How to read these examples
Each example below states a situation, names the rule being used, and then does the arithmetic in small steps. The numbers are simple on purpose, so you can redo them by hand. They are illustrations of method, not predictions about any particular game, and none of them describes a way to profit from a game with a built-in cost to the player.
Example 1: A total of seven on two dice
Two fair six-sided dice give 6 × 6 = 36 equally likely ordered results. The multiplication principle is what produces that count: six options for the first die, six for the second.
Totals of seven can be made as 1+6, 2+5, 3+4, 4+3, 5+2 and 6+1. That is 6 results out of 36.
6/36 = 1/6, about 16.7 percent.
Compare a total of twelve, which only comes from 6+6: 1/36, about 2.8 percent. The shape of two-dice sums is the reason seven turns up more than any other total.
Example 2: At least one six in four rolls
Listing every way to get a six is slow. Count the opposite instead.
- Chance of no six on one roll: 5/6.
- Rolls are independent, so multiply: (5/6)^4 = 625/1296.
- Chance of at least one six: 1 − 625/1296 = 671/1296, about 51.8 percent.
Complement counting turns a messy “one or more” question into one multiplication and one subtraction. Note what it does not say: four rolls do not guarantee a six. About 48 percent of the time there is none.
Example 3: Two aces from the top of a shuffled deck
A standard deck has 52 cards, 4 of them aces. You draw two without putting the first back.
- First card an ace: 4/52.
- Given that, the deck has 51 cards and 3 aces left, so the second is an ace with chance 3/51.
- Multiply: (4/52) × (3/51) = 12/2652 = 1/221.
1/221 is about 0.45 percent. The second fraction changed because drawing without replacement alters what remains, which is the central difference between cards and dice.
Example 4: Three heads in five coin flips
With a fair coin there are 2^5 = 32 equally likely sequences of five flips.
The number of sequences with exactly three heads is the number of ways to choose which three flips are heads. That is 5 × 4 × 3 ÷ (3 × 2 × 1) = 10.
Probability: 10/32 = 0.3125, or 31.25 percent.
This counting method is a small case of the binomial distribution, which handles any number of repeated trials with two outcomes.
Example 5: Average damage with a critical hit
A video game attack deals 10 damage. It has a 20 percent chance to be a critical hit, which doubles the damage to 20.
- Normal hit: 80 percent × 10 = 8.
- Critical hit: 20 percent × 20 = 4.
- Average per attack: 8 + 4 = 12.
Equivalently, 10 × (0.8 + 0.2 × 2) = 10 × 1.2 = 12, so the critical chance acts like a 20 percent bonus on average. Any single attack is either 10 or 20; the 12 only appears as an average over many attacks. If a hit chance of less than 100 percent also applies, multiply it in as a further factor.
Example 6: Completing a poker draw
Suppose you hold two cards and the table shows three more. You need one of 9 specific cards to complete a draw. You can see 5 cards, so 52 − 5 = 47 are unseen.
Chance the next card is one of your 9: 9/47, about 19.1 percent.
This is the basic step behind counting outs, and it assumes you have no information about the other players’ cards. In real play the chance of completing a hand is only one input; the price of staying in matters as well.
Example 7: A fair price for a simple game
A game costs 1 unit. You roll a die and receive 6 units back on a six, otherwise nothing.
- Win: 1/6 × (6 − 1) = 5/6.
- Lose: 5/6 × (−1) = −5/6.
- Expected value: 5/6 − 5/6 = 0.
Zero means the game is fair on average. If the prize were 5 units instead, the average would be 1/6 × 4 − 5/6 × 1 = −1/6, about −0.17 units per play, a loss over time. Real casino games sit on the losing side of that line, which is what the house edge measures.
Habits these examples share
- Count equally likely outcomes, or use the opposite case.
- Check whether the events are independent.
- Multiply for “and”, add for separate ways, subtract from one for “at least one”.
- Treat averages as long-run summaries, not forecasts for the next play.
Check your own working by re-doing it a second way, for example listing a small case by hand. A neat-looking answer can still be wrong, and a probability above 1 or below 0 is always a sign of a mistake.