Discrete Uniform Law

When all outcomes in a finite sample space are equally likely, probability reduces to counting.

The Law

If a sample space has $N$ equally likely outcomes, each outcome has probability:

$$P(\text{each outcome}) = \frac{1}{N}$$

For any event $A$ that contains $n$ of those outcomes:

$$P(A) = \frac{n}{N} = \frac{\text{outcomes in } A}{\text{total outcomes}}$$

Why It Works

This follows from the additivity axiom. If all $N$ outcomes have equal probability $p$, then by normalization:

$$N \cdot p = P(\Omega) = 1 \implies p = \frac{1}{N}$$

And by additivity, an event with $n$ outcomes has probability $n \cdot \frac{1}{N} = \frac{n}{N}$.

Simple Examples

Connection to Counting

The Discrete Uniform Law bridges probability and counting. When outcomes are equally likely, computing $P(A)$ is a pure counting problem:

1. Count the total number of outcomes $N$
2. Count the outcomes where $A$ actually occurs
3. Divide to get $P(A) = \frac{n}{N}$

For simple events, you can count by inspection. But as events get more complex, you need systematic counting methods — which is exactly what the Counting Methods module covers.

Worked Example: Two Dice

Roll two fair 4-sided dice. Let $X$ = first roll, $Y$ = second roll.

Sample space: $\Omega = \{(x, y) : x \in \{1,2,3,4\} \text{ and } y \in \{1,2,3,4\}\}$ — a 4×4 grid with 16 outcomes, each with probability $\frac{1}{16}$.

Event A: “The first roll is 1.”

The entire first row is favorable — 4 outcomes. Just read them off: $(1,1), (1,2), (1,3), (1,4)$.

$$P(X = 1) = \frac{4}{16} = \frac{1}{4}$$

Event B: “The sum $X + Y$ is odd.”

8 outcomes are favorable (green cells). Notice the checkerboard pattern — odd sums and even sums alternate. You can count by spotting this pattern: half the grid is odd, half is even. No case-by-case breakdown needed.

$$P(X + Y \text{ is odd}) = \frac{8}{16} = \frac{1}{2}$$

Event C: “The smaller of the two rolls equals 2.”

This means $\min(X, Y) = 2$ — at least one roll is 2, and neither roll is smaller than 2. There’s no simple pattern to spot here. Instead, we use systematic counting: break the event into non-overlapping cases and count each one.

5 outcomes are favorable (green cells). But how do we find them without scanning every cell? We break into cases:

Case 1: Both rolls equal 2.

• Only $(2, 2)$ → 1 outcome

Case 2: First roll is 2, second roll is bigger.

• $(2, 3), (2, 4)$ → 2 outcomes

Case 3: Second roll is 2, first roll is bigger.

• $(3, 2), (4, 2)$ → 2 outcomes

These cases are disjoint — no outcome appears in more than one case — so we can add them up:

$$n = 1 + 2 + 2 = 5$$

$$P(\min(X, Y) = 2) = \frac{5}{16}$$

The key difference from Event B: Event B had a visual pattern you could exploit (checkerboard). Event C has no such pattern — you must decompose the event into cases and count each case separately. This case-by-case approach is what systematic counting means, and it’s exactly what the Counting Methods module teaches you to do at scale.

When the Discrete Uniform Law Does NOT Apply

The law only works when outcomes are equally likely. A rigged dice — where some numbers are more likely than others — does not satisfy it. For example, if the number 6 comes up twice as often as the other faces, then P(6) ≠ P(1), so counting outcomes gives the wrong answer. You must assign probabilities to each outcome individually; you can’t just count.

The General Procedure

The two-dice example illustrates the general approach for any discrete uniform model:

1. Set up the sample space — list or diagram all outcomes
2. Verify equal likelihood — confirm each outcome has the same probability
3. Identify the event — find which outcomes belong to your event of interest
4. Count and divide — $P(A) = \frac{|A|}{|\Omega|}$

The hard part is rarely the probability — it’s the counting. For small sample spaces you can enumerate by hand. For larger ones, you need systematic counting techniques (permutations, combinations) to find $|A|$ and $|\Omega|$ without listing every outcome.