Pascal & the Birth of Probability
The triangle, the calculator, the wager, and the mathematics of chance
13.1 Blaise Pascal (1623–1662)
Blaise Pascal was born on June 19, 1623, in Clermont-Ferrand, France. His father, Étienne Pascal, was a tax official and talented amateur mathematician who educated his son at home according to progressive pedagogical principles. Étienne deliberately withheld mathematics from the young Blaise, intending to introduce it only when the boy was older — but the plan backfired spectacularly. At age twelve, Blaise was discovered working out the properties of triangles on his own, having independently proved that the angles of a triangle sum to two right angles (Euclid's Proposition I.32). Étienne abandoned his restrictions and gave the boy full access to mathematical texts.
By sixteen, Pascal had written an essay on conic sections (Essai pour les coniques, 1640) that contained what is now called Pascal's theorem: if a hexagon is inscribed in a conic section, the three points where opposite sides meet are collinear. This result, which Descartes could scarcely believe was the work of a teenager, established Pascal as a serious mathematician.
At nineteen, Pascal designed and built the Pascaline, one of the first mechanical calculating machines. He developed it to help his father with the tedious arithmetic of tax collection. The device used interlocking gears and could perform addition and subtraction of numbers up to six digits. Pascal built about fifty prototypes and sold some commercially, though the machines were expensive and not widely adopted.
Pascal also made important contributions to physics. His experiments on atmospheric pressure (1647–1648) demonstrated that air has weight and that the vacuum exists — contradicting Aristotle's dictum that “nature abhors a vacuum.” The SI unit of pressure, the pascal (Pa), is named in his honor.
In 1654, Pascal underwent a profound religious conversion, joining the Jansenist movement within Catholicism. He largely abandoned mathematics for theology, producing the Lettres provinciales (1656–57), a masterpiece of French prose attacking Jesuit casuistry, and the Pensées, an unfinished apology for Christianity that remains one of the great works of religious philosophy. Pascal died on August 19, 1662, at the age of just thirty-nine, likely from a combination of tuberculosis and stomach cancer.
Pascal: Key Dates
- 1623 — Born in Clermont-Ferrand
- 1640 — Essai pour les coniques (age 16)
- 1642 — Begins building the Pascaline (age 19)
- 1647–48 — Experiments on atmospheric pressure
- 1654 — Traité du triangle arithmétique; correspondence with Fermat on probability; religious conversion
- 1656–57 — Lettres provinciales
- 1658 — Studies the cycloid (his last mathematical work)
- 1662 — Dies in Paris (August 19)
13.2 Pascal's Triangle
In 1654, Pascal wrote the Traité du triangle arithmétique (“Treatise on the Arithmetic Triangle”), a systematic study of the triangular array of numbers now universally known as Pascal's triangle. The triangle had been studied centuries earlier in China, Persia, and India, but Pascal's treatment was the most thorough and rigorous to date, and he was the first to exploit its full power for combinatorics and probability.
Construction of Pascal's Triangle
The triangle is constructed as follows. Row $n$ (starting from $n = 0$) contains the numbers $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$. Each entry is the sum of the two entries directly above it:
$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$
with the convention that $\binom{n}{k} = 0$ if $k < 0$ or $k > n$.
Row 0: 1
Row 1: 1 1
Row 2: 1 2 1
Row 3: 1 3 3 1
Row 4: 1 4 6 4 1
Row 5: 1 5 10 10 5 1
Row 6: 1 6 15 20 15 6 1
Row 7: 1 7 21 35 35 21 7 1
Row 8: 1 8 28 56 70 56 28 8 1
Earlier discoveries. The arithmetic triangle was known long before Pascal:
- China: Yang Hui presented it in 1261, attributing it to Jia Xian (c. 1050). It appears even earlier in Zhu Shijie's Precious Mirror (1303).
- Persia/Iran: Omar Khayyam (c. 1100) knew the binomial coefficients, and al-Samaw'al (c. 1150) displayed the triangle explicitly.
- India: Pingala (c. 200 BCE) described a method equivalent to the triangle for counting poetic meters. Varahamihira (505 CE) and Halayudha (c. 975 CE) gave explicit descriptions.
- Europe: Petrus Apianus printed the triangle in 1527, and Tartaglia presented it in 1556.
Pascal's originality lay not in discovering the triangle but in proving its properties rigorously and connecting it systematically to the binomial theorem, combinatorics, and probability.
The binomial coefficient formula. Each entry in Pascal's triangle is a binomial coefficient:
$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
where $n! = n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1$ is the factorial. This counts the number of ways to choose $k$ objects from $n$ objects, without regard to order.
Computing Binomial Coefficients
$\binom{8}{3} = \frac{8!}{3! \cdot 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$
$\binom{10}{4} = \frac{10!}{4! \cdot 6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5040}{24} = 210$
$\binom{n}{0} = \binom{n}{n} = 1$ for all $n$ (there is exactly one way to choose nothing or everything).
13.3 The Binomial Theorem
The primary algebraic significance of Pascal's triangle is its connection to the binomial theorem, which gives the expansion of $(a + b)^n$ for any non-negative integer $n$.
The Binomial Theorem
For any non-negative integer $n$:
$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$
That is: $(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n$.
Expanding (a + b)^4
Using row 4 of Pascal's triangle: 1, 4, 6, 4, 1:
$$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$
Expanding (2x - 3)^3
Here $a = 2x$, $b = -3$, $n = 3$. Row 3: 1, 3, 3, 1.
$$(2x - 3)^3 = (2x)^3 + 3(2x)^2(-3) + 3(2x)(-3)^2 + (-3)^3$$
$$= 8x^3 - 36x^2 + 54x - 27$$
Computing (1.01)^10 Approximately
Write $1.01 = 1 + 0.01$. By the binomial theorem:
$$(1 + 0.01)^{10} = \sum_{k=0}^{10} \binom{10}{k} (0.01)^k$$
The first few terms dominate:
$$= 1 + 10(0.01) + 45(0.0001) + 120(0.000001) + \cdots$$
$$= 1 + 0.1 + 0.0045 + 0.000120 + \cdots \approx 1.10462$$
The exact value is $1.01^{10} = 1.10462212...$, confirming excellent accuracy with just four terms.
Why it works. When we expand $(a + b)^n = (a+b)(a+b)\cdots(a+b)$($n$ factors), each term in the expansion is obtained by choosing either $a$ or $b$ from each of the $n$ factors. The term $a^{n-k}b^k$ arises from choosing $b$ from exactly $k$ of the $n$ factors (and $a$ from the remaining $n - k$). The number of ways to make this choice is $\binom{n}{k}$, which is why the coefficient of $a^{n-k}b^k$ is $\binom{n}{k}$.
13.4 Properties of Binomial Coefficients
Pascal's triangle is a treasure trove of identities. Here are the most important ones:
Row Sum
The sum of all entries in row $n$ is $2^n$:
$$\sum_{k=0}^{n} \binom{n}{k} = 2^n$$
Proof: Set $a = b = 1$ in the binomial theorem: $(1+1)^n = \sum_{k=0}^n \binom{n}{k} \cdot 1^{n-k} \cdot 1^k = 2^n$. $\square$
Alternating Sum
The alternating sum of entries in row $n$ is zero (for $n \ge 1$):
$$\sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0$$
Proof: Set $a = 1, b = -1$ in the binomial theorem: $(1 + (-1))^n = 0^n = 0$. $\square$
Verification for n = 4
Row 4: $1, 4, 6, 4, 1$.
Row sum: $1 + 4 + 6 + 4 + 1 = 16 = 2^4$. Correct!
Alternating sum: $1 - 4 + 6 - 4 + 1 = 0$. Correct!
Symmetry
$$\binom{n}{k} = \binom{n}{n-k}$$
Choosing $k$ items to include is the same as choosing $n - k$ items to exclude.
Hockey Stick Identity
The sum of entries along a diagonal equals the entry at the “bottom” of the hockey stick:
$$\sum_{i=0}^{r} \binom{n+i}{i} = \binom{n+r+1}{r}$$
Equivalently: $\binom{n}{n} + \binom{n+1}{n} + \binom{n+2}{n} + \cdots + \binom{n+r}{n} = \binom{n+r+1}{n+1}$.
Hockey Stick Example
Take $n = 2, r = 3$:
$$\binom{2}{0} + \binom{3}{1} + \binom{4}{2} + \binom{5}{3} = 1 + 3 + 6 + 10 = 20 = \binom{6}{3}$$
Confirmed!
Vandermonde's Identity
$$\binom{m+n}{r} = \sum_{k=0}^{r} \binom{m}{k}\binom{n}{r-k}$$
Combinatorial proof: Consider choosing $r$ objects from a group of $m + n$ objects (split into two pools of $m$ and $n$). If we take $k$ from the first pool and $r - k$ from the second, the number of ways is $\binom{m}{k}\binom{n}{r-k}$. Summing over all possible $k$ gives the total.
Vandermonde's Identity: Numerical Check
Let $m = 3, n = 4, r = 3$:
$$\binom{7}{3} = \binom{3}{0}\binom{4}{3} + \binom{3}{1}\binom{4}{2} + \binom{3}{2}\binom{4}{1} + \binom{3}{3}\binom{4}{0}$$
$$35 = 1 \cdot 4 + 3 \cdot 6 + 3 \cdot 4 + 1 \cdot 1 = 4 + 18 + 12 + 1 = 35 \; \checkmark$$
13.5 Mathematical Induction
One of Pascal's most important contributions to mathematics was his explicit formulation and use of the principle of mathematical induction as a proof technique. While the idea of proving a statement for all natural numbers by establishing a base case and an inductive step had appeared implicitly in earlier work (notably by al-Karaji and Francesco Maurolico), Pascal was the first to state and apply the method in a fully conscious, systematic way.
The Principle of Mathematical Induction
To prove a statement $P(n)$ for all natural numbers $n \ge n_0$:
- Base case: Prove $P(n_0)$ is true.
- Inductive step: Prove that if $P(k)$ is true for some $k \ge n_0$, then $P(k+1)$ is also true.
These two steps together establish $P(n)$ for all $n \ge n_0$.
Pascal used induction to prove properties of the arithmetic triangle. Let us demonstrate the method with a classic example.
Sum of the First n Positive Integers
Claim: For all $n \ge 1$:
$$\sum_{i=1}^{n} i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$$
Proof by induction.
Base case ($n = 1$): The left side is $1$. The right side is $\frac{1 \cdot 2}{2} = 1$. They agree. $\checkmark$
Inductive step: Assume the formula holds for some $k \ge 1$:
$$1 + 2 + \cdots + k = \frac{k(k+1)}{2} \quad \text{(inductive hypothesis)}$$
We must show it holds for $k + 1$:
$$1 + 2 + \cdots + k + (k+1) = \frac{k(k+1)}{2} + (k+1) = \frac{k(k+1) + 2(k+1)}{2} = \frac{(k+1)(k+2)}{2}$$
This is exactly $\frac{(k+1)((k+1)+1)}{2}$, which is the formula with $n = k + 1$. $\square$
Induction: Sum of Squares
Claim: $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$.
Base case ($n = 1$): $1 = \frac{1 \cdot 2 \cdot 3}{6} = 1$. $\checkmark$
Inductive step: Assume true for $k$. Then:
$$\sum_{i=1}^{k+1} i^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}$$
$$= \frac{(k+1)[k(2k+1) + 6(k+1)]}{6} = \frac{(k+1)(2k^2 + 7k + 6)}{6} = \frac{(k+1)(k+2)(2k+3)}{6}$$
This is the formula with $n = k + 1$. $\square$
13.6 The Fermat–Pascal Correspondence (1654)
The founding moment of mathematical probability occurred in the summer of 1654, when Pascal and Fermat exchanged a series of letters about gambling problems posed by the Chevalier de Méré, a French nobleman and avid gambler.
The central problem was the Problem of Points (also called the “problem of the unfinished game”): two players of equal skill are playing a game of chance that consists of a series of rounds. The first player to win a certain number of rounds wins the entire stake. If the game is interrupted before either player has won, how should the stake be divided fairly?
The Problem of Points: A Specific Case
Two players, A and B, each wager 32 pistoles. The first to win 3 rounds takes the full 64 pistoles. The game is interrupted when A has won 2 rounds and B has won 1 round. How should the 64 pistoles be divided?
Pascal's reasoning: At most 2 more rounds need to be played. The possible outcomes of these 2 rounds are:
- AA: A wins (A has 3 wins)
- AB: A wins (A has 3 wins)
- BA: A wins (A has 3 wins)
- BB: B wins (B has 3 wins)
Of the 4 equally likely outcomes, A wins in 3 and B wins in 1. So A should receive $\frac{3}{4} \times 64 = 48$ pistoles and B should receive $\frac{1}{4} \times 64 = 16$ pistoles.
The general solution. If A needs $a$ more wins and B needs $b$ more wins, the game will be decided in at most $a + b - 1$more rounds. The probability that A wins is:
$$P(A \text{ wins}) = \frac{1}{2^{a+b-1}} \sum_{k=a}^{a+b-1} \binom{a+b-1}{k}$$
This is because A wins if, in $a + b - 1$ rounds (each equally likely to go either way), A wins at least $a$ of them.
Fermat solved the same problem by a different method: directly enumerating all possible sequences of outcomes. Both methods gave the same answers, and the agreement between these two independent approaches gave both mathematicians confidence that they had found the correct theory.
The Fermat–Pascal correspondence is rightly celebrated as the birth of probability theory. For the first time, chance events were treated as objects of rigorous mathematical analysis. The key innovation was the idea that even though individual outcomes are uncertain, their proportions in the long run are predictable and calculable.
13.7 Expected Value
The solution to the Problem of Points implicitly uses the concept of expected value (or mathematical expectation) — the weighted average of all possible outcomes, where each outcome is weighted by its probability.
Expected Value
If a random variable $X$ takes values $x_1, x_2, \ldots, x_n$ with probabilities $p_1, p_2, \ldots, p_n$ respectively, the expected value (or expectation) of $X$ is:
$$E[X] = \sum_{i=1}^{n} x_i p_i = x_1 p_1 + x_2 p_2 + \cdots + x_n p_n$$
The expected value represents the “fair price” of a gamble: if you played the game many times, your average payoff per game would approach $E[X]$.
Expected Value: Fair Die
A fair die has outcomes 1, 2, 3, 4, 5, 6, each with probability $\frac{1}{6}$.
$$E[X] = \frac{1}{6}(1 + 2 + 3 + 4 + 5 + 6) = \frac{21}{6} = 3.5$$
Note that 3.5 is not a possible outcome — the expected value need not be an achievable result.
Expected Value from the Fermat-Pascal Correspondence
In the Problem of Points example (A needs 1 more win, B needs 2), suppose the total stake is $S$. Player A's expected winnings are:
$$E[\text{A's share}] = S \cdot P(A \text{ wins}) + 0 \cdot P(A \text{ loses}) = S \cdot \frac{3}{4} = \frac{3S}{4}$$
This is the “fair division” that Pascal computed: A should receive three-quarters of the stake.
Properties of expected value. Expected value has several important properties that make it a powerful tool:
Properties of Expectation
For any random variables $X$ and $Y$ and constants $a, b$:
- Linearity: $E[aX + bY] = aE[X] + bE[Y]$
- Constant: $E[c] = c$ for any constant $c$
- Non-negativity: If $X \ge 0$ always, then $E[X] \ge 0$
- Independence: If $X$ and $Y$ are independent, then $E[XY] = E[X] \cdot E[Y]$
The linearity property is remarkable: it holds even when $X$ and $Y$ are not independent.
Linearity: Expected Sum of Two Dice
Let $X$ and $Y$ be the results of two dice rolls. What is $E[X + Y]$?
By linearity: $E[X + Y] = E[X] + E[Y] = 3.5 + 3.5 = 7$.
This is why 7 is the most common sum when rolling two dice — it is the expected value.
13.8 Pascal's Wager
Pascal applied his newly developed probability concepts to the most profound question he could imagine: whether to believe in God. The argument, found in the Pensées (fragment 233), is the earliest known application of decision theory and expected utility to a philosophical problem.
Pascal's reasoning proceeds as follows. Either God exists or God does not exist. You must choose whether to believe or not believe (you cannot avoid the choice — “you are already embarked”). Consider the payoffs:
| God exists | God does not exist | |
|---|---|---|
| Believe | Infinite gain (eternal bliss) | Finite loss (constrained life) |
| Don't believe | Infinite loss (damnation) | Finite gain (unconstrained life) |
Let $p$ be the probability that God exists (which Pascal argues cannot be zero, since reason alone cannot settle the question). The expected utility of believing is:
$$E[\text{Believe}] = p \cdot (+\infty) + (1-p) \cdot (-c) = +\infty$$
where $c$ is a finite cost. The expected utility of not believing is:
$$E[\text{Don't believe}] = p \cdot (-\infty) + (1-p) \cdot (+c') = -\infty$$
Since $+\infty > -\infty$, the rational choice is always to believe, regardless of how small $p$ is (as long as $p > 0$).
Pascal's Wager has been debated by philosophers and theologians for centuries. Common objections include: the “many gods” problem (which god should one believe in?), the question of whether belief can be chosen voluntarily, and doubts about the coherence of infinite utilities. But as a mathematical argument, the Wager is historically significant as the first systematic application of expected value to decision-making under uncertainty.
13.9 Christiaan Huygens
Christiaan Huygens (1629–1695) was a Dutch mathematician, physicist, and astronomer — one of the foremost scientists of the seventeenth century. He invented the pendulum clock, discovered the rings of Saturn and its moon Titan, formulated the wave theory of light, and made fundamental contributions to mechanics and optics.
Huygens learned of the Fermat–Pascal correspondence during a visit to Paris in 1655 and was immediately fascinated. In 1657, he published De Ratiociniis in Ludo Aleae (“On Reasoning in Games of Chance”), the first published textbook on probability theory. Though it was a slender work of just sixteen pages, it established probability as a legitimate branch of mathematics.
Huygens's central concept was mathematical expectation(which he called the “value of a chance”). He formulated it as follows: if a player has an equal chance of winning amounts $a$ or $b$, the “value” of the game to that player is $\frac{a + b}{2}$. More generally, if the player can win $a_1, a_2, \ldots, a_n$ with equal probability, the value is:
$$\text{Value} = \frac{a_1 + a_2 + \cdots + a_n}{n}$$
Huygens solved fourteen problems in his treatise, including several that had been posed by Pascal and Fermat. He also posed five additional problems as challenges to his readers. The work was widely read and influential, serving as the standard reference on probability until the publication of Jacob Bernoulli's Ars Conjectandi in 1713.
Huygens's Problem 1
I have 3 chances of winning 13 crowns and 2 chances of winning nothing. What is the value of my expectation?
$$E = \frac{3 \times 13 + 2 \times 0}{5} = \frac{39}{5} = 7.8 \text{ crowns}$$
This is the “fair price” one should be willing to pay to play such a game.
Huygens and the Gambler's Ruin. Among the more advanced problems Huygens considered was the “Gambler's Ruin”: two players with unequal fortunes play a series of fair games until one is bankrupt. What is the probability each player wins? If player A starts with $a$ coins and player B with $b$ coins, the probability that A wins all is:
$$P(\text{A wins}) = \frac{a}{a + b}$$
This elegant result shows that in a fair game, each player's probability of winning is proportional to their initial fortune.
13.10 Legacy
The work of Pascal, Fermat, and Huygens in the 1650s created a new branch of mathematics with extraordinary practical and theoretical consequences:
The birth of probability theory. Before 1654, chance was considered beyond the reach of reason. Gamblers relied on intuition and experience; philosophers treated randomness as an expression of divine will or cosmic caprice. The Fermat–Pascal correspondence showed that random events could be analyzed with the same precision as geometric figures or algebraic equations.
Decision theory. Pascal's Wager was the first attempt to formalize decision-making under uncertainty using mathematical expectation. This idea would evolve into modern decision theory, game theory (von Neumann and Morgenstern, 1944), and behavioral economics.
Insurance and actuarial science. The concept of expected value made it possible to price risk rationally. The first life insurance tables (John Graunt, 1662; Edmund Halley, 1693) and marine insurance calculations were direct applications of the probability theory founded by Pascal and Fermat.
Statistics. Probability theory is the mathematical foundation of statistics. The progression from Pascal and Fermat through Bernoulli (law of large numbers), de Moivre and Laplace (normal distribution, central limit theorem), to modern statistical inference all traces back to the 1654 correspondence.
Combinatorics. Pascal's systematic study of binomial coefficients, his use of mathematical induction, and his connection of the arithmetic triangle to counting problems established combinatorics as a mathematical discipline in its own right.
The Chain of Probability Theory Development
- 1654 — Fermat–Pascal correspondence (founding of probability)
- 1657 — Huygens, De Ratiociniis (first textbook)
- 1713 — Jacob Bernoulli, Ars Conjectandi (law of large numbers)
- 1733 — De Moivre, normal approximation to the binomial
- 1812 — Laplace, Théorie analytique des probabilités
- 1933 — Kolmogorov, axiomatic foundations of probability
Pascal's own life embodied a tension between the mathematical and the mystical, between the rational analysis of chance and the passionate embrace of faith. His mathematical contributions, though made in a brief burst of activity before his religious conversion, were sufficient to change the world. In the words of the mathematician Pierre-Simon Laplace: “The theory of probabilities is at bottom nothing but common sense reduced to calculus; it makes us appreciate with exactitude that which reasonable minds feel by a sort of instinct.”