THE ULTIMATE RIDDLE PART TWO.

actually i hate to tell you but it has been proven that repeating 9s do in fact equal the limit the number is approaching. look at it this way:

1/3= 0.333333333333333...

3/3= 1

but wait a minute

1/3 * 3 = 3/3

1/3= 0.33333...

0.3333333... * 3 = 0.9999999999....

since 1/3=0.3333333...

1/3 * 3 = 0.999999999...

0.99999999999... MUST EQUAL 1

thank you

when the prisoners are all assembled in the room at the beginning they should overpower the warden. that would be the easiest way out. actually, the first prisoner taken alone to the room should make the call, since every prisoner has already been to the room once in the initial meeting. eat shit warden norton.

Pamela Anderson's Left Nipple wrote:

Try as you like 99 followed by an infinite number of nines will never equal 100.

Actually 99 followed by an infinite number of 9s is EXACTLY equal to 100. It is a mathematical fact.

could they all just make a pile of hair in the room so the first time a prisoner goes in the room he plucks one hair and puts it in the pile

until there's 100

or better yet every prisoner takes a dump in the room

sucks to be #100 though

the designated person ... you jackass!

this will work!

15min wrote:

If I remember from my graph theory class, you can't.

KnowItAll wrote:

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.

Well then my 6th grade math teacher was a total b!tch for giving us that problem. I worked so many damn hours on that thing when I was a kid.

q3rqrr wrote:

or better yet every prisoner takes a dump in the room

sucks to be #100 though

LOL on that!

"when the prisoners are all assembled in the room at the beginning they should overpower the warden. that would be the easiest way out. actually, the first prisoner taken alone to the room should make the call, since every prisoner has already been to the room once in the initial meeting. eat shit warden norton."

What he didn't mention....Rexing is the warden. Think again.

[quote]Pamela Anderson's Left Nipple wrote:

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": [quote]

The original poster did ask us to try and solve the riddle, so you're right; I was trying to solve it albeit not as bad as you think I was.

Take a refresher on the theories of Johann Carl Friedrich Gauss. This problem is solved with the Normal Distribution of Random Events Theory.

Hey nip-child, listen closely. I'd be certain that your penny in a jar scenario would work if you had 100 people chosen to enter the room - randomly, repetetively, whatever - and take a penny out the 1st time he enters. When all pennies are gone (or pieces of bulb moved), all have entered.

If a person enters 99 times - which for some reason throws your brain into mal - then 98 times the person doesn't do a damn thing. Whether or not the number can be "quantified" is irrelevent. If you have to keep waiting, you do. You're not looking for the magic number of entries here. You're just looking for an end result.

Hell, you could have the prisoners drop a turd in the corner (lined up neatly, of course, so as to faciliate future counts) on their 1st visit. 100 piles = 100 prisoners = release.

The simplest solution is usually the best. Don't outthink yourselves, little Einsteins...

q3rqrr wrote:

could they all just make a pile of hair in the room so the first time a prisoner goes in the room he plucks one hair and puts it in the pile

until there's 100

or better yet every prisoner takes a dump in the room

sucks to be #100 though

This is what I get for not reading EVERY post before I type mine. Great minds think alike, apparently.

So nobody has even bothered to read the google link "i know" posted on the first page of this thread? I think it explains the problem and its possible solutions pretty well.

yeah i did, and that's what i based my answer on. screw 12 years, they can be out of there in one day.

15min wrote:

If I remember from my graph theory class, you can't.

Sure you can. The three utility companies are in a three story building.

The way againtocarthage stated the problem, I don't think there is a solution.

Stating that they are in "windowless and soundproof" cells suggests that they have no way of measuring the passage of time. They won't be able to tell if they are the 8th one picked, the 30th one picked, or most importantly, the 1st one picked.

He also stated that the light is initially off, but there is nothing to indicate that the prisoners would know that. How would they?

If the prisoners don't know if the light will be on or off initially when they figure out their plan, AND their designated counter is the 1st one picked, AND he doesn't know he's the 1st one picked,.....they have a problem.

Just Wondering wrote:

The simplest solution is usually the best. Don't outthink yourselves, little Einsteins...

Call me a little Einstein, but to presume that the warden doesn't remove or add, any or all of pennies, turds, or pieces of broken light bulb over time would be foolish, especially when doing so he could cut costs in running his prison.

Yea the google link at the very beginning of this thread explains all the possible solutions are even gives solutions for different variations of the problem. The problem the initial poster posed is actually a variation, as the original riddle had the prisoners knowing how many days had passed. It takes a long longer to reach 100% certainty than the original riddle.

Why is it the riddles always bring up ridiculous arguments on letsrun?

incrmiles wrote:

you know there is going to be one jackass prisoner with a death wish and the first chance he gets he is going to call the warden and take everyone out with him

it's a moot topic they are all going to die, face it

That's funny.

Pamela Anderson's Left Nipple wrote:

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.

Dude, I'm starting to think you're a troll. There certainly is an instance in which a prisoner can assert with 100% certainty that every prisoner has been inside the room, but, as you point out, it is not guaranteed that this instance will come to fruition. If they don't all make it into the room, then they will never be able to confirm; but if they do all make it into your room, they will be able to confirm.

To simplify: your argument is akin to saying that, because there is a chance I will never win the lottery, there is never an instance where I can assert with 100% certainty that I have won the lottery. I probably won't be able to claim that I've won the lottery in my lifetime (especially because I don't gamble), but if I DO win the lottery, well, then it will be pretty clear. Does that help?