Answer to Question #117815 in Statistics and Probability for Samuel

Question

Consider the de Mérés game which is played as follows: In Game 1, there are 4
rolls of a die and you win with at least one 6. In Game 2, there are 24 rolls of two dice and you
win by at least one pair of 6’s rolled.How many possible outcomes could be realized for
(a) Game 1
(b) Game 2
How many elements are in the event (E1) of getting an outcome without a 6 for
(a) Game 1
(b) Game 2

Answer 1

a) N(Possible outcomes for game 1)=N(all outcomes)-N(outcomes without any 6)=6⁴-5⁴=1296-625=671

b) N(Possible outcomes for game 2)=N(all outcomes)-N(outcomes without any pair of 6's)=36²⁴-35²⁴=1.10331265*10³⁷

a) N(outcomes without any 6 for game 1)=5⁴=625

b)N(outcomes without any pair of 6's for game 2)=35²⁴=1.14191312*10³⁷