The birth of probability theory is usually set in the mid-seventeenth century. At that time the two great mathematicians Blaise Pascal (1623–1662) and Pierre de Fermat (1601– 1665) discussed together some gambling problems and defined the theoretical basis of the mathematical theory of classical probability.
However, the first studies on the calculation of probabilities appeared already a century earlier, in particular in the work of Gerolamo Cardano (1501-1576), who had a great passion for gambling. He basically gave the classic definition of probability of an event as the ratio between the number of favorable cases and the total number of possible cases.
In this article we will describe some gambling problems studied by Cardano and other scholars of the period, that introduce the basic concepts of classical probability, later defined more precisely by Pascal, Fermat, Huygens and others.

## 1) Short history of gambling

The development of the calculus of probability started with the need to solve practical problems related to gambling. The word derives from the Arabic term al zahr, which means die. Gambling typically involves wagering money or other personal values on the basis of a future random event.
The passion for gambling is as old as humanity itself. In places such as China, Egypt, Greece, Rome, there are evidences that date back to thousands of years ago. The most used tool is represented by dice (astragals were used in the distant past). Playing cards appeared in the late 14th century.
The first casinos opened in Italy in the seventeenth century, for example the Ridotto of Venice in 1638, then spread throughout Europe. In the sixteenth century the game of the lotto was born in Genoa. At the beginning of the twentieth century the first mechanical slot machines were introduced, later perfected as electromechanical.
With the advent of the Internet, slot machines have become accessible online to anyone, allowing people to play from home. Thanks to the diffusion of smartphones, mobile gambling is constantly increasing.
In all ages the passion of gambling has conditioned the lives of many people for better or for worse. It has often led to the ruin of many players and for these reasons, in various countries, attempts have been made over time to ban it, without success.

A very interesting reading to understand player psychology is the famous book by the great Russian writer F. Dostoevskij [1].

## 2) The figure of Gerolamo Cardano

As we said earlier, the first studies on probability started to solve problems posed by gamblers. One of the main scholars is certainly the Italian mathematician Gerolamo Cardano (1501-1576).
Cardano is a typical Renaissance man who was interested in various sciences: mathematics, physics and mechanics, medicine, astrology, alchemy. Among other things, he was one of the first scientists to affirm the impossibility of perpetual motion, that is the creation of a machine that can operate without any loss of energy.
Cardano was a passionate gambler; from his memoirs it appears that for many years of his life he played almost every day all kinds of games of his time: dice, chess, cards, and so on.

### 2.1) Cardano and Algebra

As regards mathematics, Cardano played a primary role in the study of third and fourth degree algebraic equations. The general third degree equation can be written as follows:

$x^{3}+ Ax^{2}+Bx +C=0 \\$

Solving third-degree equations posed a challenge for Renaissance mathematicians. The mathematician Nicolò Tartaglia (1499-1557) found a method, in 1534, to solve the reduced third degree equations, that is without the second degree term:

$x^{3}+bx+c=0 \\$

The general third degree equation can be put into the reduced form by substituting $$x = y – \frac {A} {3}$$.
Cardano convinced Tartaglia to reveal to him the secret of the solution formula and after much resistance Tartaglia revealed his method, obliging him not to publish it with an oath.
In 1545 Cardano published his important work ‘Ars Magna‘ in Germany. The publication of this book represents for many the beginning of modern mathematics. In the book Cardano presented among other things the solutions of the general equations of the third and fourth degree and also the way to pass from the general equation to the reduced one. The fact that he published Tartaglia’s formula in his work caused a dispute between the two mathematicians.
However, several years had passed since the promise made to Tartaglia and during this period the latter had never wanted to publish his formula. Furthermore, together with Cardano and Tartaglia, there were two other mathematicians who made a fundamental contribution to solving the two problems in the general form: Scipione del Ferro (1465-1526) who actually discovered the solution method before Tartaglia, and Ludovico Ferrari (1522-1565), a pupil of Cardano who discovered the formula for fourth degree equations. In his ‘Ars Magna’ Cardano correctly recognized the merits of all.

The general formula for solving the reduced third degree equation is as follows:

$\displaystyle x=\sqrt[3]{-\frac{c}{2}+ \sqrt{\frac{c^{2}}{4}+ \frac{b^{3}}{27}}} + \sqrt[3]{-\frac{c}{2}- \sqrt{\frac{c^{2}}{4} + \frac{b^{3}}{27}}} \\$

Exercise 2.1
Calculate the solutions of the equation $$x^{3} = 6x + 6$$.

However, a fundamental problem remained to be solved: the so-called irreducible case could not be treated, that is when

$\dfrac{c^{2}} {4} + \dfrac {b^{3}} {27} \lt 0$

If the size in the square roots is negative, then we are in the presence of the so-called imaginary numbers, which had not yet been introduced into mathematics. The situation will be clarified later by the mathematician Raffaele Bombelli (1526-1573) who laid the foundations for the theory of complex numbers.

Exercise 2.2
Calculate the three real solutions of the equation:

$x^{3}-15x+4=0 \\$

Solution: $$4, \sqrt {3} +2, \sqrt {3} -2$$

### 2.2) Cardano and the calculation of probabilities

Cardano wrote a book called ‘Liber de Ludo Aleae‘ in which the first concepts of classical probability are introduced and discussed. Unfortunately, the booklet was not printed before 1663 and therefore did not have the attention it deserved. For the English version of the book see [2].
Cardano’s book deals in particular with the dice game, in which he himself had alternate fortunes of winnings and losses. In his booklet Cardano introduces the concept of circuit, that is the set of all possible cases, which coincides with the modern space of samples or space of events. Despite some initial errors, Cardano basically introduces the classical definition of probability as a ratio between the number of favorable and possible cases. However, in the initial part of the book Cardano also uses a second type of reasoning, called reasoning on the mean, which leads him to incorrect results.

Example 2.1 – Reasoning on the mean
By rolling a die, the probability that a single face appears, for example the number $$4$$, is equal to $$p = \frac {1} {6}$$. So with a single throw we get $$\frac {1} {6}$$ as the probability of winning a score. If we roll the die twice we get $$2 \cdot \frac {1} {6}$$, and so on. In particular, if we make three rolls we would have the $$50 \%$$ chance of a given face coming out.

If we accept this type of reasoning, $$6$$ throws would be enough to be sure that a given face comes out, while in reality a given face may never come out even after a very large number of throws.
Later Cardano became aware of the fallacy of the reasoning and from then on he used the correct method, which consisted of counting favorable and total cases.
In chapter $$XIV$$ Cardano clearly sets out the rule to be applied to calculate the probability of an event. Counting the total cases of the circuit, then counting the favorable cases and calculating the probability according to the classic formula:

$\displaystyle P(A) = \frac{\text {Number of favourable cases}}{\text {Number of possible cases}} \\$

Cardano actually uses the terms of gambling to express his general rule: for each player it’s important to know the relationship between favorable and unfavorable cases. If an event $$A$$ has probability equal to $$P (A)$$, then the odds in favor and against are as follows:

$\displaystyle \begin{array}{l} \textbf{odds in favor of A = }\dfrac{P(A)}{1-P(A)} \\ \\ \textbf{odds against A = }\dfrac{1-P(A)}{P(A)} \end{array} \\$

Problem 2.1 – Cardan
Calculate how many times you must roll a fair die to have a probability greater than $$\frac {1} {2}$$ that at least one $$6$$ comes out?

Solution
We solve with modern notation. We denote with $$A$$ the event that is true if the number $$6$$ comes out. It’s also convenient to calculate the probability of the complementary event $$\overline {A}$$, that is the event that is true if the number $$6$$ does not come out. In a launch we have $$P(\overline {A}) = \frac {5} {6}$$ and therefore $$P (A) = 1 – \frac{5} {6}$$. If we make two throws, since the throws are independent, we have $$P (\overline {A}) = \left (\frac {5} {6} \right)^{2}$$ and therefore $$P (A ) = 1 – \left (\frac {5} {6} \right)^{2}$$. So in the case of $$n$$ throws we have the following probability values:

$$\displaystyle \begin{array}{l} P(\overline{A}) = \left(\frac{5}{6}\right)^{n} \\ \\ P(A) = 1- \left(\frac{5}{6}\right)^{n} \end{array} \\$$

To solve Cardano’s problem we have to find the value of $$n$$ such that $$P (A) \ge \frac {1} {2}$$, or equivalently $$P (\overline {A} ) \lt \frac {1} {2}$$. Since

$\begin{array}{l} \left (\frac{5} {6} \right)^{3} \approx 0,578 \\ \left (\frac {5} {6} \right)^{4} \approx 0,482 \\ \end{array}$

we can conclude that $$4$$ throws are enough for it to be advantageous to bet on the exit of the number $$6$$.
Cardano initially uses reasoning on the average, and erroneously considers that the number of throws is equal to $$3$$, as he calculates

$p = 3 \cdot \frac {1} {6} = \frac {108} { 216} = \frac {1} {2}$

But then, realizing the error, he correctly calculates that the number of favorable cases for a face to come out with three throws is $$91$$ and not $$108$$, so four throws are needed to have a probability greater than $$50 \%$$ that a given face comes out.
In his book Cardano presents several examples relating to the throwing of two dice and three dice, correctly carrying out the calculations according to the classic definition of probability.

#### 2.2.1) The law of power

Another type of problem studied by Cardano is the calculation of the probabilities for a repeated event. As we saw, in a three-dice roll the odds for and against getting at least a $$4$$ are in the proportion of $$91$$ and $$125$$. Cardano tries to calculate the probability of success in each of two successive throws; he initially uses a wrong reasoning and assumes that the odds for and against are in the proportion $$91^{2}$$ and $$125^{2}$$. And so on for more trials. Cardano realizes that this method of calculation leads to absurd results. After several attempts he comes to prove the correct formula; in a set of $$2$$ independent successive tests the odds for and against obtaining two consecutive successes are in the following proportion:

$91^{2} \quad \text{e} \quad 216^{2}-91^{2}$

discovering one of the fundamental laws of the calculation of probabilities. In modern form we can affirm that the probability $$p_{n}$$ of obtaining $$n$$ successes in $$n$$ independent tests is given by the following formula:

$p_{n}=p^{n}$

where $$p$$ is the probability of success in a single test. This is an important result, which will be generalized with the Bernoulli distribution, to calculate the probability of having $$k$$ successes in $$n$$ throws.

#### 2.2.2) The law of large numbers

In the last chapter of his libretto Cardano exposes, albeit still in a rudimentary way, the content of the law of large numbers and the average of a random variable. By repeating an experiment $$n$$ times, a probability event $$p$$ will occur on average a number of times equal to $$n \cdot p$$.

## 3) Some famous problems

We describe some typical exercises of the time. From the current point of view they may seem simple but, considered in the context of the initial stage of the development of the calculation of probabilities, they have a great importance.

### 3.1) The Galileo problem

Suppose we roll three dice and add the three numbers obtained. The total scores of $$\{9,10,11,12 \}$$ can all be obtained each with $$6$$ different combinations. Why are the total scores of $$10$$ or $$11$$ more likely than the scores of $$9$$ or $$12$$ then?
This problem was posed by the Grand Duke of Tuscany to Galileo Galilei (1564-1642), who presented a solution in his treatise ‘Above the dice discoveries‘.
With three dice the total number of elementary events, i.e. triplets of numbers, is $$6^{3} = 216$$. However the sum of the three numbers can only take $$16$$ distinct values: $$\{3,4,\cdots, 18 \}$$. The following table shows the possible combinations for the four cases:

$\begin{array}{cc} 9 & 10 & 11 & 12 \\ \hline 6-2-1 & 6-3-1 & 6-4-1 & 6-5-1 \\ 5-3-1 & 6-2-2 &6-3-2 & 6-4-2 \\ 5-2-2 & 5-4-1 & 5-5-1 & 6-3-3 \\ 4-4-1 & 5-3-2 & 5-4-2 & 5-5-2 \\ 4-3-2 & 4-4-2 & 5-3-3 & 5-4-3 \\ 3-3-3 & 4-3-3 & 4-4-3 & 4-4-4 \\ \textbf{ 25} & \textbf{27} & \textbf{27} & \textbf{25} \\ \end{array}$

The last line contains the number of dispositions ordered in the various cases, which as we can see are not all the same in the $$4$$ cases. Galileo then counts the number of favorable cases and divides by the number of total cases, according to the classical definition of probability. The same type of problem had actually already been solved by Cardano and is present in chapter 13 of his book.

### 3.2) The problem of Chevalier de Méré

In 1654 Antoine Gombaud, known as Chevalier de Méré, asked Pascal the following problem concerning two ways of betting with dice:

• in the first game a single die is used and it is wagered to get at least one $$6$$ after $$4$$ consecutive throws;
• in the second type we use $$2$$ dice and we bet on the combined outcome of two $$6$$ in $$24$$ consecutive throws.

Méré used an incorrect reasoning, similar to Cardano’s initial one: in the first case according to his calculations the probability is $$\frac{1} {6} \cdot 4 = \frac{2} {3}$$. In the second case it results $$\frac{1} {36} \cdot 24 = \frac {2} {3}$$, equal to the first type of game. Although the odds were the same according to his calculations, he lost a lot of money by betting on the exit of the pair of $$6$$ in $$24$$ throws, while generally he won in the first type of game. For this reason he submitted the problem to the great mathematician Pascal.
Pascal gave the correct answer, concluding that the pair of $$6$$ on $$24$$ throws is an event with less probability than a single $$6$$ on $$4$$ throws.
Let’s see the solution using modern notation, solving the two types of games separately.
In the first game the total cases are: $$6^{4} = 1296$$; the unfavorable cases are $$5^{4} = 625$$; the favorable cases are $$1296-625 = 671$$. So the first type of game is favorable to the bettor.
In the second game the total cases are: $$36^{24}$$, the unfavorable cases are $$35^{24}$$ and the favorable cases are $$36^{24} -35^{24}$$. These are very large numbers; however, it can be seen that the number of unfavorable cases this time is greater than the favorable ones. In terms of probability, the probability of winning in the first type of game is:

$P(A_{1}) = 1 – \left(\frac{5}{6}\right)^{4} \approx 0,51775$

The first game is therefore not fair, but favorable to the bettor: on $$100$$ bets he would win on average $$52$$ times. The probability of winning in the second type of game is instead:

$P(A_{2}) = 1 – \left(\frac{35}{36}\right)^{24} \approx 0,4914$

confirming that the second game is less favorable than the first. The second game is also not fair, but this time it is unfavorable to the bettor: on $$100$$ bets he would win on average $$49$$ times.

### 3.3) The problem of the division of the stake (or problem of the points)

Two players A and B are giving away a certain stake that will be given to the first of the two who will reach a fixed number $$N$$ of points, in a sequence of independent tests. However, the game is stopped when player A has gained $$a$$ points and player B has scored $$b$$ points. The question is: how should the total stakes be divided?
In the initial version, the two players are assumed to have the same chance of winning every single point. The substance of the problem consists in calculating the probability of winning of each of the two players at the time of interruption, according to the points earned by each. In this article we will present some incorrect solution attempts proposed by the first probability scholars of the time. In a subsequent article we will present the correct and complete solutions given by Pascal and Fermat.

#### 3.3.1) Luca Pacioli’s solution

The mathematician Luca Pacioli (1447-1517) exposed the problem in his important work ‘Summa de arithmetic, geometry, proportions and proportionality (1494)‘. Pacioli was also one of the first to publish a description of the double-entry system used by book-keepers and accountants.
He considered a particular version of the problem of the points: A and B play a fair game that will be completed when one of the two wins $$6$$ games. The game is stopped when player A has won $$5$$ games and B has won $$3$$. How should the wager be distributed? Pacioli’s solution consists in dividing the stakes in proportion to the points obtained by the two players. So A receives $$\dfrac{5}{5 + 3} = \dfrac{5}{8}$$ and B receives $$\dfrac{3}{5 + 3} = \dfrac{3}{8}$$ of total stake.

#### 3.3.2) Tartaglia’s criticism

Tartaglia criticized Pacioli’s solution in his ‘General treatise on numbers and measures‘. Tartaglia pointed out that, according to Pacioli’s rule, if A wins one game and B none, then player A should take all the stakes, which is obviously not correct. Tartaglia understood that to solve the problem we must not consider the points already obtained, but we must calculate the probabilities that the two players have to earn the remaining points. Tartaglia’s method consists in paying the player who has the advantage of his initial stake plus a fraction obtained by dividing the difference of the points obtained by the total score required to complete the game.
We apply Tartaglia’s method to the following example: A and B play a ball game which is completed when one of the two wins $$60$$ games. Everyone invests $$22$$ euros. The game is stopped when player A has already won $$50$$ games and B has won $$30$$. How should the wager be distributed? Tartaglia’s method involves the following calculations:

$\begin{array}{l} 50 – 30 = 20 \text{ difference of the points obtained} \\ \dfrac{20}{60} = \dfrac{1}{3} \text{ fraction relative to the complete game} \\ 22 + \dfrac{22}{3} \text{ sum received by player A} \\ 22 – \dfrac{22}{3} \text{ sum received by player B} \\ \end{array}$

However, Tartaglia’s method is also incorrect.

#### 3.3.3) Resolution according to Cardano

Cardano also understood that the solution to the problem of the division of the stakes depends on the number of missing games and not of the ones played. Cardano proposes to divide the stakes according to the ratio of two arithmetic progressions, defined according to the points missing by the two players:

$\displaystyle \begin{array}{l} 1+ 2 + \cdots + (n-a) = \dfrac{(n-a)(n-a+1)}{2} \\ 1+ 2 + \cdots + (n-b) = \dfrac{(n-b)(n-b+1)}{2} \\ \end{array}$

So the stakes should be divided according to the relationship between the progressions of $$(n-a)$$ and $$(n-b)$$, or

$(n−b)(n−b+1) : (n−a)(n−a+1)$

Applying this rule to the first type of Pacioli’s problem $$(n = 6; \ a = 5; \ b = 3)$$, we find that the stakes should be paid to the players in the ratio $$(6: 1)$$. Cardano’s logic is based on the idea that the leading player must receive compensation proportional to the effort that the other player must make to win the game. The progressions based on the missing points are, according to Cardano, a measure of the relative effort of the two players to obtain the number of points expected by the game. Cardano’s solution is also incorrect, although the setting is fundamentally correct.
The correct solution was subsequently given by Pascal and Fermat, thanks also to the new tools made available by the developments of the Combinatorial Calculus. We will describe Pascal and Fermat’s solutions in a later article.

## Conclusion

As we have seen, Cardano made important contributions to the birth of the Calculus of Probabilities, even if his analysis on several occasions were too simplistic or even incorrect. After him Pascal, Fermat and then Huygens will lay the foundations for a rigorous definition of the concept of classical probability and the proof of the main theorems.

## Bibliography

[1]F. Dostoevsky – The Gambler and Other Stories (Penguin Classics)

[2]G. Cardano – The Book on Games of Chance (Dover)