To switch or not to switch_Monty Hall problem_Generation and intuition

Now consider the variation when there are 4 balls. The game is proceeded following the same rules. Let the ball be A,B,C,D. The true ball is A. The state space is {(A,B,C),(A,C,D),(A,B,C),(B,C,D)}. Follow the same logic.
Not Switch:
The probability of picking up the true ball is
(1/4)*0+(1/4)*0+(1/4)*0+(1/4)*1=1/4
Switch:
The probability of picking up the right ball is
(1/4)*(1/2)+(1/4)*(1/2)+(1/4)*(1/2)+(1/4)*0=3/8.

Moreover, consider the N generation of this game. There are N balls={1,2,...,n}. We have
Not switch:
The probability of picking up the true ball is 1/N.
Switch:
The probability of picking up the true ball is
(1/N)*(1/N-2)*(N-1)+0=N-1/[N(N-2)]

More generally, there are N balls, and the host marks P bad balls after the host picks up one ball.
Not switch: 1/N
Switch:
(1/N)*(1/N-1-P)*(N-1)+0=(N-1)/[N(N-1-P)]
This answer is the same with D. L. Ferguson (1975 in a letter to Selvin cited in Selvin 1975b), see http://en.wikipedia.org/wiki/Monty_Hall_problem#refSelvin1975b.