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.

Switch or not Switch? Monty Hall problem_Solution 1

My idea is as follows(I try to avoid some jargons in Bayesian probability theory).

There are 3 balls, A,B,C. The guest picks up one ball, and then host announces one "bad" ball from the remaining two balls since he knows where is the true ball. I will suppose A is the right ball.



After the actions of the guest and the host, which state of the world the guest is in? Image three balls are in a pool, and the guest picks up one from the pool. We shall have:

1. the guest picks up A, and leaving B,C in the pool.

2. the guest picks up B, and leaving A,C in the pool.

3. the guest picks up C, and leaving A,B in the pool.

Let's call these states as (B,C), (A,C), and (A,B) respectively. We have the probability that the guest is in each state is 1/3.

Now, the host names one "bad" ball from the pool, and marks it so that the guest can recognize the bad from the remaing two balls in the pool.



Not Switch case:

Suppose the guest does not switch his choice. If the guest is in (B,C) states, then the conditional probability he picks up the right ball is 1. If the guest is in (A,C) or (A,B) state, then the conditional probability he picks up the right ball is 0. Hence the probability he picks up the right ball is (1/3)*1+(1/3)*0+(1/3)*0=1/3



Switch case:

Suppose the guest switches his choice. If the guest is in (B,C) states, then the conditional probability he picks up the right ball is 0 since any ball in the pool is bad. If the guest is in (A,C) or (A,B) state, then the conditional probability he picks up the right ball is 1 since the only bad ball in the pool is marked. Therefore, the probability he picks up the right ball is (1/3)*0+(1/3)*1+(1/3)*1=2/3, and the probability he picks up the bad ball is (1/3)*1+0+0=1/3.



Bayesian Theorem

Define

event Ci=the ball i is the right ball.

event Hij=the host marks ball j after the guest picks up ball i

event CiHik=conditional on the host marks ball j after the guest picks up ball i, the ball i is the right ball.

Our target

Not switch:Pr(CiHik)

Swithtch: Pr(CjHik)

Switch or not Switch: Monty Hall problem_II

The answer is: the guest should switch his choice, anyway.

The probability the guest wins the lottery if he does not switch his choice is 1/3 and the probability he loses if he does not switch is 2/3

On the other hand, the probability the guest wins the lottery if he switches his choice is 2/3 and the probability he loses is 1/3

The reason I will state later. I have a simple solution for this question.

Now consider a generalization. There are n boxes in front of the guest. Follow the same procedure. What is the probability the guest wins the lottery if he does not swtich his choice and what is the probablity he wins if he switches his choice?

The problem is, is there a intuitive, obvious thought lead to the correct solution?

Switch? or not switch?!

There is a contest in a TV talk show. The guest can only win the lottery if he chooses the correct box which has gift inside from three boxes.
First, the host let the guest choose one box.
Second, after the guest made his choice, the host points out a box which does not have the gift inside from the remaing two boxes.
Third, the host askes if the guest wants to switch his choice.
Switch? or not switch?