The answer is 6/11. To see this, let pA = Pr[Alice wins] and pB = Pr[Bob wins]. Then note that pA = Pr[win if go first], so pB = Pr[Alice loses first roll] * Pr[win if go first]. So mathematically we have pB = 5/6 * pA, and pA + pB = 1.

Let R1 = the value of the die on roll 1, R2 = value on roll 2, etc. Since Alice rolls first, she will win if R1=6. But what if R1 is not 6? Then Alice will still win if (R1 !=6) and (R2 !=6) (so that Bob doesn't win) and (R3 = 6) (so that Alice wins). Similarly, what if R3 is not 6? Then Alice will still win if (R1 !=6) and (R2 != 6) and (R3 != 6) (by assumption) and (R4 != 6) (so that Bob doesn't win) and (R5 = 6). It should be obvious that we can repeat this pattern forever. Also, each of these events are mutually exclusive, since (for example) it's impossible for (R1 = 6) (the first possibility) and (R1 != 6 and R2 != 6 and R3=6) (our second possibility), due to the first one having R1 = 6, and the second one having R1 != 6. Therefore we can just add up the probabilities (using the rule P(A or B) = P(A) + P(B) - P(A and B)). Each time is one roll which must equal 6, which obviously has a probability of 1/6. Similarly, at the n'th step there are 2n rolls which cannot equal 6, each occurring with a probability of 5/6. Thus adding up all the probabilities gives the infinite sum: (1/6) * sum for n from 0 to infinity of (5/6)^(2*n) = (1/6) * sum for n from 0 to infinity of (25/36)^n This is an infinite geometric series. An infinite geometric series going from 0 to infinity adding up r^n has a sum equal to 1 / (1 - r). Thus in this case we get: (1/6) * (1 / (1 - 25/36)) = 6 / 11.