Riddle me this one. Completely OT!

Also, it is questionable that IF a person accepts the "challenge" of the original "you" and intimately becomes the person on the show, that at that point, is he/she privy to the remainder of the info you provide, including what the host knows. IOW, is the info about what the host knows for the benefit of the nonparticipating reader or the person who becomes the "you" on the show.

LOL

If you know the game show host knows, you switch, but if you don't know if he knows.  It doesn't matter...So it doesn't matter if the host knows, it matters if the guesser knows the host knows.

Wrong again.  It only matters if the host knows.  As long as the host knows, it is in the guesser's best interest to switch (whether the guesser knows the host knows doesn't change the probability). 

So if you MEANT the definite "you" rather than the indefinite "you", you didn't explain that well enough. 

Regardless of which "you" it is the answer doesn't change.  YOU (the guy typing on the keyboard) are the one deciding if it is better for YOU (the guy staring at Bob Barker's Babes) to switch. 

Smokey is on a game show and there are three doors, one of which there is a car behind. Smokey doesn't know which one, and doesn't know if the host knows. the host reveals there is nothing behind door two. if the host doesn't know, smokeys probability of winning the car is 1/2, if the host does, its 2/3. its smarter for smokey to switch based on this, but that doesn't guarantee the probability is 2/3.

i know the host knows, but it was unclear in the question whether or not smokey knew if the host knew.

A warden meets with 23 prisoners. He tells them the following:

- Each prisoner will be placed into a room numbered 1-23. Each will be alone in the room, which will be soundproof, lightproof, etc. In other words, they will NOT be able to communicate with each other.

- They will be allowed one planning session before they are taken to their rooms.

- There is a special room, room 0. In this room are 2 switches, which can each be either UP or DOWN. They cannot be left in between, they are not linked in any way (so there are 4 possible states), and they are numbered 1 and 2. Their current positions are unknown.

- One at a time, a prisoner will be brought into room 0. The prisoner MUST change one and only one switch. The prisoner is then returned to his cell.

- At any time t, given some N> 0, there exists a finite t_0 by which time every prisoner will have visited room 0 at least N times. (in other words, there is no fixed pattern to the order or frequency with which prisoners visit room 0, but at any given time, every prisoner is guaranteed to visit room 0 again). (If you're still confused by this statement, ignore it, and you should be ok).

- At any time, any prisoner may declare that all 23 of them have been in room 0. If right, the prisoners go free. If wrong, they are all executed.

What initial strategy is 100% guaranteed to let all go free?

As a previous poster wrote, apparently from a statistical standpoint, it is critical to the problem whether the game show participant knows whether the host knows.

Again, whether the participant knows the host knows doesn't matter at all for the sake of answering the riddle.  You (the person reading the riddle) know the host knows, so you are supplied with all the information you need to determine which decision is to the advantage of the game show participant.

lets move on and keep having some riddle fun. theres still an unsolved riddle on the board.

Prisoner 23 is the only one who is allowed to turn switch #1 OFF.

All other prisoners may turn switch #1 ON only one time.  Once they have done this, they are to toggle switch #2 on or off.

Prisoner 23 turns off switch #1 every time he sees it turned on and counts how many times he has done this.

When Prisoner 23 has counted switch #1 being turned on 22 times, all prisoners have been in the room.

(you're evil, badger).

nonetheless, i am very impressed.