Question

A French nobleman, Chevalier de Méré, had asked a famous mathematician, Pascal, to explain why the following two probabilities were different (the difference had been noted from playing the game many times): (1) at least one six in 4 independent casts of a six-sided die; (2) at least a pair of sixes in 24 independent casts of a pair of dice. From proportions it seemed to de Méré that the probabilities should be the same. Compute the probabilities of \((1)\) and \((2)\).

Short answer

The two probabilities in question are not equal. The probability of obtaining at least one six in 4 throws of a single dice is approximately 51.77%, and the probability of obtaining at least a pair of sixes in 24 throws of two dice is approximately 49.14%.

Step-by-step solution

Analyzing case 1

First, analyze the first case: the probability of getting at least one six in 4 throws of a die. The probability of getting a six in a throw of a die is \( \frac{1}{6} \). The probability of \emph{not} getting a 6 in a single throw is then \(1 - \frac{1}{6} = \frac{5}{6}\). The probability of \emph{not} getting a 6 in \emph{four} independent throws is \(\left(\frac{5}{6}\right)^4\). Therefore, the probability of getting \emph{at least} one 6 (the complement) in 4 throws is \(1 - \left(\frac{5}{6}\right)^4\).

Calculating case 1

Solve to calculate the result for the first case: one minus the probability of not getting a six in four throws. The probability of getting at least one 6 in 4 throws of a die is \(1 - \left(\frac{5}{6}\right)^4\), which is approximately 0.5177 or 51.77%.

Analyzing case 2

Second, analyze the second case: the probability of getting at least a pair of sixes in 24 throws of two dice. The probability of throwing a pair of sixes is \(\left(\frac{1}{6}\right)^2 = \frac{1}{36}\) since the events are independent. The probability of \emph{not} throwing a pair of sixes in a single throw is \(1 - \frac{1}{36} = \frac{35}{36}\). The probability of \emph{not} getting a pair of sixes in \emph{24} independent throws is \(\left(\frac{35}{36}\right)^{24}\). Therefore, the probability of getting \emph{at least} a pair of sixes (the complement) in 24 throws is \(1 - \left(\frac{35}{36}\right)^{24}\).

Calculating case 2

Compute the value for the second case: one minus the probability of not getting a pair of sixes in 24 throws. The probability of getting at least a pair of sixes in 24 throws of a pair of dice is \(1 - \left(\frac{35}{36}\right)^{24}\), which is approximately 0.4914 or 49.14%.

Compare the probabilities

Finally, compare the two probabilities calculated in the previous steps. Using the results from steps 2 and 4, it’s clear that, contrary to de Méré's thoughts, the two probabilities are not the same.

Key concepts

Chevalier de Méré Problem

The Chevalier de Méré problem is a famous historical question in probability theory that sparked significant interest in the mathematics of chance and helped to develop the concept of probability more formally. The problem itself arose from the observations of a French noble, the Chevalier de Méré, who noticed a conundrum while gambling. He observed that, even though events seemed similar, the outcomes were consistently different. Specifically, he was puzzled about why he seemed to win more often when betting on the event of rolling at least one six in four rolls of a single die, compared to the event of getting at least one double six in twenty-four rolls of two dice.

To understand this through probability theory, we need to calculate the chances for each event separately. For the first event, the important concept is that the rolls are independent; the result of one roll does not affect the others. The probability of getting at least one six in four trials is found by calculating the complement—that is, the probability of not getting a six at all in four rolls and subtracting that from one. The probability of rolling a six with one die is \(\frac{1}{6}\), and therefore, not rolling a six is \(\frac{5}{6}\). Hence, the probability of not rolling a six in four independent trials is \((\frac{5}{6})^4\), and the complement gives us the probability of at least one six which turns out to be approximately 51.77%.

Real-World Implications

The significance of this problem extends beyond gambling games to the real world where probability calculations can inform decisions in finance, science, and many other fields. It illustrates that seemingly similar events can have different probabilities and highlights the importance of precise calculation over intuitive assumptions in probability.

Independent Events

Independent events are one of the cornerstones of probability theory. Two events are considered independent if the occurrence of one does not affect the probability of the occurrence of the other. In other words, knowing the outcome of one event gives us no information about the outcome of the other event.

In gambling, the roll of a die or flipping a coin are classic examples of independent events — the outcome of one roll or flip has no bearing on the next. This concept is crucial to properly understand and solve the Chevalier de Méré's problem. When calculating probabilities for independent events, such as rolling dice multiple times, we can multiply the probabilities of each event occurring separately to find the collective probability.

For example, in the game Chevalier de Méré bet on, the probability of getting a double six in a single pair of dice is \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\), as each die is independent of the other. To find the probability of not getting a double six in 24 rolls, we use the complement principle again with the independent probability for each set of two dice. Here, understanding independent events helps us to calculate the right probabilities instead of relying on misleading intuition.

Complement of an Event

The complement of an event in probability is a fundamental concept that refers to all the possible outcomes that are not included in the original event. The probability of an event and its complement always add up to 1 because together they cover all possible outcomes.

In practical terms, often it is easier to calculate the probability of the complement event — especially when dealing with multiple trials — and then subtract it from 1 to obtain the probability of the desired event. This is what we see in both cases of the Chevalier de Méré problem. To find the probability of getting at least one six, we first calculate the probability of not getting a six in all four rolls. Then, we subtract that value from 1 to find the probability of the complementary event — getting at least one six. Similarly, to find the probability of getting at least one double six in 24 rolls of two dice, we calculate the probability of not getting a double six in any of the 24 rolls and subtract it from 1.

Importance of Understanding Complements

Understanding the concept of complement is essential in probability calculations as it often simplifies the process and helps in avoiding common pitfalls. For a student learning probability, this approach is a helpful tool, essentially providing a shortcut to finding probabilities that might otherwise require more complex calculations.