Adam and Brian alternate shooting free throws. Whoever makes one first wins the game. Adam always goes first (!)

Adam makes X% of his shots.
Brian makes Y% of his shots.

How often does Adam win? (expressed in terms of X and Y)

A little clearer---Adam and Brian alternate shooting free throws. Whoever makes one first wins the game. Adam goes first (!)

Adam rates to make X% of his shots.
Brian rates to make Y% of his shots.

How often does Adam win? (expressed in terms of X and Y)

A simple method for this. Consider only one shot for each player. Adam will hit his shot with probability X. If Adam missed, Brian will win with probability Y(1-X). If both miss then we repeat the process from the starting state. This happens with probability (1-X)(1-Y). We know that the sum of these three probabilities must be 1 since these exhaust all possibilities.

Now realize that since both missing restarts the game, we can ignore those results and simply calculate how often each one wins as a fraction of games that end with the first shot. The probability that the games ends on one of the playersÂ’ first shots is 1 - (1-X)(1-Y)
= X + Y - XY.

Now we therefore get that Adam wins with probability X/(X+Y-XY) and Brian wins with probability Y(1-X)/(X+Y-XY). As a sanity check, note that the numerators in these two expressions X and Y(1-X) give X+Y-XY when added, which is equal to the denominator. Hence the sun of Adam and BrianÂ’s win probabilities is 1 as it must be.