Probability distribution question

Question

You play a game of rolling 2 dice. You lose the game if you roll an 11 before you roll a 7. You continue to roll until either an 11 or 7 appear. What is the probability that you will lose the game given that you roll a 7 or 11

Answer 1

P(7) = 6/36, P(11) = 2/36.

 

On first roll, you could win right away by rolling 7.  If Not 7, P(not 11) is 28/36, so P(being able to continue is 28/36 = 7/9.

 

So probability of winnng on second roll, P(7 2^nd) = (7/9)(6/36); and if that's not a W (rolled 7) or L (rolled 11),

this next probability is (7/9)²(6/36), etc.

 

This constitutes an infinite geometric series with a = 6/36, r = 14/18; using the formla S = a/(1-r), we get the sum of all these ways to win as (6/36) ÷ [1-(7/9)] = (1/6)(9/2) = 3/4.  REVISED.