THE ULTIMATE RIDDLE PART TWO.

Carnivore,

His odds of not being picked the second day are still 99%. It's like the wrong sock problem...the black sock is always placed back in the drawer. N is still 100 the second day.

How do you keep track of time in solitary confinement....

deeprunxc wrote:

You make sure everyone changes the light once. If the prisoner goes into the room and the light is off, and he has never turned it on, then he turns it on. If the light is off, and he has turned it on once already, he leaves it alone. You designate one person the \"On\" person. This is the only person who can turn the light on. Every time he walks into the room and the light is on he will turn it off. Once he turns it off 99 times than every has been in the room.

Why hasn't anyone acknowledged that this solution works?

His chance of not getting picked on the second day is 99% but there is also a 99% chance from the first day (1% each day of getting picked) so it would be .99 to the D power where D = days. So multiple .99 * .99 on the second day giving him .9801 of not getting picked. .99 * .99 * .99 etc to 100 = .36603 of not being picked after 100 days.Strictly math you can never be sure since the guard could pick the same guy for 5 years then pick another guy (that would be one random possibility), that guy will think it is pretty safe to say he is the last one.That seems pretty logical although I could be wrong.

really? wrote:

Carnivore,

His odds of not being picked the second day are still 99%. It's like the wrong sock problem...the black sock is always placed back in the drawer. N is still 100 the second day.

Carnivore, you're on the right track; the answer lies in odds ratios and probability.

It comes down to how nice we want to be to the prisoners. Without doing the math, I'd give them around three years.

The most important variable is the number of times a prisoner is selected to go to the room.

Each year, all things being equal, the odds would indicated that each prisoner is selected at random an average of 3 or 4 times to go to the room (technically 3.65 times; 365 days a year, 1 in 100 chance of being selected any given day).

There is a probablility that some of them were selected zero times in that first year, however, the number decreases over time.

After the third year each prisoner would have been selected an average of 11 times (3.65 x 3 years) to go to the room.

The prisoners agree amongst themselves that on the 11th time any one of them was selected they would turn on the light. Further, they agreed that the next visitor to the room, other that the prisoner who turned on the light, who also visited the room at least 10 times previous would tell the warden that everyone has been to the room.

that's only for one specific person not getting picked.

prisoner 1 is the counter,

he turns light on,

each prisoner turns light off on their second time through and only second time through,

the prisoner 1 turns light back on each time it is on when he comes back in.

He counts the times that the light was turned off,

he can then verify when all prisoners have been through twice.

it will take a while as prisoner 1 will have to go through 100 times to make the count.

sweet wrote:

deeprunxc wrote:

You make sure everyone changes the light once. If the prisoner goes into the room and the light is off, and he has never turned it on, then he turns it on. If the light is off, and he has turned it on once already, he leaves it alone. You designate one person the "On" person. This is the only person who can turn the light on. Every time he walks into the room and the light is on he will turn it off. Once he turns it off 99 times than every has been in the room.

Why hasn't anyone acknowledged that this solution works?

This would work, but it would take an average of 33.3 years for the guy to visit the room at least 99 times. C'mon we're talking about petty criminals here not mass murderers!

I was talking about cumulative odds, not daily odds. Obviously, the daily odds are 99% of not getting picked.

academictech2 wrote:

sweet wrote:

Why hasn't anyone acknowledged that this solution works?

This would work, but it would take an average of 33.3 years for the guy to visit the room at least 99 times. C'mon we're talking about petty criminals here not mass murderers!

It does work, but I did not read the whole thread.

He did offer the solution.

flawed wrote:

How do you keep track of time in solitary confinement....

We\'ve already agreed that there is never 100% certainty that everyone has visited the central room because of the random selection process. So, we must accept an agreed upon level of certainty. For this, we need some way to measure time. There is a way to keep track of time (in an inexact manner).

In the planning phase, the prisoners should make an assumption about the expected life of the bulb (say 2000 hours). If they want to wait three years and 9 months (to reach an acceptable level of certainty), then each prisoner should follow the procedure below.

-Enter the dark room

-Turn on the light for approximately 90 minutes (they can count their heartbeats to keep track of time for that period)

-Turn off the light and spend the rest of the time in darkness

-When a prisoner attempts to turn on the light and it is burned out, then it\'s time to call the warden

If it would make you feel better, the model to determine certainty can incorporate a random distribution of the lightbulbs lifetime.

Carnivore, you're on the right track; the answer lies in odds ratios and probability.

It comes down to how nice we want to be to the prisoners. Without doing the math, I'd give them around three years.

The most important variable is the number of times a prisoner is selected to go to the room.

Each year, all things being equal, the odds would indicate that each prisoner is selected at random an average of 3 or 4 times to go to the room (technically 3.65 times; 365 days a year, 1 in 100 chance of being selected any given day).

There is a probablility that some of them were selected zero times in that first year, however, the number decreases over time.

After the third year each prisoner would have been selected an average of 11 times (3.65 x 3 years) to go to the room.

The prisoners agree amongst themselves that on the 11th time any one of them was selected they would turn on the light. Further, they agreed that the next visitor to the room, other that the prisoner who turned on the light, who also visited the room at least 10 times previous would tell the warden that everyone has been to the room.

deeprunxc wrote:

You make sure everyone changes the light once. If the prisoner goes into the room and the light is off, and he has never turned it on, then he turns it on. If the light is off, and he has turned it on once already, he leaves it alone. You designate one person the "On" person. This is the only person who can turn the light on. Every time he walks into the room and the light is on he will turn it off. Once he turns it off 99 times than every has been in the room.

sweet wrote:Why hasn't anyone acknowledged that this solution works?

Because it doesn't make any sense. Try reading and absorbing what you have read. Deeprunxc wrote "If the prisoner goes into the room and the light is off, and he has never turned it on, then he turns it on" directly clashes with "You designate one person the "On" person. This is the only person who can turn the light on".

That is why it doesn't work.

C'mon already. Most Google references to this problem indicate it was posed around 2003. If no one has come up with a published solution in three years, why is it anyone on letsrun thinks they can nail it in a few hours?

Pamela Anderson's Left Nipple wrote:

Most Google references to this problem indicate it was posed around 2003. If no one has come up with a published solution in three years, why is it anyone on letsrun thinks they can nail it in a few hours?

Sad state of intellectual affairs when one refers to Google as the gold standard.

my solution works,

each prisoner turns the light off ONCE and only once,

the counter counts how many times he has had to turn light back on.

Someone feeling slighted? It's too bad that the best you could come up with is still only a guess and not 100% certainty.

I know that you soooooooo badly want to solve this problem, but there is no solution. Chew on this "gold standard":

DUE TO RANDOM SELECTION, IT IS POSSIBLE FOR THE WARDEN TO PICK THE SAME PRISONER, EVERY SINGLE DAY, UNTIL ONE OR ALL DIE. IT IS POSSIBLE THAT NO OTHER PRISONERS WILL EVER SEE THE INSIDE OF THAT ROOM.

As long as that is possible, then there is NEVER an instance where a prisoner can assert with 100% certainty that every prisoner has been inside the room.

Please, cease being so dense.

True, but the problem there is that if the group gets unlucky, it could wait 70 years or more waiting for one unlucky guy to make it in there, at which point most of them would be dead of old age. The problem asks which solution should be used that they all can agree upon. Most would leave in under four years (with 1:1,000,000 odds of dying) versus wait to be sure (could be anywhere from 101 days to an infinite number of days).

the real NK wrote:

my solution works,

each prisoner turns the light off ONCE and only once,

the counter counts how many times he has had to turn light back on.

Nope. If a prisoner enters the room for his first visit and the light is already off, how does he turn it off again?

A riddle I got in grade school that I never found an answer for.

There are 3 buildings that provide utility services. Gas, Electrical, Water. There are 3 houses. You can not cross gas, electrical or water lines. How do you get all 3 services to all 3 houses.

This is more of a graphical riddle, so may not be easy to propose solutions on here, but I've never been able to solve the b!tch.