Suppose you cast and keep casting a die until you get either number 5 or 6? What is the probability that you will stop after even number of casts?
The probability that you will stop after an even number of casts is 1/3.
prob(5 or 6) = 2/6 = 1/3
prob(not a 5 or 6) = 2/3
prob(getting a 5 or 6) = 1/3 + (2/3)(1/3) + (2/3)^2 (1/3) + (2/3)^3 (1/3) + ...
(which of course will add up to 1)
prob (event happening on even numbered throw)
= 2nd turn + 4th turn + 6th turn + ...
= (2/3)(1/3) + (2/3)^3 (1/3) +
this is the sum of an infinite geometric series with a = (1/3)(2/3), r = (2/3)^2
= (1/3)(2/3) / (1 - (2/3)^2)
= (2/9) / (1 - 4/9)
= (2/9) / (5/9) = 2/5