Darts probability question

Question

Two darts players A and B throw alternately at a board and the first to score a bull wins the contest. The outcomes of different throws are independent and on each of their throws A has probability p_A and B has probability p_B of scoring a bull. If A has first throw, calculate the probability of A winning the contest.

Answer 1

The probability P_A,1 of A winning the first throw is p_A, and the probability P_B,1 of B winning the first throw is p_B(1-p_A) because A must fail to hit the bull so that B has the chance to throw.

If we now move to the second throw, the probability P_A,2 of A winning is p_A(1-p_A) and for B P_B,2=p_B(1-p_A)².

For the nth throw, P_A,n=p_A(1-p_A)^n-1 and P_B,n=p_B(1-p_A)ⁿ. P_A,n=p_A(1-p_A)ⁿ/(1-p_A)=(P_B,n/p_B)(p_A/(1-p_A)),

P_A,n=P_B,np_A/(p_B(1-p_A)) or P_A,n/P_B,n=p_A/(p_B(1-p_A)). Note that the expression on the right of equals is a constant, independent of the number of throws.

So the ratio is P_A,n:P_B,n=p_A:p_B(1-p_A). If p_B=p_A/(1-p_A), the ratio is 1:1 so P_A,n=½ (evens).

The probability of A winning is p_A/(p_A+p_B-p_Ap_B).