Probabilistic Methods

The most natural way to organise the sample space of two coin tosses is to place the result of the first toss along the rows and the result of the second toss along the columns. The row labelled H corresponds to "first toss is heads" and the row labelled T to "first toss is tails"; similarly the column labelled H means "second toss is heads" and column T means "second toss is tails". The four cells of the resulting 2×2 grid are in bijection with the four outcomes: cell (0,0) encodes HH, cell (0,1) encodes HT, cell (1,0) encodes TH, and cell (1,1) encodes TT. An event is simply a subset of cells — we shade or mark those cells — and its probability equals the number of marked cells divided by four.

An outcome has exactly one head when the two tosses differ: one shows H and the other shows T. Looking at the grid, this means the row label and column label are different. The cell (0,1) has row H and column T — that is HT — and cell (1,0) has row T and column H — that is TH. Exactly those two cells are marked. The two cells (0,0) = HH and (1,1) = TT are not marked because both tosses are the same there.

The two tosses agree when the outcome lies on the main diagonal of the grid — both H (top-left) or both T (bottom-right). These are cells (0,0) and (1,1). Notice that $A_2$ is the complement of $A_1$: the grid for $A_2$ marks the two cells left unmarked by $A_1$, and $P(A_1) + P(A_2) = 1$ as expected. This is a reassuring consistency check: no cell is double-counted, and together they cover the entire space.

At least one head is satisfied by every outcome except TT, which is the only cell where neither toss produced a head. Equivalently, using the complement rule: $P(A_3) = 1 - P(\text{no heads}) = 1 - P(\{TT\}) = 1 - 1/4 = 3/4$. In the grid, three of the four cells are marked, leaving only cell (1,1) = TT empty. The L-shaped region of marked cells is characteristic of "at least one" events across many sample spaces.

The condition "first toss is tails" constrains only the row: we look at the row labelled T and take both cells in that row regardless of the second toss. This gives cells (1,0) = TH and (1,1) = TT. In grid terms, the event corresponds to the entire bottom row being marked. The probability is 2/4 = 1/2, consistent with P(first toss = T) = 1/2 for a fair coin.

The condition "second toss is heads" constrains only the column: we look at the column labelled H and take both cells in that column regardless of the first toss. This gives cells (0,0) = HH and (1,0) = TH. In grid terms, the event is the entire left column. Comparing with $A_4$: that event was the entire bottom row, and this event is the entire left column; together they illustrate how row-events and column-events correspond to conditions on the first and second toss respectively.

Reading a grid in reverse: if someone hands us a grid with the entire top row (row index 0, label H) marked and nothing else, we must identify the event. Row 0 corresponds to "first toss = H". The two marked cells are (0,0) = HH and (0,1) = HT. Therefore the event is "the first toss landed heads," $A = \{HH, HT\}$. The probability is $2/4 = 1/2$. This is precisely the mirror image of $A_4$: conditions on the first toss give horizontal bands (entire rows), while conditions on the second toss give vertical bands (entire columns).

A grid with only the two diagonal cells (0,0) and (1,1) marked encodes the event "both tosses show the same face," namely $A = \{HH, TT\}$. We already computed this in part A(2): $P(A) = 1/2$. The key geometric insight is that the main diagonal of a square grid always encodes "both components are equal" — this pattern recurs for die×die grids, where the diagonal encodes "both dice show the same number." Recognising this shape enables instant identification of equality events in any Cartesian-product sample space.

We organise the 36 outcomes into a 6×6 grid. The rows are indexed 0 through 5 and labelled 1 through 6, corresponding to the value shown by the first die. The columns are similarly labelled 1 through 6 for the second die. Cell (r, c) therefore represents the outcome "first die = r+1, second die = c+1." An event is a set of such cells, and its probability equals the cell count divided by 36. The grid makes visible the geometry of events in a way that a plain list does not: sums form diagonals, one-die conditions form rows or columns, and symmetry between the two dice corresponds to symmetry across the main diagonal.

The pairs $(i,j)$ with $i + j = 8$ are: $(2,6), (3,5), (4,4), (5,3), (6,2)$. In 0-indexed grid coordinates these are (1,5), (2,4), (3,3), (4,2), (5,1). Notice these cells form an anti-diagonal line — sums of constant value always appear as anti-diagonals in the grid. There are 5 such cells. The anti-diagonal corresponding to sum 7 would have 6 cells and the highest probability among all sum-events; sum 8 has one fewer cell on each side, giving 5.

The condition $i > j$ selects all cells strictly below the main diagonal. Enumerating: when first die = 2, second can be 1 (1 cell); when first = 3, second can be 1 or 2 (2 cells); when first = 4, three choices; first = 5, four choices; first = 6, five choices. Total: $1+2+3+4+5 = 15$ cells. The lower triangular region occupies exactly $15/36 = 5/12$ of the grid, by symmetry equal to the probability that the first die is strictly less — a useful check.

The even faces are 2, 4, 6, corresponding to row/column indices 1, 3, 5 (0-indexed). We need all combinations of these rows and columns, giving a 3×3 sub-grid embedded in the 6×6 grid. There are $3 \times 3 = 9$ such cells. In the grid they form a regular pattern — every other row crossed with every other column — which is visually striking and easy to recognise.

This event is the union of "first die = 6" (entire row 5, 6 cells) and "second die = 6" (entire column 5, 6 cells). By inclusion-exclusion, the overlap is the single cell where both dice show 6, namely (5,5). So $|A| = 6 + 6 - 1 = 11$. In the grid this is the bottom row plus the rightmost column in an L-shaped pattern. The cell (5,5) is included in both arms of the L, but counted only once.

Exactly one die shows 1 means either first die = 1 and second die ≠ 1, or second die = 1 and first die ≠ 1. The first part is row 0 minus cell (0,0): five cells. The second part is column 0 minus cell (0,0): five cells. Together, $5 + 5 = 10$ cells, arranged in a cross shape with the corner cell removed. Notice this differs from "at least one die shows 1" (which would include (0,0)) by exactly one cell.

When someone presents a grid where the bottom-right 4×4 block (rows 2–5, columns 2–5) is entirely marked, the event is "both dice show at least 3." Row indices 2–5 correspond to die values 3–6, and the same for columns. The event requires both dice simultaneously in $\{3,4,5,6\}$, giving $4 \times 4 = 16$ cells. Its probability is $16/36 = 4/9$. This block pattern is characteristic of joint lower-bound events.

A grid with the six main-diagonal cells (0,0), (1,1), (2,2), (3,3), (4,4), (5,5) marked encodes the event "both dice show the same value," i.e., $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. This is exactly the event of rolling doubles. The probability is $6/36 = 1/6$. The diagonal pattern is the purest visual signature of an equality constraint between two symmetric components, and it appears identically in the coin×coin grid where the main diagonal encoded "both tosses the same."

Unlike the previous grids where both axes represented values of the same type (coin tosses or die faces), here the two axes represent fundamentally different things: rows index the weather state and columns index the day of the week. Marking cell (r, c) is a claim about the daily weather: it asserts that on day c the weather is state r. This is a different style of event grid — one that captures the behaviour of a process over time. Each column has exactly one "actual" state on any given week, but an event may allow multiple states per column (uncertainty about the state) or impose conditions on only some of the columns.

The event "Monday is sunny" constrains only Monday's state to S, leaving all other days unconstrained. In the grid, we mark only the single cell at row S (index 0) and column Mon (index 0). This is a minimal event: a single-cell mark in a 3×7 grid. It eliminates any outcome where Monday is Cloudy or Rainy, while remaining silent on the remaining six days. Provided each daily state has an equal probability of 1/3 and days are independent, $P(A) = 1/3$.

The weekend consists of Saturday (column index 5) and Sunday (column index 6). The event requires both days to be in state R (row index 2). In the grid we mark cells (2,5) and (2,6). No other cells are marked because the five weekdays are unconstrained. Assuming independence and $P(\text{day} = R) = 1/3$ per day, $P(A) = (1/3)^2 = 1/9$. The two marked cells sit in the bottom-right corner of the grid, a compact visual marker for a weekend rainy event.

The event "it rains on Wednesday OR Friday" is satisfied if at least one of those two days is rainy. This is an OR-event on two separate days. In the grid we mark (R, Wed) = (2,2) and (R, Fri) = (2,4). The word "or" here is inclusive, but since we are marking individual daily conditions (not joint outcomes), there is no double-counting issue at the grid level. The probability depends on whether independence holds: $P(A) = 1 - P(\text{Wed not R}) \cdot P(\text{Fri not R}) = 1 - (2/3)^2 = 5/9$ under independence.

The event "no day is rainy" means every day is either Sunny or Cloudy. In the grid, this corresponds to the entire row R (row index 2) being empty — and equivalently, only rows S and C are ever used. We mark all 14 cells in rows 0 and 1 (S and C) across all 7 columns. This may seem to conflict with the idea that "marking a cell means the state holds," but here we are marking the set of states the day is allowed to be. The probability is $(2/3)^7 = 128/2187$, since each of the 7 days must independently avoid rain.

The event "Thursday is not sunny" excludes state S for Thursday (column index 3) and allows C or R. In the grid we mark (C, Thu) = (1,3) and (R, Thu) = (2,3). This is a two-cell condition on a single column. The probability is $P(\text{Thu} \ne S) = 2/3$. This illustrates a complementary column pattern: the complement "Thursday is sunny" would be the single cell (0,3), and together the three cells in column Thu partition that column, summing to probability 1.

Reading the grid: cells (0,5) and (0,6) are marked, corresponding to row S and columns Sat and Sun. This encodes the event "Saturday is sunny AND Sunday is sunny," i.e., the entire weekend is sunny. The grid is the mirror image of A(2) (weekend rainy), but now in row S instead of row R. Under the equal-probability model, $P = (1/3)^2 = 1/9$, the same as the rainy-weekend probability — which makes sense because each state is equally likely and the two events are structurally identical.

When row R (index 2) is completely empty and rows S and C contain the marked cells, the event is "no rain occurs on any day of the week." This is the same event as A(4) viewed from the opposite angle: rather than marking the allowed states, we observe which row is excluded. An entire empty row in a weather grid is always a strong visual signal: it says one weather state is universally forbidden throughout the week. Under the independence model, $P = (2/3)^7 \approx 0.059$.

The six pairs that sum to 7 are $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$. In 0-indexed grid coordinates these are the anti-diagonal cells (0,5),(1,4),(2,3),(3,2),(4,1),(5,0). Sum-7 is the most probable single-sum event: it lies on the central anti-diagonal of the 6×6 grid and contains more cells than any other sum. $P(A) = 6/36 = 1/6$.

The lower-triangular region of the 6×6 grid (all cells strictly below the main diagonal) encodes the event $B: i > j$. As computed in Problem 2, this region has $1+2+3+4+5 = 15$ cells. The marked shape is a right triangle below the diagonal. $P(B) = 15/36 = 5/12$.

Event $C$ is the L-shaped union of row 5 (first die = 6) and column 5 (second die = 6). By inclusion-exclusion, $|C| = 6 + 6 - 1 = 11$. The single overlap cell is (5,5) = (6,6). $P(C) = 11/36$.

To compute $A \cup C$, take all cells in $A$ together with all cells in $C$, removing duplicates. The cells of $A$ that are also in $C$ are those where the sum is 7 and at least one die is 6: $(1,6)$ and $(6,1)$, i.e., cells (0,5) and (5,0). So $|A \cup C| = |A| + |C| - |A \cap C| = 6 + 11 - 2 = 15$. The marked region on the grid is an anti-diagonal with two cells in common with the L-shaped event C.

The intersection requires both conditions simultaneously: sum = 7 AND at least one die = 6. From the six pairs in $A$, only $(1,6)$ and $(6,1)$ involve a 6. In grid coordinates these are (0,5) and (5,0). This tiny two-cell intersection demonstrates how powerful the combination of two conditions can be: adding the constraint "at least one die = 6" to "sum = 7" reduces the event from 6 cells to 2.

We need cells in the lower triangle (first > second) that are also in the L-shape (at least one die = 6). In row 5 (first die = 6), the lower-triangle cells are those where second < 6, i.e., columns 0–4: five cells (5,0),(5,1),(5,2),(5,3),(5,4). Column 5 (second die = 6) paired with first > 6 is impossible since 6 is the maximum. So $B \cap C$ has 5 cells.

The complement $B^c$ is the event "first die is NOT strictly greater than second," i.e., first ≤ second (upper triangle plus diagonal). From the six cells of $A$, those with $i \le j$ are $(1,6),(2,5),(3,4)$: cells (0,5),(1,4),(2,3). The three pairs where $i > j$ — namely $(4,3),(5,2),(6,1)$ — are excluded by $B^c$. So $A \cap B^c$ has 3 cells, each with the second die exceeding the first.

The complement $C^c$ is "neither die shows 6." From the six cells of $A$, we remove those containing a 6: $(1,6)$ and $(6,1)$. The remaining four cells are $(2,5),(3,4),(4,3),(5,2)$, i.e., cells (1,4),(2,3),(3,2),(4,1). These are the sum-7 pairs where neither die is 6 — a fact sometimes useful because these outcomes tend to be "hidden" from one-die conditions.

Start with the 11 cells of $C$ and remove the 2 cells in $A \cap C$, namely (0,5) and (5,0). The remaining $11 - 2 = 9$ cells form the L-shape with its two tips clipped. These are outcomes where a 6 appears but the sum is something other than 7 (possible sums in $C$: 7, 8, 9, 10, 11, 12; we remove only the sum-7 ones).

Start with the 15 cells of $B$ (lower triangle) and remove those cells also in $A$. From $A$, the cells with $i > j$ are $(4,3),(5,2),(6,1)$, i.e., (3,2),(4,1),(5,0). Removing these three from the 15 lower-triangle cells gives $15 - 3 = 12$ cells. This compound event captures outcomes where the first die wins the comparison but the total is not 7.

From the 11 cells of $C$, remove those where $i > j$: the 5 cells of $B \cap C$, namely (5,0),(5,1),(5,2),(5,3),(5,4). The remaining $11 - 5 = 6$ cells are: (5,5) and the five column-5 cells above the diagonal: (0,5),(1,5),(2,5),(3,5),(4,5). These are outcomes where at least one die shows 6 but the second die is at least as large as the first.

The complement of $A \cup C$ is the event that neither condition holds: sum is not 7 AND no die shows 6. Since $|A \cup C| = 15$, its complement has $36 - 15 = 21$ cells. These are all cells in the top-left 5×5 sub-grid (no die is 6) except the anti-diagonal of sum-7. Alternatively, $|((A \cup C)^c)| = 36 - 15 = 21$. Wait — let us recount: $A$ has 6 cells, $C$ has 11, $A \cap C$ has 2, so $|A \cup C| = 15$ and the complement has $36 - 15 = 21$ cells.

The complement of $A \cap C$ is the event that at least one of the two conditions fails: either sum ≠ 7 or no die shows 6 (or both). Since $|A \cap C| = 2$, the complement has $36 - 2 = 34$ cells. This is a very large event — only two outcomes ($(1,6)$ and $(6,1)$) are excluded. The moral: intersections of two non-trivial events tend to be small, so their complements are large.

Before computing anything, let us record the six elementary frequencies. The total number of rolls is $1000$. The relative frequency of elementary outcome $\{k\}$ is $f(\{k\}) = n(k)/1000$. The resulting table shows frequencies that hover around the theoretical value $1/6 \approx 0.167$, with natural random fluctuations. A table interactive makes the data easier to read.

We compute relative frequencies by summing counts for the outcomes in each event and dividing by 1000. For event $\{2,4,6\}$ (even outcomes): $n = 154+167+170 = 491$, so $f = 0.491$. For $\{1,2,3\}$: $n = 168+154+181 = 503$, $f = 0.503$. For $\{5,6\}$: $n = 160+170 = 330$, $f = 0.330$. For $\{1,3,5\}$ (odd): $n = 168+181+160 = 509$, $f = 0.509$. For $\{1,2,3,4\}$: $n = 168+154+181+167 = 670$, $f = 0.670$. In every case the empirical frequency is close to but not equal to the theoretical probability $P(A) = |A|/6$.

Take two disjoint events $E_1 = \{1,2,3\}$ and $E_2 = \{4,5,6\}$. Their union is $\Omega = \{1,2,3,4,5,6\}$. The individual frequencies are $f(E_1) = 503/1000$ and $f(E_2) = 497/1000$. Their sum is $503/1000 + 497/1000 = 1000/1000 = 1 = f(\Omega)$. This is not a coincidence: for any two disjoint events $A$ and $B$, $n(A \cup B) = n(A) + n(B)$ exactly (no outcome is counted twice), so $f(A \cup B) = f(A) + f(B)$ always holds for disjoint events in any finite experiment. This finite additivity is a universal feature of relative frequencies.

Finite additivity is exactly what we verified above: for a finite list of disjoint events, the frequency of their union equals the sum of individual frequencies. The stronger axiom of countable additivity (or $\sigma$-additivity) extends this to countably infinite families of disjoint events. In finite experiments this distinction is invisible — you can never have infinitely many disjoint non-empty subsets of a finite sample space. But when we define probability abstractly (e.g., for continuous distributions or infinite sequences of trials), we must impose $\sigma$-additivity as an axiom, because it cannot be derived from finite experiments alone. Kolmogorov's axiom system makes this step explicit.

Let $M = \{1,2,3\}$ and $N = \{3,4,5\}$. These events are NOT disjoint: outcome 3 belongs to both. The observed counts are $n(M) = 168+154+181 = 503$ and $n(N) = 181+167+160 = 508$. Their union is $M \cup N = \{1,2,3,4,5\}$ with count $n = 168+154+181+167+160 = 830$, giving $f(M \cup N) = 0.830$. However, $f(M) + f(N) = 0.503 + 0.508 = 1.011 \ne 0.830$. The discrepancy is $0.181$, the frequency of outcome 3, which was counted once in $n(M)$, once in $n(N)$, but only once in $n(M \cup N)$. The inclusion-exclusion formula corrects this: $f(M \cup N) = f(M) + f(N) - f(M \cap N) = 0.503 + 0.508 - 0.181 = 0.830$.

The six elementary outcomes $\{1\}, \{2\}, \ldots, \{6\}$ are mutually disjoint and their union is $\Omega$. Therefore: $f(\{1\}) + f(\{2\}) + \cdots + f(\{6\}) = (168+154+181+167+160+170)/1000 = 1000/1000 = 1$. This is always exactly 1, not approximately — it is an algebraic identity that follows from the fact that every roll contributes exactly one count to exactly one elementary frequency. This identity is the empirical analogue of the normalization axiom $P(\Omega) = 1$.

The relative frequencies we computed are not the probabilities — they are estimates of the probabilities derived from a finite experiment. The theoretical probability for a fair die is $P(\{k\}) = 1/6 \approx 0.1̇6̇$ for each $k$. Our empirical values ($0.168, 0.154, 0.181, \ldots$) differ from $1/6$ by small amounts. The Law of Large Numbers guarantees that as the number of trials $n \to \infty$, the relative frequency $f(A) = n(A)/n$ converges to $P(A)$ for every event $A$ (under suitable conditions). Probability is therefore the mathematical idealisation that the empirical frequencies approach in the long run.

There is a fundamental difference between a relative frequency and a probability. A frequency is an observed, concrete, finite quantity: it depends on the particular experiment we ran, and two repetitions of the same experiment will give different frequencies. A probability is a theoretical, abstract, timeless quantity: it is a property of the model, not of any particular run. The probability model (i.e., a probability measure $P$ on a measurable space $(\Omega, \mathcal{F})$) is constructed to match the observed frequencies in the limit. Once constructed, it obeys the three Kolmogorov axioms exactly — not approximately. This conceptual leap from the empirical to the theoretical is the foundation of modern probability theory: we take the properties we observe to hold approximately in large samples and elevate them to exact axioms in the abstract model.

A probability space in Kolmogorov's framework is a triple $(\Omega, \mathcal{F}, P)$, where $\Omega$ is the sample space, $\mathcal{F}$ is a $\sigma$-algebra of events (subsets of $\Omega$ that are "measurable"), and $P: \mathcal{F} \to [0,1]$ is a function satisfying three axioms. Axiom 1 (Non-negativity): $P(A) \ge 0$ for every event $A \in \mathcal{F}$. Axiom 2 (Normalisation): $P(\Omega) = 1$. Axiom 3 (Countable additivity): if $A_1, A_2, \ldots$ is a countable sequence of mutually disjoint events, then $P\bigl(\bigcup_{k=1}^{\infty} A_k\bigr) = \sum_{k=1}^{\infty} P(A_k)$. These three axioms, together with the specification of the measurable space $(\Omega, \mathcal{F})$, completely determine the structure of a probability model. Everything else in probability theory — conditional probability, independence, random variables, expectations, characteristic functions — follows from these axioms by pure mathematical deduction.

Axiom 1 is the easiest to motivate empirically. A relative frequency is always a ratio of two non-negative integers: $f(A) = n(A)/n$, where $n(A)$ is the count of outcomes in $A$ (never negative) and $n$ is the total number of trials (strictly positive). Therefore $f(A) \ge 0$ always. If probability is the idealisation of frequency in the limit, it must inherit non-negativity. We cannot have a negative probability any more than we can have a negative count. Axiom 1 thus reflects the most basic intuition that probability measures "how often" something occurs, and frequencies are intrinsically non-negative.

Axiom 2 is equally natural. In every trial of any experiment, exactly one elementary outcome occurs. The entire sample space $\Omega$ is therefore the "certain event" — something in $\Omega$ always happens. Empirically, every die roll produces a face in $\{1,2,3,4,5,6\}$, so $f(\Omega) = n(\Omega)/n = n/n = 1$ exactly, in any finite experiment without exception. The normalisation $P(\Omega) = 1$ is therefore not an approximation: it is an exact identity at every sample size. Axiom 2 simply elevates this identity to the abstract model.

Axiom 3, in its finite form, states that $P(A_1 \cup \cdots \cup A_k) = P(A_1) + \cdots + P(A_k)$ whenever the events are mutually disjoint. As we showed in Problem 5 Part B, this holds exactly for relative frequencies: if $A$ and $B$ are disjoint, then no outcome is counted in both $n(A)$ and $n(B)$, so $n(A \cup B) = n(A) + n(B)$ and thus $f(A \cup B) = f(A) + f(B)$. The same argument extends to any finite collection of disjoint events by induction. The finite version of additivity is therefore forced upon us by the way counting works — it is not an assumption but a theorem about frequencies, elevated to an axiom for the abstract model.

The full Axiom 3 is stronger than finite additivity: it applies to countably infinite sequences of disjoint events. Why do we need this? Consider the experiment of tossing a fair coin infinitely many times. The sample space $\Omega$ is uncountably infinite, and interesting events — like "heads appears infinitely often" — cannot be expressed as finite unions of elementary outcomes. To assign probabilities to such events, we need the measure $P$ to be continuous with respect to monotone limits of event sequences. It turns out that this continuity is exactly equivalent to countable additivity. Without it, we cannot handle limits, cannot define distributions on the real line, cannot state the Law of Large Numbers or the Central Limit Theorem. Countable additivity is the minimal axiom that allows probability theory to interface with analysis.

The axiomatic approach gives probability theory three enormous advantages. First, generality: the same framework covers discrete spaces (coin tosses, die rolls), continuous spaces (Buffon's needle, Gaussian random variables), and abstract spaces (random processes, random graphs). Second, rigour: every probabilistic result can be proved from the axioms, and any apparent paradox must either be resolved within the system or signal a violation of the axioms. Third, unification: conditional probability, expectation, independence, stochastic processes, and statistical inference all emerge from the same three axioms, giving the entire theory a coherent logical structure. Without the axiomatic foundation, probability would remain a collection of ad hoc rules derived from intuition — useful in practice but fragile under scrutiny.

The axiomatic framework does not tell us what the probabilities should be — it only constrains the structure they must have. The assignment of probabilities to events is a modelling choice, external to the axioms. For a fair die we choose $P(\{k\}) = 1/6$; but the axioms do not compel this — they would be equally satisfied by $P(\{1\}) = 1/2$ and $P(\{2\}) = \cdots = P(\{6\}) = 1/10$. Verifying that a model is appropriate requires empirical evidence (as in Problem 5) or physical reasoning. Moreover, the requirement that $\mathcal{F}$ be a $\sigma$-algebra, rather than all subsets of $\Omega$, is a technical necessity: on uncountable spaces, it is mathematically impossible to define a countably additive probability on every subset (Vitali's theorem). The $\sigma$-algebra $\mathcal{F}$ specifies which events are "measurable" and thus have a well-defined probability — a subtlety invisible in finite sample spaces but essential for the general theory.

The intellectual journey of this section can be summarised as follows. We started with concrete experiments — coin tosses, die rolls, weekly weather — and represented their outcomes and events as grids. We computed relative frequencies of events and observed that frequencies are always non-negative, always sum to 1 over $\Omega$, and always satisfy finite additivity for disjoint events. We then asked: what is the minimal set of rules that captures these properties abstractly, extends to infinite sample spaces, and supports rigorous mathematical deductions? The answer is Kolmogorov's three axioms. The grid representation, which made events visible as geometric shapes, is a stepping stone between the intuitive enumeration of outcomes and the formal measure-theoretic definition of a probability space. In the formal theory, the grid is replaced by a $\sigma$-algebra and a measure, but the spirit is the same: an event is a subset of $\Omega$, and its probability is a number assigned to that subset in a consistent, additive way.