All the numbers below are illustrations with simple, assumed rules, not figures from any real game. Each is calculated step by step so you can check the arithmetic and reuse the method.
Example 1: two dice, both even
Question. Roll two fair dice. What is the chance both show an even number?
Each die shows an even number (2, 4 or 6) with probability 3/6 = 1/2. The dice are independent, so multiply: 1/2 × 1/2 = 1/4.
Check by counting. There are 3 × 3 = 9 pairs of even faces among 36 outcomes, and 9/36 = 1/4. ✓
Example 2: a total of 7 or 11
Question. With two dice, what is the chance the total is 7 or 11?
A total of 7 can occur in 6 ways and 11 in 2 ways ((5,6) and (6,5)). They cannot happen together, so add: 6/36 + 2/36 = 8/36 = 2/9, about 22.2%.
This is the addition rule for mutually exclusive events. The 7 on its own is the most common total, which is explained in dice probability.
Example 3: two cards of the same suit
Question. Draw two cards from a standard 52-card deck. What is the chance they are the same suit?
The first card can be anything. Whatever suit it is, 12 of the remaining 51 cards match it. So the chance is 12/51 = 4/17, about 23.5%.
Common error. Using 13/52 for the second card ignores that one card of the suit has already gone. That is an example of a dependent draw, as discussed under dependent events.
Example 4: a spinner
Setup. A spinner has 8 equal sectors: 3 red, 1 gold and 4 blue.
Question A. What is the chance of red on one spin? 3/8 = 37.5%.
Question B. What is the chance of red and then gold in two spins? The spins are independent, so 3/8 × 1/8 = 3/64, about 4.7%.
Question C. What is the chance of at least one gold in three spins? Use the complement. No gold on one spin is 7/8, and on three spins it is (7/8)³ = 343/512. So at least one gold is 1 − 343/512 = 169/512, about 33.0%.
Check: 7 × 7 × 7 = 343 and 8 × 8 × 8 = 512, and 512 − 343 = 169. ✓
Example 5: a loot drop
Setup. An item drops with 5% probability on each kill, and each kill is independent.
Question A. What is the chance of at least one drop within 10 kills?
- No drop on one kill: 0.95.
- No drop in 10 kills: 0.95¹⁰, about 0.599.
- At least one: 1 − 0.599 = 0.401, so about 40.1%.
Question B. What is the chance of still having nothing after 20 kills? 0.95²⁰ is the square of 0.95¹⁰, so about 0.599² ≈ 0.359, or 35.9% (the exact value is about 35.8%).
Question C. On average, how many kills until the first drop? For a 5% chance the long-run average is 1/0.05 = 20 kills. But the chance of having a drop by 20 kills is only about 64%, so a substantial minority of players will need more than the average, which shows how an average can mislead about an individual’s experience. Many games address this with pity timers, though how any particular game does so is outside what can be assumed here.
After 20 misses the next kill is still 5%. The earlier failures do not make a drop due, which is the point behind the gambler’s fallacy.
Example 6: dealt a pair of aces
Question. From a standard deck, what is the chance that your two starting cards are both aces?
First card an ace: 4/52. Second card an ace given the first was: 3/51. Product: 12/2652 = 1/221, about 0.45%.
Check by counting. There are C(52, 2) = 1,326 two-card hands and C(4, 2) = 6 of them are ace pairs, giving 6/1,326 = 1/221. ✓
Example 7: expected value of a simple game
Setup. You pay 1 unit to roll a die. A roll of 6 pays back 4 units (so a net gain of 3); any other roll pays nothing (net −1).
Expected net result: 3 × (1/6) + (−1) × (5/6) = 3/6 − 5/6 = −2/6 = −1/3, about −0.33 units per roll.
So on average the player loses a third of a unit per roll. Over 6 rolls the average loss is 2 units, though real results vary widely. A game with a negative expected value for the player is the normal situation in casino-style games, where this gap is the house edge, and no sequence of rolls makes the average positive.
Example 8: percentage change versus percentage points
Question. A drop rate rises from 2% to 3%. How large is the change?
It is 1 percentage point. In relative terms it is 1/2 = 50% higher. Both statements are true, but they sound very different, so check which one is meant when a claim says “50% better”.
Method summary
| If the question says | Think |
|---|---|
| both / and | multiply, adjusting for dependence |
| either / or | add, subtracting any overlap |
| at least one | 1 minus the chance of none |
| exactly k in n tries | count the arrangements, then multiply |
| average result | weight each outcome by its probability |
Reminder
These illustrations show how the arithmetic works. They are not suggestions for how to play, and they do not offer a way to win. Real outcomes are uncertain, casino games carry a house edge, and gambling can cause harm. If it is affecting you or someone close to you, please contact a local support service.