Coin Tosses

Question from http://wiki.xkcd.com/irc/Puzzles

Sue and Bob take turns rolling a 6-sided die. Once either person rolls a 6 the game is over. Sue rolls first, if she doesn’t roll a 6, Bob rolls the die, if he doesn’t roll a 6, Sue rolls again. They continue taking turns until one of them rolls a 6.

Bob rolls a 6 before Sue.

What is the probability Bob rolled the 6 on his second turn?

Note: it is not (5/6)*(1/6).

My Solution:

Looks like a pretty simple question using Bayes Theorem and Conditional Probability.

We are required to find P(Bob rolled 6 in his second turn | Bob Won)
P(Bob winning) = (5/6)*(1/6) + (5/6)^3*(1/6) + (5/6)^5*(1/6) + …..

   = 1/6 * (5/6 ) / (1 – 25/36)

  =( 5/36) / 11/36

  = 5/11

P(Bob winning | Bob rolled 6 in his second turn) = 1

P(Bob rolled 6 in second turn) = (5/6)^3*(1/6) 

Therefore, P(Bob rolled 6 in his second turn | Bob Won) = P(Bob winning | Bob rolled 6 in his second turn) * P(Bob rolled 6 in his second turn) / P(Bob winning)

    = 1 * (5/6)^3*(1/6) / (5/11)

   = 25 * 11 / 6^4

   = 275 / 1296