Probability of winning at craps

[hholzer]

(The link goes to books.google.com)
http://tinyurl.com/2a48rlv at the bottom paragraph
how do they get 1/3 for the probability of winning
for rolling a sum of 4 twice?
(the table mentioned in the paragraph is on the previous page)

I'm not seeing how the P( roll a sum that is 4 | roll a sum that is 4) = 1/3

 

[mathman]

Once you roll a 4, you win if you get a 4 before a 7 and you lose if you get a 7 before a 4. Since nothing else matters, you need to consider only the ratio of probabilities of 4 rolls to 7 rolls, which is 1:2. Therefore the win probability is 1/3.