Blaise Pascal

Blaise Pascal

Mechanical arithmetic and reasoning about chance · 1623–1662 · France

Pascal's work offers two complementary lessons: an arithmetic rule can be embodied in a mechanism, and an uncertain outcome can be analyzed by a precisely stated model. Neither lesson means that a machine is infallible or that real-world uncertainty disappears once a probability is calculated.

Historical foundations

Pascal was born in Clermont in 1623 and died in Paris in 1662. His early work included projective geometry; his essay on conics was published in 1640, following a presentation in 1639. He developed a calculator in the 1640s in connection with his father's tax work. The familiar childhood geometry story comes from biographical tradition and should not be treated as a controlled record of an independent discovery. MacTutor biography

Surviving Pascalines provide evidence beyond written descriptions. They used input wheels and a carry mechanism, the sautoir. Some were adapted to currency, requiring 12 deniers per sol and 20 sols per livre rather than decimal rollover at every position. Musée des Arts et Métiers object record

The royal privilege for Pascal's calculator dates to 1649, not 1645. The earlier year belongs to his public presentation and explanatory writing. Contemporary promotional claims of freedom from error are claims to evaluate, not laboratory measurements. The exact numbers of prototypes, finished machines, and sales vary across accounts, so this reading does not turn them into a precise production total. Primary machine writings and privilege, Pascal research center on the 1649 privilege

His 1654 correspondence with Pierre de Fermat is a landmark in mathematical probability. It was not the beginning of all reasoning about chance. Pascal's arithmetical triangle also had predecessors. His contribution includes systematic arguments and applications, rather than exclusive ownership of the triangular array. Pascal–Fermat correspondence, Pulskamp translation

Worked mechanism: mixed-radix carry

The following is an original arithmetic model, not an operating manual for an individual Pascaline. Represent money as (livres, sols, deniers), with 0 ≤ deniers < 12 and 0 ≤ sols < 20. Add (3,18,9) and (2,4,7).

1. Deniers: 9+7=16. Store 4 and carry 1 sol.
2. Sols: 18+4+1=23. Store 3 and carry 1 livre.
3. Livres: 3+2+1=6.

The result is (6,3,4). Check by converting to the smallest unit. One livre is 240 deniers, so the inputs are 3×240+18×12+9=945 and 2×240+4×12+7=535. Their sum is 1480, equal to 6×240+3×12+4.

Each carry preserves value because twelve deniers equal one sol and twenty sols equal one livre. A decimal-only algorithm would be wrong for this representation even if every gear movement matched its instructions. Correct execution and correct interpretation are separate requirements.

Pascal's carrying mechanism used falling components rather than simply forcing every higher wheel through a single uninterrupted gear train. The existence of an arithmetic invariant does not guarantee physical reliability: manufacturing, wear, and operation remain empirical questions. Computer History Museum comparison

Worked probability: divide an unfinished game's stake

Assume independent rounds between equally skilled players, with no draws. The first player to three wins receives a pot. At interruption, A has two wins and B has one. A needs one more; B needs two.

If the next round is A's, A wins immediately. If B wins it, the players are tied with one win needed, so each has probability one-half of taking the pot. Therefore A's eventual winning probability is:

(1/2) × 1 + (1/2) × (1/2) = 3/4.

A receives three-quarters of the pot under the stipulated expected-payoff rule. This is a mathematical justification for that division given the assumptions, not a universal definition of fairness.

An equivalent method enumerates two-round strings AA, AB, BA, BB. The first two strings represent the same early victory by A, extended by a hypothetical unused round. A wins in three strings. Padding works because each length-two string has probability one-quarter; stopping early and then treating terminal histories A, BA, BB as equally likely would be wrong. Their probabilities are one-half, one-quarter, and one-quarter.

For unequal round-winning probability p for A, the same state gives p+(1-p)p=2p-p². The answer depends on the model. A table of possible outcomes alone does not establish equal probabilities.

Exercises

1. Add (7,19,11) and (0,0,1) in the currency model. Then find the difference between (4,2,3) and (1,5,8) by borrowing. Verify both answers in deniers.
2. At interruption A needs two wins and B needs three, with independent fair rounds. Find A's share of a 160-unit pot. Explain why counting four-round strings remains valid if the actual game can end earlier.
3. At the original state where A needs one win and B needs two, suppose A wins each round with probability 0.6. Calculate A's eventual win probability. Identify two observations that would make the independent, constant-probability model questionable.

Compare your reasoning with the worked solutions

1. Adding one denier produces 12 deniers, which becomes one sol; the resulting 20 sols become one livre. The sum is (8,0,0), or 1920 deniers. For subtraction, borrow one sol to make 15 deniers, leaving one sol. Then borrow one livre to make 21 sols, leaving three livres. Subtracting gives (2,16,7). Check: the inputs are 987 and 308 deniers; their difference is 679, equal to 480+192+7.
2. By four rounds one player must have reached the required wins. A wins if at least two of the four hypothetical outcomes are A wins. There are C(4,2)+C(4,3)+C(4,4)=6+4+1=11 such strings out of 16. A's share is 110. Extending an early terminal history with all possible unused suffixes preserves its probability; it does not change the actual winner. The suffixes are a counting device, not additional played rounds.
3. A wins immediately with probability 0.6 or loses once and then wins with probability 0.4×0.6=0.24. The total is 0.84. Fatigue changing skill over time would challenge constant probability. Dependence caused by learning, strategic adaptation, or persistent environmental conditions would challenge independence. The calculation can remain internally correct while its assumptions fit the actual game poorly.