I took it from 2+2, and it looks interesting.

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Ok, so this isn't THE prisoner dilemma, but it is a problem in which prisoners face a dilemma. I got this from my friend, who was asked this problem in some job interview.

You and 30 other prisoners (the # doesn't matter) are being held in a common jail cell. You know that shortly you will all be taken to solitary confinement, at which point you will no longer be able to communicate, but you are free to discuss strategy now.

You also know that after you are separated, the guards will sporadically take individuals to a room with two switches, labeled "switch A" and "switch B," and that each switch is either in the "up" or "down" position, and that right now they are both down. When you are taken to this room, you have to switch exactly one of the switches, and then you go back to solitary.

You don't know the order that the guards will take you; you might visit the room 10 times in a row while everyone else sits in their cell, or you might sit in your cell while the guards cycle through everyone else 1000 times, etc. (I.e., assume the guards act randomly in choosing who visits the room next.)

Your task as a group is to at some point have someone declare factually that every prisoner has visited the room. For the sake of argument, say that if someone guesses that everyone has visited, and that person is wrong, then you all are put to death, whereas if that person is correct, then you all go free.

So what is your group's strategy?

I think what you do is you assign a value to one of the switch positions.

Using statistics, you figure out what the probability is that all have visited the room if one has visited the room X times.

For example, if you have 30 people, maybe the odds that are once 1 person has been in the room 20 times it is 99.9% that all have been in once. (disclaimer, I have not done the math, I am just throwing out an example)

So, once 1 person has been in the room 20 times, he/she puts both swtiches in the up position. The next prison declares that thay have all been in the room.

All live 99.9% of the time.

I assure you that there is a definite solution to this problem.

Brag: It took me about 5 minutes to figure it out.

Edit: It's not a statistical one like TripleB proposes. Though I would like to see the details of his suggestion . I am not sure that such a setup would work.

I just got out of the shower and am not even dressed yet, but thinking about it for a minute I think it has something to do with one switch being designated as a "new person" switch, that is only flipped on peoples first visit to the room. The other switch is for people who have already been to the room once and have already flipped the "new person switch."

So one person (leader) is in charge of counting and that person tells the warden once each has visited.

You leave switch A as a counter. Each person will be instructed that the first time they enter the room to turn switch A up once when they are in the room when switch A is in the down position. Every one else will just change switch B. No one else but the leader will change the switch down (or off). Once the leader goes into the room and sees switch A up, he should switch it down and keep track of how many times he does it. He should then know factually once all the others have switched the switch.

Sound about right?

I had the same idea - about leaving one switch for first time switches and the other for every other time but couldn't make a connection to a logical answer.

Mac's sounds right.