the most likely sum.
What is the probability of at least one 6 in 4 rolls?
51.77% (about 51.8%). P(no 6 in four rolls) is (5/6)^4 = 48.23%, so at least one 6 is 1 − 48.23% = 51.77%.
How do you calculate the probability of two events?
Multiply for AND, add-minus-overlap for OR. P(two heads) = 1/2 × 1/2 = 25%. P(heads or a 6) = 1/2 + 1/6 − 1/12 = 7/12 = 58.33%.
What is conditional probability?
The chance of A given B already happened: P(A|B) = P(A and B)/P(B). P(6 | even) = 1/3. After one ace, P(second ace) = 3/51 = 5.88%.
What is the probability of exactly 2 heads in 3 coin flips?
3/8 = 37.5%. Three of the eight outcomes have exactly two heads (HHT, HTH, THH).
Try It Yourself
Enter any favourable and total counts into a free probability calculator and compare with your hand calculation — including at-least-one-6-in-4-rolls (51.77%) and the Monty Hall switch (2/3). The Fraction Calculator reduces 6/36 to 1/6, and the Exponent Calculator checks (5/6)^4 = 0.4823.