Question

There is an interesting story involving Blaise Pascal, Pierre Fermat, and the Chevalier de Mere. Consider the events A = {at least one ‘one’ would appear when four six- sided dice are rolled} and B = {a pair of ‘ones’ would appear at least once when two six-sided dice are rolled 24 times}. In the 1600s, French gamblers bet on whether event A will occur. They also bet on whether event B will happen. Many thought that these two events were equally likely. The Chevalier de Mere was one among probably many who thought these two games had an equal chance of success. Was he right? Compute Pr(A) and Pr(B) to justify your answer.

Answer 1

No he was not right

Explanation:

Probability of not getting one in one roll in dice is = 5/6

Therefore probability of not getting one in four roll of a dice is

= (5/6) ⁴ = 0.4823

Since the roll of dice is independent events

Now probability of getting at least one 'one' when four six sided dice are rolled

= 1 - probability of not getting one in four roll of a dice

= 1 - 0.4823

= 0.5177

Therefore Pr(A) = 0.5177

Now , we examine the probability of getting no double-six in a roll of a pair of dice. When rolling two 6 sides dice there are possible 6 ² = 36 outcomes .Only one outcome is getting one in both dices .Therefore probability of not getting one on both dice in one roll is = 35/36

Therefore probability of not getting double one in 24 rolls is

= (35/36) ²⁴

= 0.5086

Probability of at least getting one double one in 24 rolls is

= 1 - probability of not getting double one in 24 rolls

= 1 - 0.5086

= 0.4914

Therefore Pr(B) = 0.4914

From the above calculations we can say probabilities are different andthe second probability is smaller.Therefore this two games had not an equal chance of success.