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Click and Claps, the famed Tapit brothers of the PBS program CarTalk, have put forth a real brain twister of a puzzle.
So far no luck with me, I'm still trying, but my ideas are getting more and more farfetched. 
Here is the puzzle...New Puzzler: Prison Switch ScenarioRAY: I got this from my pal, Stan Zdonik, who teaches at Brown University. It was given to him by a colleague who failed to provide him with the answer -- maybe because he didn't know it?
This puzzle has been making the rounds of Hungarian mathematicians' parties. Here it is... The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another. "In the prison is a switch room, which contains two light switches labeled A and B, each of which can be in either the on or the off position. I am not telling you their present positions. The switches are not connected to anything. "After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell. "No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. "But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time anyone of you may declare to me, 'We have all visited the switch room.' "If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators." What strategy should the prisoners devise? 
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Carpal Tunnel
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Since the prisoners will have not chance to communicate, their only safe strategy is to
agree never to say all have been to switch room. They will never get free, but they will
never get fed to the alligators.
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Maybe they could all visit the switch room before they are quarantined and then he will have to let them go. However, there is a rider to that, the warden says, "If this is true"; how on earth is he going to confirm that? In reality, I think there is no hope. For, whatever strategy they plan, the prisoner picker is the warden and he could go on picking the same guy, or the same ten or the same twenty two, but never ever twenty three. methinks, the puzzle itself evadeth me!
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Carpal Tunnel
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The warden could easily keep records to know when every prisoner had been to switch room.
The prisoners could not.
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i think that the solution is somehow dependent that there is an uneven number of them they should agree on an initial position of the switches, which the 1st prisoner will make, say A and B switched off. discovering this position, the next prisoner changes this to A and B are on. if anybody visits the switch room second, third etc time, he doesn't touch anything... or even more complicated plan A B 1. off off 2. on off 3. off on ..... don't ask me will it work or not. this is the end of the day. i am capable of inventing an insane plane but not seeing its consequences 
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Dear vika: I'm not good at games, but I see no way the prisoners could know initial position,
nor communicate status of switches to others. The warden could send same guy 23 times
and the others would never know.
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vika, such a scheme is not doable: This prisoner will select one of the two
switches and reverse its position. He must move one, but only one of the switches. He
can't move both but he can't move none either.
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ok, this just came to me:
if they agree to only flip the left switch, for example, until someone goes for the second time; then he flips the other switch. they continue to swith that switch until a second person goes twice. then he goes back to flipping the first switch. etc.… they could count the number of times they have reversed. when they get to 23, they would know that they all have gone once.
have I missed something?
formerly known as etaoin...
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OP
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I feel like I'm being led down a primrose path by the creator of this puzzle, but the careful wording below seems critical, but cryptic, wording to me... "But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time anyone of you may declare to me, 'We have all visited the switch room'. "This seems to indicate that the total number of visits (or switches thrown) would have to be either 23, or a multiple of 23, that is, 46, 92, 184, etc. All even numbers, naturally, except 23. Now say, for example, if the prisoners could distinguish between the two switches beforehand * and assign beforehand a separate meaning for the switch position of each...eh??? (odd/even)  Well, you see what I'm getting at. And if you do will you please post it, because I sure as heck don't. * (Post Edit: Of course they can, Milo-you-big-dummy, they are labeled A and B!)  Second Post Edit: Etaoin, how is the counting known to the isolated prisoners? But if, let's say, that if eleven prisoners have pre-agreed to always switch the A switch and the other twelve have agreed to switch the B ???
And remember the prisoners are selected at random (or whim, if you perfer,) and then their name, so to speak, goes back into the hat, which is the crux of the quiz: Randomness tamed by order.
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I'm not sure the problem has a solution as stated. It contains internal inconsistencies.
--I will select one prisoner at random...
--I'm going to choose prisoners at random...
--I may choose the same guy three times in a row, or I may jump around and come back...
--given enough time, everyone will eventually visit the switch room as many times as everyone else...
If he is picking, and can elect not to choose prisoner X ever, it isn't random. And if the problem is inconsistent in its presentation, can it still have a logical answer?
I thought of it might involving odd/even parity, but if any one can visit the room multiple times that won't work...
Can our collective self-esteem wait a week for the answer?
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