When a gambler asked a mathematician for help

When a gambler consistently loses large sum of money, what can he or she do? When one particular gambler, Chevalier de Méré (1607-1684), was losing a big fortune, he called a “mathematical help line”. In fact, his correspondence with Blaise Pascal (1623-1662) earned him a place in the history book. The problems that were presented by de Méré, jointly worked on by Pascal and Pierre de Fermat (1601-1665), are regarded as the beginning of the emerging academic field of probability. Chevalier de Méré was in need of help for two problems – the problem of points and on the dice problem that now bears his name. In this post we discuss the dice problem. The problem of points is discussed in the next post.

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The Dice Problem

There were two dice problems from Chevalier de Méré. The first game involves four rolls of a fair die. In this game, de Méré made bet with even odds on rolling at least one six when a fair die is rolled four times. His reasoning was that since getting a six in one roll of a die is \frac{1}{6} (correct), the chance of getting a six in four rolls of a die would be 4 \times \frac{1}{6}=\frac{2}{3} (incorrect). With the favorable odds of 67% of winning, he reasoned that betting with even odds would be a profitable proposition. Though his calculation was incorrect, he made considerable amount of money over many years playing this game.

The second game involves twenty four rolls of a pair of fair dice. The success in the first game emboldened de Méré to make even bet on rolling one or more double sixes in twenty four rolls of a pair of dice. His reasoning was that the chance for getting a double six in one roll of a pair of dice is \frac{1}{36} (correct). Then the chance of getting a double six in twenty four rolls of a pair of dice would be 24 \times \frac{1}{36}=\frac{2}{3} (incorrect). He again reasoned that betting with even odds would be profitable too.

But experience showed otherwise. As he lost a lot of money, he realized something was not quite right with the second game. In 1654, he challenged his renowned friend Blaise Pascal to find an explanation. The solution emerged in a series of letters between Pascal and Fermat. Out of this joint effort, a foundation was laid for the idea of probability as an academic subject. One particular idea that emerged was the Pascal triangle. Another one was the binomial distribution. In fact, anyone who understand the binomial distribution can very quickly see the faulty reasoning of de Méré.

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The Simulation

Before we get to the calculation, let’s simulate the games played by de Méré. Using random numbers generated from using the Rand() function in Excel, we simulated 100,000 iterations of each of the games. In our 100,000 simulations of the first game – rolling a die four times, there are 51,380 iterations with at least one six. This suggests that de Méré’s position would win more than half of the time, though not the \frac{2}{3} odds that he believed. But it was profitable for him nonetheless.

In our 100,000 simulations of the second game – rolling a pair of dice 24 times, there are only 49,211 iterations with at least one double six. This seems to support that de Méré’s position is a losing proposition, that he would be losing his bets more than half the time.

Of course, de Méré could have done similar simulation, though in a much smaller scale, by rolling the dice himself (say, 100 times). He could have seen the light much sooner.

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The Calculation

Let’s see why the first game was profitable for de Méré 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 Méré 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 Méré 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 Méré would only win about 49 games out of 100 and his opposing side would win about 51 games out of 100 games. Out of 1000 games, he would on average win 491 games (the opposing side would win on average 509 games). With each bet as one franc, the opposing side of de Méré would win 18 francs for each 1000 francs wagered (thus the opposing side having the house’s edge of about 1.8%).

The odds indicated by the simulations discussed above are in line with the calculated results. It would be interesting to known what action did de Méré take after learning the answers. Maybe he stopped playing the second game and only played the first game. Maybe he modified the second game so that the odds of winning for him was at least even (or better).

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\copyright \ 2016 \text{ by Dan Ma}

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4 thoughts on “When a gambler asked a mathematician for help

  1. Pingback: The problem of points | A Blog on Probability and Statistics

  2. Pingback: Every character is known but password is hard to crack | A Blog on Probability and Statistics

  3. Pingback: Five probability problems to help us think better | Math in the Spotlight

  4. Pingback: The Dice Problem – Daniel Ma

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