Question

You throw a die until you get 6. What is the expected number of throws conditioned on the event that all throws, before 6 was reached, were even numbers. You have to show direct calculation,

Answer 1

Given that all throws before 6 were reached resulted in only even numbers

==> Only possible outcomes of the die are 2,4,6

Probability of getting 6 = 1/3

Probability of not getting 6 = 2/3

Let X denote the number of throws required before 6 was reached where the die can result in only 2,4,6

X follows a geometric distribution with parameter p = 1/3

Here r = 2/3

Therefore the expected number of throws conditioned on the event that all throws, before 6 was reached, were even numbers is 3