Very cool tough riddle site

[theeye]

Actually, I'd call them puzzles rather than riddles

A couple of examples:

this is a magic trick performed by two magicians, A and B, with one regular, shuffled deck of 52 cards. A asks a member of the audience to randomly select 5 cards out of a deck. the audience member -- who we will refer to as C from here on -- then hands the 5 cards back to magician A. after looking at the 5 cards, A picks one of the 5 cards and gives it back to C. A then arranges the other four cards in some way, and gives those 4 cards face down, in a neat pile, to B. B looks at these 4 cards and then determines what card is in C's hand (the missing 5th card). how is this trick done?

100 prisoners are imprisoned in solitary cells. Each cell is windowless and soundproof. There's a central living room with one light bulb; the bulb is initially off. No prisoner can see the light bulb from his or her own cell. Each day, the warden picks a prisoner equally at random, and that prisoner visits the central living room; at the end of the day the prisoner is returned to his cell. While in the living room, the prisoner can toggle the bulb if he or she wishes. Also, the prisoner has the option of asserting the claim that all 100 prisoners have been to the living room. If this assertion is false (that is, some prisoners still haven't been to the living room), all 100 prisoners will be shot for their stupidity. However, if it is indeed true, all prisoners are set free and inducted into MENSA, since the world can always use more smart people. Thus, the assertion should only be made if the prisoner is 100% certain of its validity.

Before this whole procedure begins, the prisoners are allowed to get together in the courtyard to discuss a plan. What is the optimal plan they can agree on, so that eventually, someone will make a correct assertion?

[hymie[$]]

The card trick: Four cards can be arranged in 24 ways, which just happens to be one-half of the 48 cards which remain in the deck. To gain that extra one bit, A can hand the packet to B either face up or face down.

[hymie[$]]

The prisoners: In the preparation meeting, each prisoner is assigned a number between 1 and 100, and they begin counting days in repeating cycles of 100. In the simple version, if a prisoner is in the living room on a day which matches his number, he leaves or turns the bulb on, and otherwise he leaves or turns the bulb off. When a prisoner arrives in the living room and sees that the bulb is lit on day N, he knows that prisoner N - 1 has been there. In this way, each prisoner slowly accumulates knowledge of who has been to the living room, and eventually one of them will know that all the others have been there.

For a speedup, once a prisoner knows that some other prisoner numbered N has been to the living room, he can take the part of that prisoner should he himself visit the living room on day N, and turn the light on. Even with this speedup, the universe is likely to expire before the prisoners go free. Perhaps we are talking about imprisoned angels