The probability PA,1 of A winning the first throw is pA, and the probability PB,1 of B winning the first throw is pB(1-pA) 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 PA,2 of A winning is pA(1-pA) and for B PB,2=pB(1-pA)2.
For the nth throw, PA,n=pA(1-pA)n-1 and PB,n=pB(1-pA)n. PA,n=pA(1-pA)n/(1-pA)=(PB,n/pB)(pA/(1-pA)),
PA,n=PB,npA/(pB(1-pA)) or PA,n/PB,n=pA/(pB(1-pA)). Note that the expression on the right of equals is a constant, independent of the number of throws.
So the ratio is PA,n:PB,n=pA:pB(1-pA). If pB=pA/(1-pA), the ratio is 1:1 so PA,n=½ (evens).
The probability of A winning is pA/(pA+pB-pApB).