Rediscovering 1657: The Starting Point of Huygens and the Dawn of Modern Probability

A Partial New Translation & Commentary on De Ratiociniis in Ludo Aleae

Pinlin [Calvin] Xu, Nov 2023

Foreword

Growing up fascinated by astronomy, I knew Christiaan Huygens for discovering Titan and Saturn’s rings and for his inspiring quotes in Carl Sagan’s Cosmos. It turns out Huygens was also a talented physicist and engineer, inventor of the pendulum clock, and even the author of the first printed text on probability, De Ratiociniis in Ludo Aleae (On Reckoning in the Game of Chance). The 14-pape treatise came as an appendix to the 1657 Exercitationum Mathematicarum Libri Quinque, the 4th book in a series by his former professor at the University of Leiden, Frans van Schooten (who is most known for popularizing René Descartes’ analytical geometry).

The dice game problem Huygens examines in the text is known as the problem of points, about fairly dividing the prize pool amongst several players before anyone has actually won. Famous names in mathematics such as Tartaglia and Cardano had all worked on it, and in 1654 Fermat and Pascal exchanged a series of letters discussing the problem, which Huygens heard about and became a basis for his investigation. In total Huygens wrote 14 propositions and 5 problems for the reader, using modern algebraic notation developed by Viète and Descartes that makes the text surprisingly accessible today. Jacob Bernoulli, of Bernoulli random variable fame we learn in CS 109, would incorporate Huygens' work and his commentary in the first part of his opus magnum on probability Ars Coniectandi (Art of Guessing) published in 1713.

Below is a new literal translation and commentary by me for fellow students. Let’s see where it all began:

Misc notes for course staff

Thank you for reading this!

As I am also enrolled in CS 221 this quarter (which also has an optional final project!), I end up feeling like coding up something for the CS 109 challenge inevitably becomes implementing something from CS 221 slides. As such I decided to go with the history person inside me and write a commentary.

The following text consists of a new translation by me, with the help of all the dictionaries and ChatGPT. I have aimed the style to be literal (preserving sentence structures) but modern, as the 1714 Keimer and Woodward translation is very difficult to read despite being more idiomatic.

While trying to preserving the syntax of the Latin original makes it less readable, I feel it is necessary for me to minimize the risk of misinterpreting what Huygens means (in the way scripture is sometimes translated), especially as his definitions are very different from modern ones & he uses words like expect and expectation quite loosely. I paraphrase and summarize in my commentary whenever needed.

I also think it is kind of cool that a literal translation preserves exactly the way and style the original. My translation really makes you feel like you can read Latin & appreciate the distinct way early modern scientific texts were written in (just like academic writing in English now has a certain style).

Original Latin text I transcribed with OCR and some miscellaneous notes are kept in this document inside inline details HTML5 tags. I wish to provide a checked transcription of the text in accordance to modern Latin orthographic conventions for my reference (and hopefully for others’).

This document was originally typeset in MarkDown such that it can be easily converted html and to other formats using pandoc.

(This turns out to be a poor decision, as recursively rendering Markdown inside embedded HTML is only well-supported by Github Flavored Markdown spec, and Github’s renderer has so many bugs! I never thought that would be the case; the final html artifact will likely have collapsible sections inserted by hand)

(This took most of my productive time during Thanksgiving Break. I transcribed and translated the book, and could not fall asleep thinking what Huygens could possibly mean.)

CHRISTIANI HUGENII, DE RATIOCINIIS IN LUDO ALEAE

Even if the outcomes of games, which chance alone governs, tend to be uncertain, nevertheless in these, how much closer one is to winning than to losing, always has a certain determination. As, if someone manages to throw a three with a single die on the first throw, it's uncertain indeed whether he will win; but how much more likely it is that he will lose rather than win is in fact determined, and found out through calculation. Thus likewise, if I compete with someone in this manner, that the victory consists of three games, and I have already won one game, it is still uncertain which of us will emerge as the first victor of the third [i.e., first to win the third game]. But how much my expectation (expectatio), & on the other hand how much his, should be valued, it is appropriate to obtain with very certain reasoning, and hence to define, if it were agreeable for us to leave the game as it is unfinished, how much larger a portion of what has been staked should be allotted to me than to my opponent; or even if someone wishes to take over my place and lot, what price it would be fair for me to sell it to him for. And hence countless questions can arise among two, three, or more players. And since such computation is by no means common[ly known], & is often usefully applied, I will briefly set forth here by what reasoning or method it should be carried out, and then also explain what properly pertains to the game of dice.

Latina

Etsi lusionum, quas sola fors moderatur, incerti solent esse eventus, attamen in his, quanto quis ad vincendum quam perdendum propior sit, certam semper habet determinationem. Ut si quis primo iactu una tessera se ternarium iactere contendat, incertum quidem an vincet; at quanto verisimilius sit eum perdere quam vincere, reipsa definitum est, calculoque subducitur. Ita quoque, si cum aliquo certem hac ratione, ut ternis lusibus constet victoria, atque ego iam unum lusum vicerim, incertum adhuc uter nostrum prior tertii victor sit evasurus. Verum quanti expectatio mea, & contra quanti illius, aestimari debet, certissimo rationcinio consequi licet, atque hinc definire, si ludum uti est imperfectum linquere inter nos conveniret, quanto maior portio eius quod depositum est mihi quam adversario meo tribuenda esset; vel etiam si quis in locum sortemque meam succedere cupiat, quo pretio me eam ipsi vendere aequum sit. Atque hinc innumerabiles quaestiones exoriri possunt inter duos, tres, pluresve collusores. Cumque minime vulgaris sit huiusmodi supputatio, & saepe utiliter adhibeatur, breviter hic qua ratione aut methodo expedienda sit exponam, ac deinde etiam, quae ad aleam sive tesserias proprie pertinent, explicabo.

The opening paragraph summarizes what the problem of points is: in a dice game, given the current game state (history of rolls for each player), can you calculate how likely a player is going to win? This has the practical application of dividing the prize pool if a game cannot be finished. Huygens gives the inspiring and unequivocal answer yes, that this abstract notion should be quantifiable and derivable. He also feels the need to justify the novelty of his research and why gambling is worth researching.

You would only figure out the general rules of the dice game as you keep reading; Huygens probably expects the reader to already know them (I guess it was popular back then). Spoilers: each player pays an entry fee that gets added to the prize pool (the stake). After many rolls of dice, a winner is determined however the specific rules are agreed upon. They would get all the money in the prize pool but would need to pay some pre-negotiated amount to each of the other players who lost. We will model the game a bit later.

History & language notes

Huygens is quite consistent in using alea to refer to the game of dice, and tessera to refer to the physical die (six-sided as we know today; long, four-sided ones were known as talus, also meaning the ankle bone which they were made from). I have considered Caesar’s famous line “The die is cast” (Alea iactus est), reported by Plutarch and Suetonius, to be attestation that alea refers to a physical object that can be thrown. However Huygens’ usage seems very in line with Classical Latin upon further examination. Thus Caesar’s words have also been translated as “Let the game be ventured.”

Moreover I shall use this foundation on both sides: evidently, in a game of chance, the lot (sors) or expectation of each person towards obtaining something should be valued as much as if he has it, he might finally arrive at a similar lot or expectation, competing under equal conditions. For example, if someone, without my knowledge, hides 3 coins in one hand, 7 coins in the other, and gives me the choice of which hand I would prefer to take the coins from; I say that this is of the same value to me as if 5 coins were given to me. Because having five coins, I can again reach the point where I obtain the same expectation for obtaining 3 or 7 coins: and this [is also] competing in an equal game.

Latina

Hoc autem utrobique utar fundamento: nimirum, in aleae ludo tanti aestimandam esse cuiusque sortem seu expectationem ad aliquid obtinendum, quantum si habeat, possit denique ad similem sortem sive expectationem pervenire, aqua conditione certans. Ut, exempli gratia, si quis me inscio altera manu 3 solidos occultet, altera 7 solidos, mihi que optionem det ex utra manu solidos accipere malim; hoc tantundem mihi valere dico, ac si 5 solidi mihi dentur. Quoniam quinque solidos habens, denuo eo pervenire possum, ut aequam expectationem nanciscar ad 3 vel 7 solidos obtinendos: idque aequo lusu contendens.

This is really a postulate. Note how Huygens’ definition of expectation is completely different from ours in CS 109! Recall that we define the expected value / expectation of a random variable to be the average of all the values the random variable can take on, weighted by their respective probabilities. Suppose $X$$X$ is the payoff of a player, our modern definition for the expectation of $X$$X$ is:

$E[X]=\sum _{x}x\cdot P(X=x)$$E[X] = \sum\limits_x x \cdot P(X = x)$

Huygens, on the other hand, makes claim about the value of the expectation. He essentially says that, given a game state consisting of different payoff outcomes that have different probabilities of happening (the “lot”), the intuition for the expected payoff of the game state (how much it is worth) is:

1. suppose we construct another game where each player pays the same amount to the stake and has the same chance of winning (“fair game”, justus lusus, as he calls it)
2. if that game can replicate the exact same game state
3. the entry fee to that game is the expectation of the game state.

This definition is somewhat recursive. It makes sense when you consider the goal of the problem of points, which is to fairly split the prize pot for a game that has not terminated. Huygens thinks that it should be obvious the that the fair expected value for a non-terminal game state is one that can get you to an identical state in another fair game if you want to. Huygens does not say whether the original game you played was fair, just that you get a game state from it—which is really a random variable in disguise, as you have all the values the payoff can take on, and the probability of it happening.

I must stress that Huygens means “the same” whenever he writes “similar” (similis) in this text. I choose to keep it this way because it is his conscious usage. He uses “the same thing” idem and “equal” (aequus, aequalis) throughout, and I take it that he uses “similar” to equate things that are not easily quantified (i.e., custom types if you think programming) like game states.

Later I will show why Huygens’ expectation is compatible with our modern definition, but let us first read how he did it. We notice he did not actually construct the fair game in the example above where he claims the expectation is 5 for two equally likely outcomes, 3 and 7, just claiming it exists. He will do that in the proposition below:

History & language notes

The coin Huygens refers to is a solidus, a late Roman gold coin weighing about 4.5 grams. English "soldier" came from solidus, the currency soldiers were paid in. 10 actual solidi would be pretty close to a soldier's annual salary in the Eastern Roman Empire in the 10th century.

It is not very clear what Huygens means by using the foundation on both sides, but I think he means he would both show how each result follows from the intuition in a walkthrough kind of word proof, and also prove the result algebraically (and demonstrate a numerical example). The 1714 translation simply omits this part and says “AS a Foundation to the following Proposition, I shall take Leave to lay down this Self-evident Truth”.

PROPOSITIO I. (expected value of 2 equally likely outcomes)

Si a vel b expectem, quorum utrumvis aeque facile mihi obtinere possit, expectatio mea dicenda est valere $\frac{a+b}{2}$$\frac{a+b}{2}$.

If I expect a or b, both or which I can obtain equally easily, my expectation should be said to be valued at $\frac{a+b}{2}$$\frac{a+b}{2}$.

This result follows from the definition of expectation as we learned in CS 109. But Huygens kind of did not define expectation the way CS 109 did; he gave an intuition claiming what it should be, and he needs to show his intuition is correct (or at least consistent). The following proofs all try to show the same thing: there exists a fair game where each player pays the expected amount in the proposition to enter, and one player will encounter the same outcomes with the same probabilities in the proposition.

To not only demonstrate but also derive this rule from first principles, having put $x$$x$ for that which equals my expectation, I must be able, when I have $x$$x$, to again arrive at a similar lot, competing under equal conditions. Let the game therefore be such that I compete with another under this condition, that each person stakes $x$$x$, and the winner will give $a$$a$ to the loser. This game is also fair, and it is clear that in this way I have an equal chance of obtaining $a$$a$ if I lose the game; or $2x-a$$2x - a$ if I win: for then I obtain $2x$$2x$, namely what was staked, from which I must pay a to the other. But if $2x-a$$2x−a$ were worth as much as $b$$b$, I would have an equal chance of obtaining $a$$a$ as of $b$$b$. I therefore put $2x-a=b$$2x−a = b$, and fix $x=\frac{a+b}{2}$$x = \frac{a + b}{2}$, for the value of my expectation. The demonstration of this is easy. Indeed, having $\frac{a+b}{2}$$\frac{a + b}{2}$, I can compete with another who also wishes to stake $\frac{a+b}{2}$$\frac{a + b}{2}$, under the condition that the winner will give a to the loser. By such reasoning, I obtain a similar expectation to obtain $a$$a$ if I lose, or to obtain $b$$b$ if I win; for then I obtain $a+b$$a+b$, namely what was staked, and from it concede a to the other.

Latina

Ad hanc regulam non solum demonstrandam, verum etiam primitus eruendam posito $x$$x$ pro eo quod aequivalet expectationi meae, oportet me, quum x habeo, rursus ad similem sortem pervenire posse, aequa conditione certantem. Ponatur itaque lusus esse talis, ut cum altero certem hac conditione, ut quisque deponat x, ac ut victor victo traditurus sit a. Hic autem lusus justus est, & patet me hac ratione aequam habere sortem ad obtinendum $a$$a$, si lusum perdam scilicet; aut $2x-a$$2x−a$, si vincam: tum enim obtineo $2x$$2x$, id nempe quod depositum est, de quo alteri erogandum est a. Quod si autem $2x-a$$2x−a$ tantundem valeret atque $b$$b$, aequa mihi fors obtinere $a$$a$ ad quam ad $b$$b$. Pono itaque $2x-a=b$$2x−a = b$, & fixo $x=\frac{a+b}{2}$$x = \frac{a + b}{2}$, pro valore meae expectationis. Cuius demonstratio facilis est. Etenim habens $\frac{a+b}{2}$$\frac{a + b}{2}$ possum cum alio certare, qui etiam $\frac{a+b}{2}$$\frac{a + b}{2}$ deponere volet, hac conditione ut vincens victo sit traditurus a. Qua ratione similiis expectatio mihi obtinere ad obtinendum a, si perdam, aut ad obtinendum b, si vincam; tum enim obtineo $a+b$$a + b$, id nempe quod depositum est, alterique inde concedo a.

Huygens’ “symbolic” proof is done with algebra that is entirely familiar to us today. Though somewhat confusingly to us, when he states obtaining $a$$a$ and obtaining $b$$b$ are equally likely outcomes, he says there is “a similar expectation (similiis expectatio) to obtain $a$$a$ if I lose, or to obtain $b$$b$ if I win”. Here expectation really means probability to us. Huygens really uses the word quite loosely in different contexts.

He then gives a numerical demonstration that importantly constructs the fair game that has an entry fee of 5 and a game state where there is equal likelihood to gain 3 or 7 coins, which is missing from his postulate:

In numbers. If by equal chance I obtain either 3 or 7, then my expectation according to this Proposition is worth 5; & it is certain that having 5, I can again arrive at the same expectation. For if, competing with another, I stake, and he similarly stakes 5, under the condition that the winner is to give 3 to the other: this game will be entirely fair, & it is clear to me to have an equal chance of obtaining 3, if I lose, or 7, if I win: since then I obtain 10, of which I concede 3 to the other.

Latina

In numeris. Si ad 3 vel 7 aequa fors mihi obtinat, tum expectation mea per hanc Propositionem valet 5; & certum est me 5 habentem rursus ad eandem expectationem pervenire posse. Si enim cum alio certans deponam, atque ille similiter 5 deponat, hac conditione, ut, qui vincit, alteri sit daturus 3: erit hic lusus omnino justus, & patet mihi aequam obtinere sortem ad obtinendum 3, si perdam, aut 7, si vincam: quoniam tunc obtineo 10, de quo alteri concedo 3.

Hopefully this makes what Huygens claims clearer. Let us quickly go over the next two propositions that follow from the modern definition expectation but require proofs in his formulation:

PROPOSITIO II. (expected value of 3 equally likely outcomes)

Si $a,b$$a, b$, vel $c$$c$ expectem, quorum unumquodque pari facilitate mihi obtinere possit, expectatio mea aestimanda est $\frac{a+b+c}{3}$$\frac{a + b + c}{3}$.

If I expect $a,b$$a, b$, or $c$$c$, each one of which I can obtain with equal ease, my expectation should be estimated at $\frac{a+b+c}{3}$$\frac{a + b + c}{3}$.

This proposition involves 3 possible outcomes, which Huygens derives by constructing a 3-player game. Recall that in the dice game one can still only win or lose. Huygens creates more outcomes with different payoffs by constructing a game where one makes separate deals with different (groups of) players, mutually promising to give the losing party varying amounts in each deal, such that the number of outcomes will be the number of deals (if you lose) plus one (if you win; you get the prize pool and pay a constant total amount to all losing parties).

To find this again, let it be supposed, as before, that $x$$x$ stands for the value of my expectation. Therefore, when I have $x$$x$, I must be able to arrive at the same expectation by a fair game. Let the game be such that I play with two others under the condition that each of the three of us stakes $x$$x$; & that I make this agreement with one, if he himself emerges the victor, he will give me $b$$b$, & I will give him $b$$b$, if the same happens to me. But with the other, I enter into this condition, that if he wins the game, he will give me $c$$c$, or I will give him $c$$c$, if I win. And it is clear that this game is fair. Moreover by this reasoning, I have an equal chance of obtaining $b$$b$ if indeed the first wins, or $c$$c$ if the second wins, or even $3x-b-c$$3x - b - c$ if I win; for then I obtain $3x$$3x$, which is staked, from which I give $b$$b$ to one, & $c$$c$ to the other. But if $3x-b-c$$3x−b−c$ were equal to $a$$a$ itself, the same expectation would befall me for obtaining $a$$a$ as for $b$$b$ or $c$$c$. Therefore I set $3x-b-c=a$$3x - b - c = a$, & fix $x=\frac{a+b+c}{3}$$x = \frac{a + b + c}{3}$, for the value of my expectation. In the same way it is found, if by equal chance I may obtain $a,b,c$$a, b, c$, or $d$$d$, it is of such value as $\frac{a+b+c+d}{4}$$\frac{a + b + c + d}{4}$. And so forth.

Latina

Ad quod rursus inveniendum, ponatur, ut ante, $x$$x$ pro valore expectationis meae. Oportet ergo me, cum $x$$x$ habeo, ad eandem expectationem pervenire posse justo lusu. Ponatur lusus esse talis, ut cum duobus aliis ludam hac conditione, ut quisque nostrum trium deponat $x$$x$; & ut cum uno hoc pactum aggrediar, si ipse victor evadat, mihi sit daturus $b$$b$, & ego ipsi traditurus sim $b$$b$, si idem mihi obtinat. Cum altero autem hanc ineam conditionem, ut ille ludum vincens mihi traditurus sit $c$$c$, aut ego ipsi sim daturus $c$$c$, si ego vincam. Et patet hunc ludum justum esse. Aequam autem hac ratione sortem habeo ad obtinendum $b$$b$, si nimirum primus vincat, aut $c$$c$, si secundus vincat, aut etiam $3x-b-c$$3x - b - c$ si ego vincam; tunc enim obtineo $3x$$3x$, quod depositum est, de quo uni concedo $b$$b$, & alteri $c$$c$. Quod si $3x-b-c$$3x - b - c$ aequale fuerit ipsi $a$$a$, eadem mihi obtingeret expectatio ad obtinendum $a$$a$, quae ad $b$$b$, aut ad $c$$c$. Pono itaque $3x-b-c$$3x - b - c$ ob $a$$a$, & fixo $x=\frac{a+b+c}{3}$$x = \frac{a + b + c}{3}$, pro valore meae expectationis. Eodem modo invenitur, si ad $a,b,c,$$a, b, c,$ aut $d$$d$ aequa fors mihi obtingat, id tanti valoris esse, quanti $\frac{a+b+c+d}{4}$$\frac{a + b + c + d}{4}$. Atque ita porro.

I just want to say it is kind of cute he says “or even” (aut etiam) getting $3x-b-c$$3x - b - c$ if he wins. At the end he claims that this result can be extended to 4 outcomes “and so forth”, implying that it is a general solution for the expectation of equally likely outcomes, which we know it is.

The next and final proposition on expectation will really stretch the construction proof idea:

PROPOSITIO III. (expected value of 2 outcomes with different likelihoods)

Si numerus casuum, quibus mihi eveniet $a$$a$, sit $p$$p$, numerus autem casuum quibus mihi eveniet $b$$b$ sit $q$$q$, sumendo omnes casus aeque in proclivi esse: expectatio mea valet $\frac{pa+qb}{p+q}$$\frac{pa + qb}{p + q}$.

If the number of cases in which I will receive $a$$a$ is $p$$p$, and the number of cases in which I will receive $b$$b$ is $q$$q$, assuming all cases to be equally likely: my expectation values at $\frac{pa+qb}{p+q}$$\frac{pa + qb}{p + q}$.

Again, this one follows from our modern definition, but Huygens needs to construct a more complex game for his proof:

To uncover this rule, let again $x$$x$ be put for the value of my expectation: therefore, when I have $x$$x$, I must be able to reach the same expectation, as before, by a fair game. And for this, I will take so many players that together with me they make up the number $p+q$$p + q$, each of whom stakes $x$$x$, so that the total staked is $px+qx$$px + qx$, & with each of the players, as many as the number $q$$q$ indicates, I will individually enter this pact, that each of them who wins will give me $b$$b$, or I will give the same $b$$b$ to him, if I win. Similarly with the remaining players, making up $p-1$$p - 1$, I will individually undertake this condition, that each of them who wins the game will give me $a$$a$, & I the same amount (namely $a$$a$) to him, if I win. And it is clear that this game under this condition is fair, with obviously no one suffering injustice. Then it is clear that I now have $q$$q$ expectations for $b$$b$, & $p-1$$p - 1$ expectations for $a$$a$, & 1 expectation (namely me winning) for $px+qx-bq-ap+a$$px + qx - bq - ap + a$, for then I obtain $px+qx$$px + qx$, what is staked, from which I must give $b$$b$ to each of the $q$$q$ players, & $a$$a$ to each of the $p-1$$p - 1$ players, which together make up $ab+pa-a$$ab + pa - a$ [error, should be $qb+pa-a$$qb + pa - a$]. Therefore, if $qx+bx-bq-aq+a$$qx + bx - bq - aq + a$ [error, should be $px+qx–bq–ap+a$$px + qx – bq – ap + a$] were equal to $a$$a$ itself, I would have $p$$p$ expectations for $a$$a$ (since I already had $p-1$$p - 1$ expectations for it) and $q$$q$ expectations for $b$$b$, and thus I would have returned to my former expectation. Therefore, I set $px+qx-bq-ap+a=a$$px + qx - bq - ap + a = a$, and fix $x=\frac{ap+bq}{p+q}$$x = \frac{ap + bq}{p + q}$, for the value of my expectation, just as we posited at the beginning.

Latina

Ad hanc regulam eruendam, ponatur rursus $x$$x$ pro valore expectationis meae: ergo oportet me, cum $x$$x$ habeo, ad eandem expectationem pervenire posse, ut ante, justo lusu. Ad hoc autem tot collusores sumam, ut una mecum numerum ipsum $p+q$$p + q$ efficiant, quorum deponat quisque x, ita ut depositum sit $px+qx$$px + qx$, et quisque sibi lusoribus, quot indicat numerus q, sigillatim hoc pactum inibo, ut eorum qui vincat mihi sit daturus b, aut ego contra ipsi idem b, si vincam. Similiter cum reliquis collusoribus, constitutentibus $p-1$$p - 1$ sigillatim hanc conditionem aggrediar, ut eorum quisque, qui ludum vincit, mihi sit daturus a, et ego tantundem ($a$$a$ scilicet) ipsi, si ego vincam. Et patet hunc lusum hac conditione justum esse, nemine videlicet injuriam patiente. Deinde patet me nunc q expectationis habere ad b, et $p-1$$p - 1$ expectationes ad a, et 1 expectationem (me nempe vincere) ad $px+qx-bq-aq+a$$px + qx - bq - aq + a$, tunc enim obtineo $px+qx$$px + qx$, id quod depositum est, de quo tradere debeo b unicuique q lusorum, & a unicuique $p-1$$p - 1$ lusorum, quae simul conficiunt $ab+pa-a$$ab + pa - a$ [err. $qb+pa-a$$qb + pa - a$]. Si ita que $qx+bx-bq-aq+a$$qx + bx - bq - aq + a$ [err. $px+qx–bq–ap+a$$px + qx – bq – ap + a$] aequale esset ipsi $a$$a$, haberem $p$$p$ expectationes ad $a$$a$, (quandoquidem iam $p-1$$p - 1$ expectationes ad id habebam) et q expectationes ad b, et sic ad priorem meam expectationem rursus pervenissem. Quocirca pono $px+qx-bq-ap+a=a$$px + qx - bq - ap + a = a$, & fixo $x=\frac{ap+bq}{p+q}$$x = \frac{ap + bq}{p + q}$, pro valore expectationis meae, omnino ut in initio posuimus fuit.

With the algebra typos in this proof (probably committed by the typesetter; more discussion in the notes section below), one cannot help but feel this style of proofs is getting tedious. It has been said “In his own understanding the method he applied for these solutions served only the purpose to demonstrate the power of the new algebra created by Viète and Descartes. The truth of Viète's statement that the new algebra leaves no sensible problem unsolved could now be demonstrated by Huygens' success to apply algebra to the realm of chance which hitherto seemed inaccessible for mathematics.” (Schneider, “Huygens, Christiaan - Encyclopedia of Mathematics.”) I will skip the “in numbers” demonstration he does after as that is also very long and is just substituting in $p=3$$p = 3$, $q=2$$q = 2$, $a=13$$a = 13$, $b=8$$b = 8$. Perhaps people did not trust symbolic math enough back then.

Notes on the typos

The consecutive typos I highlighted and corrected here were already silently corrected in the 1714 English translation because the wrong variables clearly don’t belong there. The commentary by Otten says there is another typo at “Then it is clear that I now have $q$$q$ expectations for $b$$b$, & $p-1$$p - 1$ expectations for $a$$a$, & 1 expectation (namely me winning) for $px+qx-bq-ap+a$$px + qx - bq - ap + a$” and writes “This is a mistake. The Expectation is $a$$a$, not $1$$1$”. The original text is actually correct here. Recall that in the fair game defined by Huygens, there is only one payoff outcome for winning. The Latin text agrees with this as expectationem is singular here.

Indeed, the expectation cannot be $a$$a$ or $b$$b$, as those are the values of the outcomes, not the numbers of outcomes. Notice that the other expectations have counts $q$$q$ and $p-1$$p - 1$ respectively, such that the total number of outcomes sum to $p+q$$p + q$ as required.

The root of the issue is that Huygens for some reason uses “expectation” to mean outcome here. Although Otten also notes that Huygens’ usage is fairly loose, it still trips up a modern reader.

Although Huygens generalized his previous proposition to an arbitrary number of equally likely outcomes, he does not state whether this result works for more than two outcomes with different likelihood of happening. It should be possible to inductively construct fair games with more negotiating parties, and we know that his formula is correct.

Looking back at the first three propositions, Huygens for some reason uses different ways to say what the expectation should be equal to: namely “should be said to be” (dicenda est valere), “should be estimated at” (aestimanda est), and “values at” (valet). If you read every time expectation is mentioned, it it very evident that Huygens’ idea of “expectation” is less of a number that is a property of a random variable, and more akin to the random variable itself, the numerical value of which is the expectation per our definition (when he is not using the term loosely to also mean likelihood or event/outcome anyways). Of course Huygens never talks about random variables, only a collection of outcomes and their respective probabilities, which he uses like the probability mass function (functions in modern math had not been invented yet).

Why compatible: generalizing Huygens’ approach

As you probably have felt by now, Huygens’ game construction scheme is a roundabout way to calculate the weighted average of a random variable’s values. But looking at the historical context and motivation for early probability work, we see it made sense for him for formulate an idea in terms of the mechanics of gambling games. Let us actually prove why Huygens’ approach works by generalizing Proposition 3 to an arbitrary number of outcomes with different probabilities:

Bridging expectation

Definition 1. Consider a random variable $X$$X$ that has $n\in \mathbb{R}$$n \in \mathbb{R}$ possible values $x_1,\dots ,x_n$$x_1, \dots, x_n$, each with probability $\frac{n_i}{N}\in \mathbb{Q}$$\frac{n_i}{N} \in \mathbb{Q}$ of happening, where $N=\sum _{i=1}^n{n}_i$$N = \sum\limits_{i = 1}^n n_i$. The Huygenian expectation of $X$$X$, $E_H[X]$$E_H[X]$, is the entry fee of a fair game where the player can have $n$$n$ possible payoff outcomes of receiving $x_1,\dots ,x_n$$x_1, \dots, x_n$, each of which has probability $\frac{n_i}{N}$$\frac{n_i}{N}$. Corollary 1. The prize pool of the fair game is $N\cdot E[X]=\sum _{i=1}^n{x}_i\cdot n_i$$N \cdot E[X] = \sum\limits_{i = 1}^n x_i \cdot n_i$, as it should be equal to the total expected payout. Theorem 1. $E_H[X]=E[X]=\frac{1}{N}\sum _{i=1}^n{x}_i\cdot n_i$$E_H[X] = E[X] = \frac{1}{N} \sum\limits_{i = 1}^n x_i \cdot n_i$ Proof. Let $G$$G$ be a fair game with the following properties:

1. $G$$G$ has a main player; without the loss of generality we examine their game state
2. $G$$G$ has $n$$n$ negotiating parties, each with $n_i$$n_i$ players, with whom the main player negotiates a unique amount $x_i$$x_i$ that the winner needs to pay to each loser in the party
3. $G$$G$ has $N=\sum _{i=1}^n{n}_i$$N = \sum\limits_{i = 1}^n n_i$ total players who all have an equal probability of winning; without the loss of generality the main player is in the first negotiating party

I claim $G$$G$ is the game we are looking for. Although there seem to be $n+1$$n + 1$ different outcomes:

• $n$$n$ outcomes when the main player loses

  • a player from negotiating party $i$$i$ is the winner; main player gets paid $x_i$$x_i$;

    • for $i=1$$i = 1$, has probability $\frac{n_i-1}{N}$$\frac{n_i - 1}{N}$ of happening
    • for $1<i\le n$$1 < i \leq n$, has probability $\frac{n_i}{N}$$\frac{n_i}{N}$ of happening

• 1 outcome when the main player wins

  • the main player gets the entire prize pool $\sum _{i=1}^n{x}_i\cdot n_i$$\sum\limits_{i = 1}^n x_i \cdot n_i$, but needs to pay all other players their negotiated amount

    • has probability $\frac{1}{N}$$\frac{1}{N}$ of happening

we see in the outcome where the player wins, the amount of money they actually get after paying all the losers is $\sum _{i=1}^n{x}_i\cdot n_i-x_1\cdot (n_1-1)-x_2\cdot n_2-\cdots -x_n\cdot n_n=x_1$$\sum\limits_{i = 1}^n x_i \cdot n_i - x_1 \cdot (n_1 - 1) - x_2 \cdot n_2 - \dots - x_n \cdot n_n = x_1$, the same as the first of the $n$$n$ outcomes. Thus the outcome for $i=1$$i = 1$ also has probability $\frac{n_i-1}{N}+\frac{1}{N}=\frac{n_i}{N}$$\frac{n_i - 1}{N} + \frac{1}{N} = \frac{n_i}{N}$ of happening (as the event where the player wins is mutually exclusive from the event the player loses). We have found a game where there are $n$$n$ different monetary outcomes with payoffs $x_1,\dots ,x_n$$x_1, \dots, x_n$, and probabilities $\frac{n_1}{N},\dots ,\frac{n_n}{N}$$\frac{n_1}{N}, \dots, \frac{n_n}{N}$ of happening, as required. Thus $E_H[X]=\frac{1}{N}\sum _{i=1}^n{x}_i\cdot n_i$$E_H[X] = \frac{1}{N} \sum\limits_{i = 1}^n x_i \cdot n_i$ as the entry fee of $G$$G$.

Note that since the prize pool and the total number of players are fixed, it is impossible to have another game that replicates the game state but has a different entry fee. ◻

PROPOSITIO IV. — PROPOSITIO IX. (solving the problem of points)

Most of Propositions 4 to 9 are not really propositions but questions. They basically each ask the solution to the problem of points with different parameters, i.e., if player A needs to win $x$$x$ more games to win, and player B needs to win $y$$y$ more games to win, etc. how should the prize pool be divided right now. For example:

PROPOSITIO VI.

Ponamus mihi deficere duos lusus & collusori meo tres lusus.

Let us assume that I am short of winning two games and my playmate is short of winning three games.

PROPOSITIO VII.

Ponamus mihi deficere duos lusus & collusori meo quatuor.

Let us assume that I am short of winning two games and my playmate is short of winning four.

Huygens also answers the question when there are three players. But why would such solutions starting with small parameters be useful? To us, these problems are all quite easy with Core Probability knowledge, as the fraction of the prize pool each player should have is the probability of them winning. Today we know that we can use the PMF of a binomial to find the probability of winning the required number of games, and as programmers, we can generate all sequences of more games to be played until someone wins, and count in how many of them each player wins. The enumeration of different scenarios Huygens did in his word proofs is not too different from that.

It turns out the time & space complexity of this approach grows exponentially when the number of games needed to win gets larger, and computation was very expensive for Huygens. In Proposition 9, Huygens presents a general solution to an arbitrary number of players and came up with something like dynamic programming table (something Pascal also did with his triangle):

PROPOSITIO IX.

Ut tot collusorum, quot quis voluerit, ex quibus uni plures et alii pauciores lusus deficiunt, cuiusque pars inveniatur, considerandum est, quid illi, cuius partem invenire volumus, deberetur, si vel ipse, vel quislibet reliquorum primum sequentem ludum vinceret. Horum autem partes si in unam summam colligantur, et aggregatum per numerum collusorum dividatur, quotiens ostender unius quesitam partem.

In order to find the share of as many players as one wishes, among whom one lacks more and others fewer games, it must be considered what would be owed to that person, whose share we wish to find, if either he or any of the others were to win the next game. Moreover if their shares are collected into one sum, and the aggregate is divided by the number of players, it will show the sought-after share of one.

Huygen’s proposition appeals to the idea of total probability, even though he did not formulate it that way. Consider there are $n$$n$ players who all need win some more rounds to win the whole game. We want to find without the loss of generality the probability of Player A winning the game. Since in each round each player has an equal chance of winning and each player winning is mutually exclusive:

$\begin{array}{rl}P(\text{Player A wins the game})& =\sum _i^{n}P(\text{Player A wins the game }|\text{ Player i wins the next round})\cdot P(\text{ Player i wins the next round})\\ & =\frac{1}{N}P(\text{Player A wins the game }|\text{ Player i wins the next round})\end{array}$$\begin{align}P(\text{Player A wins the game}) &= \sum\limits_i^n P(\text{Player A wins the game } | \text{ Player i wins the next round}) \cdot P(\text{ Player i wins the next round})\\&= \frac{1}{N}P(\text{Player A wins the game } | \text{ Player i wins the next round})\end{align}$

Moreover, we should have already calculated what $P(\text{Player A wins the game }|\text{ Player i wins the next round})$$P(\text{Player A wins the game } | \text{ Player i wins the next round})$ is, since when Player $i$$i$ wins the next round, they need one fewer rounds needed to win the game, and Huygens started calculating the solutions from when the number of rounds needed are small. Huygens was transforming a hard problem into multiple subproblems he had already computed before, and saving the result in a table:

Title: “Table for 3 playmates”

Each row on top: “Games that they lack [how many more they need to win]”; on the bottom: “Their parts”. For every cell there are three fractions with a common denominator that sum to 1. They represent the probability of that player winning in that state.

PROPOSITIO X. — PROPOSITIO XIV. (solving harder problems)

In the last 5 propositions, Huygens tackles increasingly difficult probability problems. In Proposition 10 he tries to find the probability of rolling a six in a certain number of throws, and in Proposition 12 he tries to find the probability of rolling two sixes, which we would model using the geometric and negative binomial respectively.

Huygens derives everything painstakingly from first principles. In the 11 times he cites previous propositions in this part, 8 of them are to Proposition 3, 2 of them to Proposition 2, and only 1 to a later proposition (Proposition 10). I encourage anyone interested in how he did it to refer to the 1714 translation. As I think the important foundation of this text is in the first 3 propositions and in trying to solve the problem of points, I shall excuse myself to put down the pen here.

Afterword

I deeply respect the work Huygens did after spending so much time trying to understand it. Our predecessors had so little to work with and attempted to solve hard problems whenever they could. Such courage inspires us. I recall one of my favorite quotes from Seneca the Younger’s Natural Questions, one that I also first read in Sagan’s Cosmos:

Veniet tempus quo ista quae nunc latent in lucem dies extrahat et longioris aevi diligentia. Ad inquisitionem tantorum aetas una non sufficit, ut tota caelo vacet … Itaque per successiones ista longas explicabuntur. Veniet tempus quo posteri nostri tam aperta nos nescisse mirentur … multa saeculis tunc futuris, cum memoria nostri exoleverit, reservantur: pusilla res mundus est, nisi in illo quod quaerat omnis mundus habeat … Rerum natura sacra sua non semel tradit;

The time will come when diligent research over long periods will bring to light things which now lie hidden. A single lifetime, even though entirely devoted to the sky, would not be enough for the investigation of so vast a subject … And so this knowledge will be unfolded only through long successive ages. There will come a time when our descendants will be amazed that we did not know things that are so plain to them … Many discoveries are reserved for ages still to come, when memory of us will have been effaced. Our universe is a sorry little affair unless it has in it something for every age to investigate … Nature does not reveal her mysteries once and for all.

Sen. QNat. 7.25-30

And this is why I love science and history.

Acknowledgements

I thank the support of all CS 109 course staff and Professor Chris Piech. Thank you for asking the Latin question in lecture that awoke something in me again.

ChatGPT 4 was invaluable throughout my writing. This project would not have been possible without it. As the first step I transcribed the book scan with it, and it is superbly accurate even when faced with antique ligatures compared to traditional OCR. I then asked it to check for typos from transcription and create an interlinear translation, at which point I only need to check its output. While I had not been impressed by Latin text ChatGPT generates, it is very capable at translating Latin to English, and I usually only need to make minor or stylistic changes (and proofread when it hallucinates math).

This project was also greated aided by “Huygens and The Value of all Chances in Games of Fortune” by Nathan Otten from University of Missouri – Kansas City, who won the HOM-SIGMAA student paper contest with his commentary. I found it early on during writing and it was deeply helpful to hear another person’s insights. If you liked reading this document, I recommend you to check out Nathan's, as our focuses are quite different. Moreover Nathan’s commentary examines the 1714 English translation, which has some differences from the Latin version that you would notice.

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