Prison

Originally posted by XanthosNZ
There are 100 prisoners. In one hour the warden will pick one of them at random and kill that person (I'm assuming no other prisoner knows which prisoner was killed). They also know that tomorrow (and every day from here on) one prisoner will be selected at random and led into a room where they have the option of toggling a light switch. If while in this r ...[text shortened]... t isolated so they must plan a strategy in the next hour.

What strategy should they employ?

How many light bulbs are in the room... Say there were fifty, Each person is assigned a number and a state of the light switch. When in the room they remove/break the number of light bulbs that they were assigned as a number and flip the switch in the correct position.

Prisonner
1 Down, 1 bulb
2. Up, 1 bulb
3. Down, 2 bulbs
4. Up, 2 bulbs
Etc...If this were the case then the prisoners would be free after just two people visiting the room.


But on the downside if there was only 1 bulb I think the max number of people you could assign numbers and state of the light switch would be 5

Prisoner
1. Up, Unbroken
2. Down, Broken
3. Up, Broken
4. Down, Bulb Removed
5. Up, Bulb Removed
Down, unbroken would be the sign that none of the "chosen" ones had been in the room.

Does anyone see a better solution, or is this one flawed extremely bad...

When they have to say if all others have been in the room it goes like this:

Designate a prisoner to be the one to tell the guard that all others have been there. Name him prisoner A.

Prisoner A will turn the light off when he finds it on. And leaves it off when it's off. He remembers how many times he had to turn the light off.

All others leave the light on if they find it on. If they find it off, they turn the light on if they haven't done so earlier. They leave it off otherwise.

Once prisoner A reaches 99, he knows 99 different prisoners have turned the light on. At that moment he knows all prisoners have been there.

The process might be sped up a little if prisoner A isn't fixed beforehand, but designated as the third prisoner to come out of his cell. The first prisoner has to turn the light on. The second prisoner will turn it off, unless it's the same as on day 1. The third prisoner then knows how many different prisoners have been before him, and he can start the count on 1, 2 or 3. Then continuing as above it should save a few days 🙂

Yes. Now find a solution that works even faster. 😉

That plan takes 20-30 years on average, as I recall. It's possible to get down to around 10.