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 four 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 probabilities for both scenarios are different which shows that even though on the surface both scenarios look similar, the underlining probabilistic models are different. The probability for event A (at least one six in four independent casts of a die) is approximately 0.5177 (52% chance) and for event B (at least a pair of sixes in 24 independent casts of a pair of dice) is approximately 0 (almost no chance).

Step-by-step solution

Define the events

Event A: getting at least one six in four throws of a six-sided die.\n Event B: getting at least a pair of sixes in 24 throws of two dice.

Calculate the probability of Event A

The probability of not getting a six in one throw of a die is \(\frac{5}{6}\). So, the probability of not getting a six in four throws is \(\left(\frac{5}{6}\right)^4\). Therefore, the probability of getting at least one six in four throws is 1 - \(\left(\frac{5}{6}\right)^4\).

Calculate the probability of Event B

The probability of not getting a six on a single dice throw is \(\frac{5}{6}\) and the probability of not getting a six on two dice throws is \((\frac{5}{6})^2\). Thus, the probability of not getting a six on 24 throws of two dice is \((\frac{5}{6})^{48}\) and the probability of getting at least a pair of sixes is 1 - \((\frac{5}{6})^{48}\). It should be noted here that, this would be an approximate probability and not entirely accurate due to possibility of more than one pair sixes occurring in 24 throws.