Prison

Excellent solution. The only potential problems with it (practically) are as follows.

1) The guard may opt to only select a handful of prisoners ever to send into the room, save 5 of them. In this case, there is a high likelihood nobody would ever take the guess, since I am assuming no contact is made between the prisoners (which would be necessary to avoid someone slipping a note that they were there yesterday).

2) It is possible one or more of the prisoners may get off on their count of days. If that person comes in the day after the light is turned on (or off), their guess could be wrong, and the result bad. If it is one of the two, then they may make a mistake and either turn on the light when it's not his day, or leave it be when he should turn it on.
The former could be disasterous, and the latter a delaying move.

All in all, still a darn good solution though.

Originally posted by TheMaster37
This is not the problem first stated. However, I think this might work;

Two prisoners are labelled A and B. All prisoners need to remember who A and B are. All prisoners need to keep track of the days, more precisely if the number of days passed since this day is odd or even. I assume the light is off at the start.

Prisoner A turns the light on on ...[text shortened]...
If the light was on at the start, the prisoners will have to turn the light off, instead of on.

What if A gets sent into the room on an odd day and B on an even day???

Also it's the prisoner who toggles the switch that has to say who was in there the previous night.

Originally posted by luskin
Also it's the prisoner who toggles the switch that has to say who was in there the previous night.

Yes, I couldn't find any way around that other than the plan being to wait until a prisoner was selected twice in a row (as it was stated the selection was random, it should happen eventually)

I never thought of that. In that case there is no need for any labelling. The only plan they need is that nobody touches the switch until someone(anyone) gets selected twice in a row. Then on the second night he toggles the switch and "guesses" that he himself was there the previous night.

Originally posted by geepamoogle
Excellent solution. The only potential problems with it (practically) are as follows.

1) The guard may opt to only select a handful of prisoners ever to send into the room, save 5 of them. In this case, there is a high likelihood nobody would ever take the guess, since I am assuming no contact is made between the prisoners (which would be necessary ...[text shortened]... disasterous, and the latter a delaying move.

All in all, still a darn good solution though.

1 He can't. He picks prisoners randomly,so there is no possibility of choosing,say 10 of them, and keeping rest in cells.

2. I don't suppose that if you would have nothing better to do than count days,and you would have known that your live depends on it you would make a mistake. 😛

What if A gets sent into the room on an odd day and B on an even day???

Nothing,they just leave the room like any other prisoner,and wait for their day to come.

Has this similar problem ever been posted before?

"There are one hundred prisoners about to go into solitary confinement. Each day after today, one prisoner will be let into a room with a lightbulb, which is initially turned off. The prisoner in the room can turn the lightbulb on or off as he wishes. He can also announce to the guard that all one hundred prisoners have been in the room before. If he is correct, everyone will go free, and if he is incorrect, all the prisoners will be killed. The prisoners have decided that they are not willing to accept any risk of guessing wrong and being executed. What plan should the prisoners adopt to minimize the expected time they will remain imprisoned?"

Originally posted by Choreant
This one is nice,however might be known to You...

There is a prison with 100 captives in the solitary cells. Every night the guard will pick one of the prisoners(at random) and allow him to spend a night in a living room with a lightbulb(which is not visible from the cells). Prisoner can toogle the bulb,if he wishes,but if he does,he must guess who spend ...[text shortened]... night,one of them(picked intentionally by the guard) will be killed.

What shalll they do ???

You could also just use the bulb to burn a part of your body. Each prisoner is assigned a part of the body so just memorize which person is supposed to burn which part of their body.

Originally posted by GregM
Has this similar problem ever been posted before?

"There are one hundred prisoners about to go into solitary confinement. Each day after today, one prisoner will be let into a room with a lightbulb, which is initially turned off. The prisoner in the room can turn the lightbulb on or off as he wishes. He can also announce to the guard that all one hundred pr ...[text shortened]... at plan should the prisoners adopt to minimize the expected time they will remain imprisoned?"

This answer allow prisoners to leave,but not so fast,I presume, so if the question is really how to minimize time in jail,it might not be that.

However...

They divide time in 100 days periods. During every period no one touches the bulb,until one is in a bulbroom for a second time(within the period). Every prisoner in the room, at the last day of the period,checkes the bulb. If it is off,he annouces that all prisoners spend a night there,if the bulb is on he must switch it off,thus begging the next period. Finally,they shall find the bulb off and leave safely.

Originally posted by uzless
You could also just use the bulb to burn a part of your body. Each prisoner is assigned a part of the body so just memorize which person is supposed to burn which part of their body.

That won't work. They can't see each other, so burning one's body is useless .

Originally posted by Choreant
This answer allow prisoners to leave,but not so fast,I presume, so if the question is really how to minimize time in jail,it might not be that.

However...

They divide time in 100 days periods. During every period no one touches the bulb,until one is in a bulbroom for a second time(within the period). Every prisoner in the room, at the last day of the ...[text shortened]... itch it off,thus begging the next period. Finally,they shall find the bulb off and leave safely.

Indeed, that will take a very, very long time. It requires there to be a hundred-day period in which each prisoner is selected with no repeats. That is an extremely unlikely occurrence, and they wouldn't make it out within a billion years.

There are much faster ways. 🙂

one thing,is the light of the bulb visible from the cells???

Originally posted by Choreant
one thing,is the light of the bulb visible from the cells???

No.

The prisoners agree to select 1 person. When he gets picked he spends the entire time quickly switching the light on and off till it breaks. If a prisoner is selected and the light is working they say nothing and everyone lives. If the light is broken they guess the agreed guy and everyone is free.
Could take a while, but they will get out eventually, noone gets killed. Of course if a prisoner is ever selcted twice in a row they guess themselves and are freed