## One gambling problem that launched modern probability theory

Chevelier de Mere (1607-1684) was a gentleman gambler in France who made it to the history book by turning to Blaise Pascal, an eminent mathematician of his time, for help in finding a mathematical answer for why he consistently lost money in a certain game of dice. Unlike other gamblers who might just chalk it up to bad luck, he pursued the cause of the problem with the help of Pascal. As a result of Pascal’s effort combined with that of Pierre de Fermet, the area of probability subsequently emerged as an academic field of study.

In this note, we discuss the errors in de Mere’s games of dice and the analysis offered by Blaise Pascal.

Chevelier de Mere’s predicament involved two games of dice. In the first one, de Mere made bet with even odds on getting at least one six on four rolls of a fair die. He reasoned correctly that the chance of getting a six in one roll of a die is $\frac{1}{6}$. He then incorrectly thought that in four rolls of a die, the chance of getting one six would be $\frac{4}{6}=\frac{2}{3}$. Though his reasoning was faulty, he made considerable money over the years in making this bet.

With the success of the first game, de Mere modified the game by betting with even odds on getting at least a double six on 24 rolls of a pair of dice. He reasoned correctly that the chance of getting a double six in rolling a pair of dice is $\frac{1}{36}$. However, he erred in thinking that in 24 rolls of a pair of dice, the chance of getting one double six would be $\frac{24}{36}=\frac{2}{3}$.

Based on empirical data (i.e. he lost a lot of money), he knew something was not quite right in the second game of dice. So he challenged his renowned friend Blaise Pascal to help him find an explanation. In a series of letters between Pascal and Pierre de Fermet, the problem of de Mere was solved. Out of this joint effort, a foundation was laid for the theory of probability. Nowadays, any one with a good understanding of the binomial distribution would be able to spot the faulty probability reasoning of de Mere.

Let’s see why the first game was profitable for de Mere and why the second game was not.

The First Game
In a roll of a die, there are six possible outcomes: 1, 2, 3, 4, 5, 6. If the die is fair, the probability of getting a six is $\frac{1}{6}$. Likewise, the probability of getting no six in one roll of a fair die is $\frac{5}{6}$.

The probability of getting no six in four rolls is: $\displaystyle P(\text{no six in four rolls})=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\biggl(\frac{5}{6}\biggr)^4=0.482253$.

Thus in four rolls of a fair die, the probability of getting at least one six is: \displaystyle \begin{aligned}P(\text{at least one six in four rolls})&=\displaystyle 1 - P(\text{no six in four rolls})\\&=1 - 0.482253\\&=0.517747\end{aligned}

Thus the probability of getting at least one six in four rolls of a fair die is 0.517747. Out of 100 games, de Mere would on average win 52 games. Out of 1000 games, he would on average win 518 games. Suppose each bet is one French franc. Then de Mere would gain 36 francs for each 1000 francs in wagered. Thus he had the house’s edge of about 3.6%.

The Second Game
In a roll of a pair of dice, there are a total of 36 possible outcomes (i.e. the six outcomes of the first die combined with the six outcomes of the second die). Out of these 36 outcomes, only one of them is a double six. So, the probability of getting a double six is $\frac{1}{36}$ in rolling a pair of dice. Likewise, the probability of not getting a double six is $\frac{35}{36}$.

The probability of getting no double six in 24 rolls of a pair of dice is: \displaystyle \begin{aligned}P(\text{no double six in 24 rolls})= \biggl(\frac{35}{36}\biggr)^{24}=0.5086\end{aligned}

Thus the probability of getting at least one double six in 24 rolls is: \displaystyle \begin{aligned}P(\text{at least one double six in 24 rolls})&=\displaystyle 1 - P(\text{no double six in 24 rolls})\\&=1 - 0.5086\\&=0.4914\end{aligned}

Thus the probability of getting at least one double six in 24 rolls of a pair of fair dice is 0.4914. On average, de Mere would only win about 49 games out of 100 and his opposing side would win about 51 games out of 100 games. With each bet as one franc, the opposing side of de Mere would win 2 francs for each 100 francs wagered (thus having the house’s edge of about 2%).

1. carnotcycle says: