Hang on, the cells clearly have no walls! Since you never said they did. They can just walk away. Dumb guards. :D
It was not the guards, but rather the Organians, who came down and explained the rules.
In Urdu and in Chocktaw? Or only in Basque?
One prisoner need only promise the Organians a glass of water, whereupon he can choose the Big Gun or the Good Package. If he gets the gun, well, he can easily destroy the prison. The Good Package doesn't free anyone, but it saves the world from destruction, so there's that. As long as disgruntled CIA employees dressed as clowns don't catch on, the plan is foolproof... especially if one of the prisoners is really John Ritter.
RPG Superstar 2009 Top 8
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Randall Munroe, the creator of the xkcd webcomic, maintains a wiki of very clever puzzles. This is one of my favorites, because the answer is difficult to arrive at, but blatantly obvious in retrospect. Which is, I suspect, a plausible operating definition of "genius".
100 people are being held prisoner in a jail. They are told that in one hour, they will all be taken to separate windowless, soundproof cells. One at a time, and in a random order, they will be taken from their cells, interrogated, and then sent back to their cells. All interrogations will take place in the same room, which contains one light bulb and the switch that operates it. The light is initially off, but the inmates are free to toggle the switch as often as they want, whenever they are in the interrogation room, and the prison guards will not toggle the switch at all. No prisoner can see the light from his cell. Only one prisoner is interrogated at a time, each prisoner can be interrogated multiple times, and they have no way of communicating besides the light switch. The length and amount of time between interrogations is random, so no help there.
At any time, any prisoner under interrogation may state, "Everyone has been interrogated at least once." If this statement is true, everyone will be released. If it is false, all of the prisoners will be executed.
The prisoners have one hour to work out their strategy before they're isolated for good. How do they get released?
Note: The selection process for interrogations is random and fair; some prisoners may be interviewed multiple times before another prisoner is interrogated at all, and after any point in time, every prisoner will be interrogated an infinite number of times more.
The more I think about this one, the more I think it is impossible as written. You could say that the first prisoner to enter the room turns the light on, if it is not already on, and then becomes the 'counter' and that subsequent prisoners turn the light off the first time they see the light on and do nothing all other times. The counter then announces to the guards "Everyone has been interrogated at least once" when he has turned the light on 100 times. This assumes that the first prisoner knows he is the first prisoner. Since length of interrogations are random and we have no guarantee that the guards begin calling prisoners immediately, the first prisoner cannot know that he is the first prisoner.
I just don't see a 100% certainty being possible. I give up.
EDIT: Unless, on the first day, any prisoner entering the room turns the light off and leaves it off. The prisoners do not count visits on the first day. The first prisoner entering the room on the second day turns it on and becomes the counter. This assumes the prisoners know what time it is or can see the sun come up. Since you said the rooms are windowless, that wouldn't work either. The first prisoner still cannot be sure they are the first prisoner.
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RPG Superstar 2010 Top 16
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Hi, Tarren.
The prisoners need to decide ahead of time, during their hour-long consultation, which of them is going to be the Counter. (Let's call her Prisoner #1.) All the other prisoners are going to try to signal the Counter that they've been interrogated at least once.
Prisoner # 12 - turns the light on.
Prisoner # 38 - leaves the light alone.
Prisoner # 39 - leaves the light alone.
Prisoner # 01 - receives the signal from someone (Prisoner #12, but she doesn't know that), and turns the light off.
Prisoner # 08 - turns the light on.
Prisoner # 85 - leaves the light alone.
Prisoner # 16 - leaves the light alone.
Prisoner # 06 - leaves the light alone.
Prisoner # 01 - receives the signal from Prisoner #8, and turns the light off.
Prisoner # 12 - has already signaled, so leaves the light alone.
Prisoner # 67 - turns the light on.
Prisoner # 51 - leaves the light alone.
Prisoner # 60 - leaves the light alone.
Prisoner # 01 - receives the signal from Prisoner 67, and turns the light off.
Prisoner # 67 - has already signaled, so leaves the light alone.
Prisoner # 12 - has already signaled, so leaves the light alone.
Prisoner # 01 - enters the room and the light is still off. She leaves the light alone, because she hasn't received a new signal.
Prisoner # 09 - turns the light on.
Prisoner # 95 - leaves the light alone.
...and so on
Eventually, Prisoners 2 through 100 will have had the opportunity to signal the Counter, at which point she makes her announcement.
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RPG Superstar 2009 Top 8
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Hi, Tarren.
The prisoners need to decide ahead of time, during their hour-long consultation, which of them is going to be the Counter. (Let's call her Prisoner #1.) All the other prisoners are going to try to signal the Counter that they've been interrogated at least once.
Prisoner # 12 - turns the light on.
Prisoner # 38 - leaves the light alone.
Prisoner # 39 - leaves the light alone.
Prisoner # 01 - receives the signal from someone (Prisoner #12, but she doesn't know that), and turns the light off.
Prisoner # 08 - turns the light on.
Prisoner # 85 - leaves the light alone.
Prisoner # 16 - leaves the light alone.
Prisoner # 06 - leaves the light alone.
Prisoner # 01 - receives the signal from Prisoner #8, and turns the light off.
Prisoner # 12 - has already signaled, so leaves the light alone.
Prisoner # 67 - turns the light on.
Prisoner # 51 - leaves the light alone.
Prisoner # 60 - leaves the light alone.
Prisoner # 01 - receives the signal from Prisoner 67, and turns the light off.
Prisoner # 67 - has already signaled, so leaves the light alone.
Prisoner # 12 - has already signaled, so leaves the light alone.
Prisoner # 01 - enters the room and the light is still off. She leaves the light alone, because she hasn't received a new signal.
Prisoner # 09 - turns the light on.
Prisoner # 95 - leaves the light alone.
...and so onEventually, Prisoners 2 through 100 will have had the opportunity to signal the Counter, at which point she makes her announcement.
How does prisoner 1 know that the light was not already on and that he is not in fact the first prisoner to enter the room? Maybe the guards have just been carefully rereading John Yoo's memo on Military Interrogation of Alien Unlawful Combatants Held Outside the United States to make sure they didn't go overboard.
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RPG Superstar 2010 Top 16
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All interrogations will take place in the same room, which contains one light bulb and the switch that operates it. The light is initially off, but the inmates are free to toggle the switch as often as they want, whenever they are in the interrogation room, and the prison guards will not toggle the switch at all.
Somewhere on the first page is the follow-up question: "what do the prisoner's do if they don't know the initial value of the light?" The answer is a little more complicated.
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RPG Superstar 2009 Top 8
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All interrogations will take place in the same room, which contains one light bulb and the switch that operates it. The light is initially off, but the inmates are free to toggle the switch as often as they want, whenever they are in the interrogation room, and the prison guards will not toggle the switch at all.Somewhere on the first page is the follow-up question: "what do the prisoner's do if they don't know the initial value of the light?" The answer is a little more complicated.
Okay, now I've read the spoilers on the first page and I'm still not seeing 100% certainty if they don't know the initial value of the light. Even if they do count to 198 or 297 or 396 etc., there is still a small chance that one of those counts is for the initial value of the light. The longer you wait, the less uncertainty you have but it is always there.
