Equally Likely Outcomes

May 2026 · 16 min read

Many real experiments — rolling a fair die, drawing a card from a well-shuffled deck, picking a ball from a bag — come with strong physical symmetry that makes every elementary outcome equally likely. The axioms then collapse into a single formula:

This is the classical definition of probability, and it explains why combinatorial analysis is the lifeblood of an introductory probability course: probabilities reduce to counting.

Why It Works (From the Axioms)

Suppose $|S| = N$ and every elementary outcome has the same probability $p$. By Axiom 3 applied to the disjoint singletons,

$1 = P(S) = \sum_{s \in S} p = N p \quad\Longrightarrow\quad p = \frac{1}{N}.$

For an event $E$ with $|E| = k$, again by Axiom 3,

$P(E) = \sum_{s \in E} p = k \cdot \frac{1}{N} = \frac{|E|}{|S|}.$

So the classical formula is not a separate definition: it is a theorem that follows from the axioms once we add the modelling assumption "all outcomes equally likely". The whole game becomes counting $|E|$ and $|S|$, which is exactly what the Combinatorial Analysis series taught us to do.

A Five-Step Recipe

Almost every classical probability problem follows the same workflow.

1. Identify the experiment. What is the underlying random process?
2. Choose a sample space in which the outcomes are equally likely. Often this means modelling outcomes as ordered tuples even if order doesn't seem to matter, because ordering creates symmetry.
3. Compute $|S|$ using a counting tool (multiplication rule, permutations, combinations, ...).
4. Express the event $E$ as a subset of $S$ and compute $|E|$ by the same kind of argument.
5. Divide. $P(E) = |E| / |S|$.

Example 1. Sum of Two Dice

Roll two fair dice. What is $P(\text{sum} = 7)$?

• $S = \{(i, j) : 1 \le i, j \le 6\}$, $|S| = 36$. We use ordered pairs to make the outcomes equally likely.
• $E = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$, $|E| = 6$.
• $P(E) = 6/36 = 1/6$.

If we instead used unordered pairs, $\{1,6\}$ and $\{2,5\}$ would each appear once, but $\{3,3\}$ and $\{4,4\}$ would also appear once — yet a $(3,3)$ outcome is half as likely as a $(1,6)$ in the real experiment. The unordered sample space is not equally likely. This is the classical “ordered vs unordered” trap.

Example 2. Poker Hands

Deal 5 cards from a standard 52-card deck. What is $P(\text{full house})$?

A "full house" is three of a kind plus a pair (e.g. three jacks and two sevens).

• $|S| = \binom{52}{5}$ — the number of unordered 5-card hands. The deck is shuffled, so all hands are equally likely.
• For $|E|$: choose the rank of the triple ($13$), then 3 of its 4 suits ($\binom{4}{3} = 4$); choose the rank of the pair ($12$ remaining ranks), then 2 of its 4 suits ($\binom{4}{2} = 6$).
• $|E| = 13 \cdot 4 \cdot 12 \cdot 6 = 3744$.
• $P(\text{full house}) = 3744 / \binom{52}{5} = 3744 / 2598960 \approx 0.00144$.

Example 3. The Urn (Balls and Boxes)

An urn contains $r$ red and $b$ blue balls. Draw $k$ without replacement. What is $P(\text{exactly } j \text{ red})$?

This is the hypergeometric distribution.

• $|S| = \binom{r+b}{k}$ if we treat order as irrelevant (and balls of the same colour as distinguishable).
• $|E|$: choose which $j$ of the $r$ reds you got and which $k - j$ of the $b$ blues, $\binom{r}{j}\binom{b}{k-j}$.
• $P(\text{exactly } j \text{ red}) = \dfrac{\binom{r}{j}\binom{b}{k-j}}{\binom{r+b}{k}}$.

This single formula recovers, as special cases, "what's the probability the first card is an ace?", "what's the probability of drawing 2 hearts in 5 cards?", and lots more.

Example 4. The Birthday Problem

$n$ people are in a room. What is the probability that at least two share a birthday?

Assume $365$ equally likely birthdays and that the $n$ birthdays are independent — so $|S| = 365^n$ ordered birthday tuples are equally likely.

Computing “at least one collision” directly is awful (lots of overlap). Use the complement (Proposition 2 from Part 4).

The number of birthday tuples with all birthdays distinct is $365 \cdot 364 \cdots (365 - n + 1)$. So

$\displaystyle P(\text{all distinct}) = \frac{365 \cdot 364 \cdots (365 - n + 1)}{365^n} = \prod_{k=0}^{n-1} \left(1 - \frac{k}{365}\right),$

and

$P(\text{at least one shared}) = 1 - P(\text{all distinct}).$

The surprise: $n = 23$ already gives $P(\text{shared}) \approx 0.507$. By $n = 50$ it's $\approx 0.97$. Most people guess much higher than $23$.

Visual: Birthday Probabilities Climbing

Step through values of $n$ and watch $P(\text{at least one shared birthday})$ rise. The crossing of $1/2$ at $n = 23$ is a famous counter-intuitive moment.

Why the Birthday Number Feels Wrong

People guess huge values for $n$ because they unconsciously ask “how many people do I need to share a birthday with me?”, which has answer roughly $365 \log 2 \approx 253$. But the question is “some two people share”, which involves $\binom{n}{2}$ pairs — quadratically many opportunities. A rough heuristic: collisions become likely once $\binom{n}{2} \approx 365$, i.e.\ $n \approx \sqrt{2 \cdot 365} \approx 27$. Spot-on.

Connection to Combinatorial Analysis

Every classical probability problem in this section dropped one of these counting tools from the previous series:

• Two dice problem — the multiplication rule from Part 2.
• Poker full house — combinations from Part 4.
• Hypergeometric urn — combinations and the multiplication rule.
• Birthday problem — permutations $P(365, n)$ from Part 3 (in the numerator).

If a counting formula confuses you here, revisit that post first.

When the Outcomes Are Not Equally Likely

Most real experiments have non-uniform weights. Two strategies:

1. Refine the sample space until equality holds. Drawing a card from a deck where some cards have been removed is non-uniform on the rank, but uniform on the underlying card; pick the right $S$ and the formula returns.
2. Switch to weighted sums. If $S = \{s_1, s_2, \ldots\}$ with non-uniform $p_i$, then $P(E) = \sum_{s_i \in E} p_i$. The axioms still hold; the classical formula doesn't.

Common Mistakes

• Counting $|S|$ as the number of unordered configurations and $|E|$ as ordered, or vice versa. Pick a convention and stick to it for both.
• Treating non-uniform outcomes as uniform. Sum-of-two-dice $\{2, 3, \ldots, 12\}$ has 11 outcomes — but they are not equally likely.
• Forgetting the complement trick on "at least one" problems — especially the birthday problem.
• Off-by-one in the birthday product (using $365 - n$ instead of $365 - n + 1$).

Practice

1. A bowl contains 6 white and 5 black balls. Draw 4 without replacement. Find $P(\text{2 white and 2 black})$.
2. $5$ cards are dealt from a standard deck. Find $P(\text{flush})$ and $P(\text{straight})$ (5 cards in sequence; allow A as low or high).
3. Toss a fair coin 10 times. Find the probability of exactly 3 heads.
4. $n = 30$ people in a room. Estimate $P(\text{shared birthday})$ to two decimal places.
5. Three balls are drawn at random from an urn containing 8 white, 4 black, and 2 orange balls. Find the probability that the colours are all different.
6. Show that the matching-problem probability from the previous post can also be computed via $1 - |D_n|/n!$ where $D_n$ is the number of derangements of $\{1, \ldots, n\}$.