Suppose your desk asks you to price an option — a contract that pays out only if a stock finishes above a strike. Its value today clearly depends on what the stock does tomorrow, and what the stock does tomorrow is random. So before you can write down a single pricing formula, you must answer a question that sounds almost philosophical: what, precisely, does "random" mean?

Mathematics answers with a contract of its own. To speak of chance at all, you must put three things on the table:

• the possible worlds — a set \Omega listing every way the experiment could turn out, one outcome \omega per world;
• the askable questions — a collection \mathcal{F} of subsets of \Omega, the events: "did the stock go up?", "did at least one coin land heads?" — each question is the set of worlds in which its answer is yes;
• the odds — a function \mathbb{P} attaching a number in [0,1] to every askable question.

Formally, a probability measure is nothing more than a measure with one extra requirement: the whole space has measure one. Write it \mathbb{P}, so that \mathbb{P}(\Omega) = 1 — something is certain to happen. The complete object,

is a probability space: \Omega the sample space of outcomes, \mathcal{F} the σ-algebra of events, and \mathbb{P} the probability assigned to each event. This triple is the stage on which every random phenomenon in this course plays out — every stock path, every default, every roll of the die is a point \omega moving through this machinery. Learn to build the triple explicitly for small examples now, and the continuous-time models later will feel like the same construction with a bigger \Omega.

Kolmogorov's axioms

Stripped to essentials, a probability measure obeys three axioms — written down by Andrey Kolmogorov in 1933, they are the foundation on which the rest of probability is built entirely:

• Non-negativity: \mathbb{P}(A) \ge 0 for every event A.
• Normalisation: \mathbb{P}(\Omega) = 1.
• Countable additivity: for pairwise-disjoint A_1, A_2, \dots, \mathbb{P}\!\left(\bigcup_{i} A_i\right) = \sum_{i} \mathbb{P}(A_i).

Notice what is not on the list: no mention of fairness, symmetry, frequencies, or beliefs. The axioms are deliberately agnostic about what probability "really is" — they only pin down how it must behave. That neutrality is a feature: the same three rules govern a casino die, a default probability calibrated from bond spreads, and the artificial "risk-neutral" odds used to price derivatives.

From these alone come the everyday rules. Let us derive the three workhorses line by line, using only the axioms above.

The complement rule. An event A and its complement A^{c} share no outcome and between them cover every outcome, so they form a disjoint union of the whole space:

Take \mathbb{P} of both sides. On the left, normalisation gives \mathbb{P}(\Omega) = 1; on the right, additivity over the two disjoint pieces splits the probability into a sum:

Rearranging the equation 1 = \mathbb{P}(A) + \mathbb{P}(A^{c}) isolates the complement:

The empty set has probability zero. The empty set is the complement of the whole space, \emptyset = \Omega^{c}, so feed A = \Omega into the rule we just proved:

Inclusion–exclusion. For two events A and B that may overlap, additivity does not apply directly — but it does once we rewrite the pieces as disjoint unions. First split A \cup B into A together with the part of B lying outside A; these are disjoint and cover the union:

Likewise split B itself into the part inside A and the part outside it:

Apply additivity to each disjoint split:

The second equation hands us the leftover piece, \mathbb{P}(B \setminus A) = \mathbb{P}(B) - \mathbb{P}(A \cap B). Substitute that into the first:

The - \mathbb{P}(A \cap B) is precisely the overlap, which \mathbb{P}(A) + \mathbb{P}(B) would otherwise count twice.

Building a probability space by hand: one coin toss

The definition earns its keep only when you can construct the triple explicitly. Start with the smallest interesting experiment: toss one fair coin.

Step 1 — list the possible worlds. Two ways it can land:

Step 2 — choose the askable questions. For a finite \Omega the natural choice is every subset — the power set:

Read each event as a question: \emptyset asks "did nothing happen?" (never yes), \{H\} asks "heads?", \{T\} asks "tails?", and \Omega asks "did the coin land at all?" (always yes). Check the σ-algebra requirements by eye: it contains \Omega, it is closed under complements (\{H\}^{c} = \{T\}, \Omega^{c} = \emptyset), and closed under unions (\{H\} \cup \{T\} = \Omega). ✓

Step 3 — assign the odds. Fairness means

Step 4 — verify the axioms. Non-negativity: all four values are \ge 0. ✓ Normalisation: \mathbb{P}(\Omega) = 1. ✓ Additivity: the only non-trivial disjoint union is \{H\} \sqcup \{T\} = \Omega, and indeed \tfrac{1}{2} + \tfrac{1}{2} = 1 = \mathbb{P}(\Omega). ✓ All three axioms hold, so (\Omega, \mathcal{F}, \mathbb{P}) is a genuine probability space — the smallest one worth having.

Note the division of labour: nothing about \Omega or \mathcal{F} knew the coin was fair. A biased coin lives on the same stage (\Omega, \mathcal{F}) with a different \mathbb{P} — say \mathbb{P}(\{H\}) = 0.6, \mathbb{P}(\{T\}) = 0.4. Same worlds, same questions, different odds. Keep that separation in mind; finance leans on it hard.

Two tosses: the space grows fast

Now toss the fair coin twice, recording the ordered pair of results. The possible worlds multiply:

Take \mathcal{F} = 2^{\Omega} again — with four outcomes the power set already holds 2^4 = 16 events (a coin tossed n times has 2^n outcomes and a power set of 2^{2^n} events; by n = 5 that is over four billion askable questions). Fairness and independence make every outcome equally likely, so the measure is uniform counting:

The axioms are quick to verify: |A| \ge 0 gives non-negativity; \mathbb{P}(\Omega) = 4/4 = 1 gives normalisation; and disjoint events have |A \sqcup B| = |A| + |B|, so additivity is just counting.

A compound event, two ways. What is the probability of at least one head? Call that event C.

Via the complement rule. The complement of "at least one head" is "no heads at all", the single world C^{c} = \{TT\}:

Via inclusion–exclusion. Write C = A \cup B where A = \{HH, HT\} ("first toss heads") and B = \{HH, TH\} ("second toss heads"). They overlap on A \cap B = \{HH\}, so

Two independent routes through the axioms, one answer — exactly the sort of consistency the three rules guarantee. When a probability computation ever disagrees with itself, an axiom has been violated somewhere, and the axioms tell you where to look.

The one-step binomial stock: finance's favourite probability space

Here is the probability space that launches every derivatives course. A stock is worth S_0 today; by tomorrow it will have either risen or fallen. The possible worlds could not be simpler:

In the world \omega = \text{up} the stock moves to uS_0; in \omega = \text{down}, to dS_0 (with d < 1 < u). The measure needs only one number, the up-probability p:

Normalisation and additivity hold by construction: p + (1 - p) = 1 = \mathbb{P}(\Omega). Structurally this is the coin-toss space wearing a suit — which is precisely the point. A call option struck between dS_0 and uS_0 pays out exactly on the event \{\text{up}\} \in \mathcal{F}, so "what is the option worth?" is an askable question the moment the triple is written down.

And now the payoff of separating stage from odds: the deepest trick in derivative pricing is to keep (\Omega, \mathcal{F}) fixed but swap the measure — replace the real-world \mathbb{P} with an artificial risk-neutral measure \mathbb{Q} under which discounted prices behave like fair bets. That manoeuvre only makes sense because the triple keeps the worlds, the questions, and the odds as three separate dials. Chain many of these one-step spaces together and you are most of the way to the binomial tree, and from there to Black–Scholes.

The fair die, end to end

One more finite space, this time with room to explore. Take \Omega = \{1,2,3,4,5,6\}, \mathcal{F} = 2^{\Omega} the power set (all 2^6 = 64 events), and \mathbb{P}(A) = |A| / 6 — each outcome equally likely. Pick an event below; its faces light up and its probability shows as a fraction. Every choice obeys 0 \le \mathbb{P}(A) \le 1.

Let us actually compute with this space, checking each rule against the formula \mathbb{P}(A) = |A|/6.

• An even face. Let E = \{2, 4, 6\}. It has three outcomes, so \mathbb{P}(E) = \tfrac{|E|}{6} = \tfrac{3}{6} = \tfrac{1}{2}.
• Its complement, two ways. An odd face is E^{c} = \{1, 3, 5\}, so directly \mathbb{P}(E^{c}) = \tfrac{3}{6} = \tfrac{1}{2}. The complement rule must agree: \mathbb{P}(E^{c}) = 1 - \mathbb{P}(E) = 1 - \tfrac{1}{2} = \tfrac{1}{2}. ✓
• Inclusion–exclusion with overlap. Let A = \{2, 4, 6\} (even) and B = \{4, 5, 6\} (at least 4). They share A \cap B = \{4, 6\}, so \mathbb{P}(A) = \tfrac{3}{6}, \mathbb{P}(B) = \tfrac{3}{6}, \mathbb{P}(A \cap B) = \tfrac{2}{6}. Inclusion–exclusion predicts

and counting directly confirms it: A \cup B = \{2, 4, 5, 6\} has four outcomes, so \mathbb{P}(A \cup B) = \tfrac{4}{6} = \tfrac{2}{3}. The -\,\tfrac{2}{6} removed exactly the double-counted overlap \{4, 6\}.

As you switch events in the figure, watch how the machinery never changes — the six worlds and the sixty-four questions are fixed once and for all; only which question you are asking moves. That is the probability-space discipline in miniature.