Setup: Urn with 6 red balls and 2 green balls. Player A selects balls (without replacement) from the urn until he obtains a green ball. Player A's score will be the number of red balls he draws. Once Player A obtains a green ball, Player B starts drawing from the urn in the same manner and his score will also be the number of red balls he obtains before drawing the final green ball. What is the probability that Player B's score will be greater than player A's score?
I know that Player B can't have a higher score than Player A unless A scores 0,1, or 2. Therefore I think one strategy may be to write the probability as a sum of 3 mutually exclusive events conditioned on A drawing less than 3:
$P(B>A | A_{0}) + P(B>A | A_{1}) + P(B>A | A_{2})$
Is it correct to write $B>A$ in the first term as $P(B_{1}\cup B_{2} \cup B_{3} \cup B_{4} \cup B_{5} \cup B_{6} | A_{0})$ ? Seems like this approach would blow up and become a lot of work on larger problems. Interested in your thoughts. Thanks!
Hint: $$P(B_1\cup B_2\cup B_3\cup B_4\cup B_5 \cup B_6|A_0)=1-P(B_0|A_0)$$