Please correct me on this:
Check on host's logic, when he has to choose between 2 doors and when he doesn't has to.
Car is behind A.
P(win) = P(1/3 Ben chooses A, 1/2 host shows C, 1/2 Ben stays) + P( 1/3 Ben chooses A, 1/2 host shows B, 1/2 Ben stays) + P(1/3 Ben chooses B, 1/1 host shows C, 1/2 Ben switches) + P(1/3 Ben chooses C, 1/1 host shows B, 1/2 Ben switches)
P(win) = P(1/3*1/2*1/2) + P(1/3*1/2*1/2) + P(1/3*1/1*1/2) + P(1/3*1/1*1/2) = 1/2
and:
P(loose)= P(1/3 Ben chooses A, 1/2 host shows C, 1/2 Ben switches) + P(1/3 Ben chooses A, 1/2 host shows B, 1/2 Ben switches) + P(1/3 Ben chooses B, 1/1 host shows C, 1/2 Ben stays) + P(1/3 Ben chooses C, 1/1 host shows B, 1/2 Ben stays)
P(loose) = P(1/3*1/2*1/2) + P(1/3*1/2*1/2) + P(1/3*1/1*1/2) + P(1/3*1/1*1/2) = 1/2
Car has 1/3 chances to be at A; So repet it for car in B and C and you'll get to
P(win)=1/3 * 1/2 + 1/3 * 1/2 + 1/3 * 1/2 = 1/2
P(loose)= 1/3 * 1/2 + 1/3 * 1/2 + 1/3 * 1/2 = 1/2