a The Chevalier de Mere used to bet that he would get at lea

(a) The Chevalier de Mere used to bet that he would get at least one 6 in four rolls of a die. Was this a good bet?

(b) He also bet that he would get at least one pair of 6’s in 24 rolls of two dice. What was his probability of winning this bet?

(c) Compare the probability of at least one 6 when six dice are rolled with the probability of at least two 6’s when 12 dice are rolled.

1) Probability of getting a six when a die is rolled is 1/6

So probability of not getting a 6 is 1 - 1/6 = 5/6

So P(Getting at least one 6 in 4 rolls) = 1 - P(Not getting a single 6 in 4 rolls) = 1 - (5/6)⁴

= 1 - (625/ 1296) = 671/1296 = 0.5177

As this probability is greater than 0.5 it is a good bet

b) Getting a pair of a 6 in a single roll of two dices is 1/36

So probability of not getting a pair of 6\'s in a single roll of two dices is 1- 1/36 = 35/36

So P(Getting at least one pair of 6\'s in 24 rolls) = 1 - P(Not getting a single pair of 6\'s in 24 rolls) = 1 - (35/36)²⁴

= 1 - 0.5086 = 0.4914

So probability of winning the bet is 0.4914

c) Probability of atleast one 6 is a six die is 1(5/6)⁶ = 10.3349 = 0.6651.

Probability of atleast two 6\'s when 12 dies is rolled = 1 - P(No sixes when 12 dies are rolled) + P(1 six when 12 die is rolled)

= 1 - [ (5/6)¹² + ¹²C₁ * (1/6) * (5/6)¹¹]

= 1 - [ 0.112156 + (12*0.1667 * 0.13459)]

= 1 - [0.112156 + 0.26918]

= 1 - 0.381336

= 0.618664

As, Probability of atleast one 6 in a six die is greater than Probability of atleast one 6 is a six die