Question

Chevalier de Méré won money when he bet unsuspecting patrons that in 4 rolls of 1 die, he could get at least one 6; but he lost money when he bet that in 24 rolls of 2 dice, he could get at least a double 6. Using the probability rules, find the probability of each event and explain why he won the majority of the time on the first game but lost the majority of the time when playing the second game.

Answer 1

P(At least one 6 in 4 rolls) = 1 – P(No 6 in 4 rolls)

                                              = 1 – (5/6)4

                                             = 1 – 625/1296

                                             = 671/1296

                                             = 0.518

 

P(At least on double 6 in 24 rolls) = 1 – P(No double 6 in 24 rolls)

                                                            = 1 – (1 – 1/36)24

                                                           = 1 – (35/36)24

                                                           = 1 – 0.5086

                                                           = 0.4914

 

As probability of first event is greater than probability of second event, so majority of time win was recorded in most of the cases in second game.

As probability of first event is greater than probability of second event, so majority of time win was recorded in most of the cases in second game.