How to Figure Out Probability: A Step-by-Step Method With Dice, Roulette and Mines Examples
Quick answer
To figure out probability, list every possible outcome, check they are equally likely, then divide the number of favourable outcomes by the total. For 'at least one', subtract the chance of 'none' from 1. For several independent events all happening, multiply their probabilities. For draws without replacement, count with combinations.
| Basic formula | favourable outcomes ÷ total outcomes |
|---|---|
| At least once | 1 − P(never) |
| A and B (independent) | P(A) × P(B) |
| Combinations | C(n, r) = n! ÷ (r! × (n − r)!) |
| Red on a European wheel | 18 ÷ 37 = 48.65% |
| At least one 6 in 4 dice rolls | 1 − (5/6)^4 = 51.77% |
To figure out probability, count the outcomes you want, count all the possible outcomes, and divide the first by the second. That works whenever each outcome is equally likely. Red on a European roulette wheel is 18 red pockets out of 37, so its probability is 18 ÷ 37 = 48.65%. Harder questions, like "at least one" or "two in a row", use a few extra rules that build on the same idea.
Below is a six-step method that covers almost every probability question you'll meet at a casino table or in a game, with worked examples from dice, roulette and Mines. Each step comes with a formula you can reuse. If you'd rather skip the arithmetic, our chance calculator does the "at least once" and losing-streak math for you.
The six steps at a glance
- Define the experiment and list the outcomes.
- Check the outcomes are equally likely.
- Divide favourable outcomes by total outcomes.
- Use the complement for "at least one".
- Multiply for "and".
- Count with combinations, then sanity-check.
Step 1: Define the experiment and list every outcome
Be precise about what you're doing once. "Spin a European roulette wheel" has 37 outcomes: the numbers 0 to 36. "Roll one six-sided die" has 6. "Roll two dice" has 36 if you treat the dice as different (first die, second die), because 6 × 6 = 36. The full list is called the sample space.
Getting this step wrong is the most common source of errors. "Roll two dice and look at the total" has 11 possible totals (2 to 12), but those 11 totals are not the outcomes you should count, as step 2 explains.
Step 2: Check the outcomes are equally likely
The divide-and-count method only works if every outcome in your list has the same chance. Each pocket on a fair wheel and each face of a fair die qualifies. Dice totals don't: a total of 7 can be made six ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), while 12 can only be made one way (6+6).
The fix is to list outcomes at a finer level, the 36 ordered pairs, where every outcome is equally likely, and then group them. Our craps odds guide shows the full two-dice table built this way.
Step 3: Divide favourable outcomes by total outcomes
This is the core formula. Britannica defines probability in gambling as the number of favourable outcomes divided by the total number of possible outcomes:
P(event) = favourable outcomes ÷ total outcomesEuropean wheel, red: 18 ÷ 37 = 48.65%. American wheel (38 pockets), red: 18 ÷ 38 = 47.37%.
More examples:
| Question | Favourable | Total | Probability |
|---|---|---|---|
| One number, European roulette | 1 | 37 | 2.70% |
| A dozen bet, European roulette | 12 | 37 | 32.43% |
| Even number on one die | 3 | 6 | 50% |
| Doubles with two dice | 6 | 36 | 16.67% |
| Two dice total 7 | 6 | 36 | 16.67% |
| First pick safe in Mines, 5 mines on 25 tiles | 20 | 25 | 80% |
When favourable outcomes can't happen together, you can also add their probabilities: red or zero on a European wheel is 18/37 + 1/37 = 19/37 = 51.35%. That's the same as counting 19 favourable pockets.
Step 4: Use the complement for "at least one"
Questions with "at least one" have many ways to succeed (once, twice, three times…). It's far easier to work out the one way to fail, which is "never", and subtract:
P(at least one) = 1 − P(none)At least one 6 in four rolls of a die: P(no 6 on one roll) = 5/6, so P(none in four) = (5/6)^4 = 0.4823 and P(at least one) = 1 − 0.4823 = 51.77%.
The complement also explains why "1 in N" doesn't mean "within N tries". One roulette number in 37 spins: P(never) = (36/37)^37 = 36.3%, so the number appears at least once only 63.7% of the time.
| Spins | No red at all | At least one red |
|---|---|---|
| 1 | 51.35% | 48.65% |
| 2 | 26.37% | 73.63% |
| 3 | 13.54% | 86.46% |
| 5 | 3.57% | 96.43% |
| 10 | 0.13% | 99.87% |
Step 5: Multiply for "and"
When several things must all happen, multiply. If the events are independent (one result doesn't affect the next), OpenStax gives the multiplication rule:
P(A and B) = P(A) × P(B) (independent events)Red twice in a row, European wheel: 18/37 × 18/37 = 23.67%. Two dice both showing 6: 1/6 × 1/6 = 1/36 = 2.78%.
When events are not independent, multiply the chance of each step given the steps before it. Mines is the classic case: each safe tile you uncover is gone, so the next pick has fewer tiles left.
Mines, 5 mines, 2 safe picks: 20/25 × 19/24 = 380/600 = 63.33%First pick: 20 safe tiles of 25. Second pick: 19 safe tiles of the 24 left.
Spins and dice rolls are independent; draws from a fixed set without putting anything back (cards, Mines tiles, keno balls) are not. Treating independent events as if they were linked, so that a loss makes a win "due", is the gambler's fallacy.
Step 6: Count with combinations, then sanity-check
When outcomes are too many to list, count them. The OpenStax counting principles give three tools:
Multiplication principle: m ways then n ways = m × n waysThree dice: 6 × 6 × 6 = 216 outcomes.
Permutations (order matters): P(n, r) = n! ÷ (n − r)! Combinations (order doesn't): C(n, r) = n! ÷ (r! × (n − r)!)C(25, 2) = 25 × 24 ÷ 2 = 300 ways to choose 2 tiles from 25. C(20, 2) = 20 × 19 ÷ 2 = 190 ways to choose 2 safe tiles from 20.
The Mines example from step 5 again, this time by counting:
P(2 safe picks, 5 mines) = C(20, 2) ÷ C(25, 2) = 190 ÷ 300 = 63.33%Same answer as multiplying step by step. Counting is easier when there are many picks.
Then sanity-check. Every probability is between 0 and 1. The probabilities of all outcomes add up to 1. And the answer should match what happens over many trials, because probability is the long-run relative frequency of an outcome, as OpenStax puts it.
Checking the method against our simulations
The best test of a probability calculation is a long run of random trials. Our data includes both exact figures and simulations for dice, roulette and Mines:
| Event | Method | Calculated | Our check |
|---|---|---|---|
| Two dice total 7 | 6 ÷ 36 | 16.67% | 16.62% (1M rolls) |
| Two dice total 2 | 1 ÷ 36 | 2.78% | 2.75% (1M rolls) |
| Crypto dice, roll under 10 | 10 ÷ 100 | 10.00% | 9.999% (1M rolls) |
| Crypto dice, roll under 75 | 75 ÷ 100 | 75.00% | 75.002% (1M rolls) |
| Mines: 5 safe picks, 3 mines | C(22, 5) ÷ C(25, 5) | 49.57% | exact count |
The simulated results land within a few hundredths of a percentage point of the counted ones. That's the law of large numbers at work: over a million trials, observed frequencies settle very close to the true probability.
From probability to the price of a bet
Once you can figure out probability, you can price any bet. Multiply the chance of winning by the total return and compare with 1:
house edge = 1 − P(win) × (payout + 1)Red on a European wheel pays 1 to 1: 1 − (18/37) × 2 = 1 − 36/37 = 2.70%.
Our odds vs probability guide explains how payout odds and true odds relate, the roulette house edge guide applies the formula to every wheel, and which casino games have the best odds compares edges across games. For crypto dice, the dice odds calculator turns any win chance into a multiplier.
Common mistakes
- Counting outcomes that aren't equally likely. Two-dice totals 2 to 12 are 11 outcomes, but 7 is six times as likely as 2.
- Adding when you should multiply. Two 6s in a row is 1/6 × 1/6, not 1/6 + 1/6.
- Forgetting removed items. In Mines, cards and keno, each draw changes what's left.
- Expecting "1 in N" within N tries. One roulette number in 37 spins shows up only about 64% of the time.
- Thinking past results change the next one. Independent spins and rolls have no memory.
More guides on odds, house edge and betting systems are on our odds and strategy math hub.
Frequently asked questions
How do you compute probability?
Count the outcomes that count as a success and divide by the total number of equally likely outcomes. One number on a European roulette wheel is 1 favourable pocket out of 37, so the probability is 1 ÷ 37 = 2.70%. When outcomes aren't equally likely, break them into smaller outcomes that are, such as the 36 combinations of two dice.
What is the formula for outcome probability?
P(outcome) = number of favourable outcomes ÷ total number of possible outcomes, assuming each outcome is equally likely. The result is always between 0 (impossible) and 1 (certain). Multiply by 100 for a percentage: 18 red pockets out of 37 gives 0.4865, or 48.65%.
When do you multiply probabilities and when do you add them?
Multiply when you need several things to all happen and they are independent: red twice in a row is 18/37 × 18/37. Add when you need any one of several outcomes that can't happen together: red or zero on a European wheel is 18/37 + 1/37 = 19/37.
How do you work out the probability of something happening at least once?
Work out the chance that it never happens, then subtract from 1. The chance of no 6 in four rolls of a die is (5/6)^4 = 0.482, so the chance of at least one 6 is 1 − 0.482 = 51.77%. This is usually far easier than adding up every way it could happen.
How can I check a probability I calculated?
Make sure it's between 0 and 1, and that all outcomes add up to 1. Then compare it with a long run of trials: probability is the long-run share of times an outcome occurs. Our simulations of a million rolls or spins land within a few hundredths of a percentage point of the exact figures.
Sources
- Two Basic Rules of Probability — OpenStax, Rice University. Multiplication rule; independent events
- Terminology — OpenStax, Rice University. Probability as long-run relative frequency; law of large numbers
- 9.5 Counting Principles — OpenStax (College Algebra). Multiplication Principle: m ways then n ways gives m x n; P(n,r) = n!/(n-r)!; C(n,r) = n!/(r!(n-r)!)
- Gambling: Chances, probabilities and odds — Encyclopaedia Britannica. Probability = favourable/total; odds = unfavourable to favourable
- Doctrine of the maturity of the chances — Encyclopaedia Britannica. Gambler's fallacy: falsely assumes plays are dependent