Originally Posted by jhan523

OMG! I understand! That is not simple Greg. Definitively not an "ooooooooooooooo" moment >.>

How about and "ahhhhhh" moment.

LMAO someone actually solved it

Originally Posted by PTGregD

Joseph got it, so I'll explain it a bit clearer:

My idea is a *very slight* variation of Joseph's solution, just changing the beginning to make it easier to understand.

So:

During the meeting before hand, the prisoners elect one person to be the leader; Let's call this person Greg.

Now, everyday the warden calls a prisoner in. If it is the prisoner's first time, they turn the light on. If the light is already on, they just leave it that way, but don't count it as their first time (basically pretending that they never were in the room).

Eventually, Greg will be called to go into the room. Unless Greg was the first person called, the light will be on. Greg will switch the light off and in his head keep track that a new prisoner had entered the room.

This continues until every single prisoner has entered the room and had a chance to flick the lightbulb on. Eventually, Greg will have turned off the light 99 times, meaning that all 99 prisoners and himself have been in the room.

That is assuming that he can keep a count for several months if not years. Very impractical. I would opt for the finger up the bum solution I proposed.

Originally Posted by PTGregD

How about and "ahhhhhh" moment.

No, not an "aaaaaaaaaah" moment either. Lol

Edit: Unless the "aaaaaaaaaaaaaaaaaah" is out of anger and disgust toward the very impractical riddle.

Originally Posted by ytpos

:O I get it! Except, holy crap that would take a long time. They might as well just wait out their sentences.

With a pretty strong statistical certainty, it would take about 26 years. That being said, there are alternate solutions that have been proposed that rely on probability to bring the time it takes to 5 years. But in those cases, the prisoners aren't 100% accurate; they are 99.999% accurate.

How do you even solve those types of riddles? :S It was pretty hard...

Originally Posted by jhan523

That is assuming that he can keep a count for several months if not years. Very impractical. I would opt for the finger up the bum solution I proposed.

You're in solitary confinement! What else is there to think about?

Originally Posted by Matthew K

How do you even solve those types of riddles? :S It was pretty hard...

Well this was a pretty tough one; like I said, it's rated in the hardest category by Mensa.

But I mean, in the end, all you needed to do was formulate a logical system that the prisoners could use.

Originally Posted by jhan523

No, not an "aaaaaaaaaah" moment either. Lol

Edit: Unless the "aaaaaaaaaaaaaaaaaah" is out of anger and disgust toward the very impractical riddle.

lmao. hahaha

Originally Posted by PTGregD

With a pretty strong statistical certainty, it would take about 26 years. That being said, there are alternate solutions that have been proposed that rely on probability to bring the time it takes to 5 years. But in those cases, the prisoners aren't 100% accurate; they are 99.999% accurate.

Yeah 26 years isn't long, atleast not in Canada. You can get parole after 25 years for a life sentence in Canada. So you can theoretically be out before the stupid counting is done.

I still don't get how you could say 26 years. You can cycle through 98 people (the leader excluded) for eternity.

Originally Posted by PTGregD

You're in solitary confinement! What else is there to think about?

Solitary confinement for several years without communication to anyone would cause some serious brain damage. You become manic.

That would suck if Greg lost count.

Originally Posted by jhan523

I still don't get how you could say 26 years. You can cycle through 98 people (the leader excluded) for eternity.

Because the warden chooses @ random and not out of spite (thereby never choosing all 100 prisoners at least once).

Because it's random, you can use statistical distribution probabilities to create a distribution that will tell you how long it will take for the leader to be in the room enough times. And that's approximately 26.2 years.

Originally Posted by PTGregD

Because the warden chooses @ random and not out of spite (thereby never choosing all 100 prisoners at least once).

Because it's random, you can use statistical distribution probabilities to create a distribution that will tell you how long it will take for the leader to be in the room enough times. And that's approximately 26.2 years.

And what's the standard deviation? It must be pretty big...

Originally Posted by jhan523

And what's the standard deviation? It must be pretty big...

The variance on the problem is approximately 2 years. So with pretty good confidence, you're looking @ a range of 24.2-28.2 years for you to be set free.

But like I said, there are some more complicated methods of solving, some which settle for 99.999% accuracy to get out in 5 years, and some which settle for 100% accuracy and can get out in 9-11 years. I just gave the simplest answer so that more people would understand it.