I found this question in a book: Two palyers A and B alternatively roll a pair of unbiased die. A wins if on a throw he obtain exactly 6 points, before B gets 7 points, B wining in the opposing event. If A begins the game prove that the probability of A winning is 30/61 and that the expected number of trials need for A's win is approximately 6. I was trying to solve the problem using conditional probability and expectation. To get 30/61 I conditioned on the sum on the dice on the 1st roll and then on the sum on the dice on the second roll. However I am not sure whether we can calculate the required expectation through conditioning. I know that the exact answer is 371/61=6.0819. Through conditioning I only can 5.57 which however can also be approximated to 6. Also if X denotes "the number of trials needed for A to win" And Yi be the sum on the ith roll Conditioning on the first roll, E[X]= E[X/Y1=6)P(Y1=6) + E[x/Y1 is not equal to six]P[Y1 is not =6) = 1x5/36 + E[X/Y1 not =6)x31/36 Then conditioninf on the 2nd roll. E[X/Y1 not = 6]= E[X/Y1 not=6, Y2 not = 9)P(Y2 not equal 9) + E[X/Y1 not=6, Y2=9)P(Y2=9) That's kind of where I am stuck for what is the expected number of trials for A to win given that B wins on the second trial?? Or should I try to condition on the 3rd or last event? Or should I approach the problem without using condition. Man I am so confused now! :rofl: thanks!!