Answer:
Player A has a probability 2/3 of making each of her shots, then she has a probability 1/3 of missing each shot.
Player B has a probability 1/2 of making each of his shots, then he also has a probability 1/2 of missing each shot.
Let's separate each case.
Let's define:
P(x) = probability of winning at the "x" shot.
Player A wins on the first shot.
Because she has a probability 2/3 of making each of her shots, the probability of winning at the first shot is
P(1) = 2/3
Now let's see the next case, player A wins at her second shot.
This means that first, she misses her first shot, with a probability of:
p₁ = 1/3
Player B must miss his shot, the probability is:
p₂ = 1/2
Now player A must make her shot, so the probability is:
p₃ = 2/3
The joint probability is the product of the individual probabilities, so we have:
P(2) = (1/3)*(1/2)*(2/3) = 1/9
Now we can see the pattern, for P(3) we have
A misses: p₁ = 1/3 (first shot of A)
B misses: p₂ = 1/2
A misses: p₃ = 1/3 (Second shot of A)
B misses: p₄ = 1/2
A makes the shot: p₅ = 2/3
P(3) = (1/3)*(1/2)*(1/3)*(1/2)*(2/3) = 1/54
We already can see the pattern.
P(n) = (1/3)^(n - 1)*(1/2)*(n - 1)*(2/3)
Player A has a probability P of winning, and we can write P as:
P = P(1) + P(2) + P(3) + ...
Then we will have:
P = 2/3 + 1/9 + 1/54 + 1/324 + ... ≈ 0.8
