Logic Puzzles

[simulater10]

Originally Posted by pntballa18

stick it in her but and pee.

The warden, of course.

seconded

[PsYcHoZeRoSk8eR]

Everyone goes in at once and no one leaves until everyone has their number. Unless it was stated somewhere that this is not a possibility.

[Doug31]

Ok, we all give up. What's the answer? Most likely that it's impossible, or there's something you really didn't tell us.

[MiniNeo]

Freaking.

I thought of a way that involves rearranging the ID cards but that doens't really work since the room resets =|

[Shashakiro]

ok come on kit, if this took you two hours no one here is going to get it at all unless like AP or Reach come along and really try on it.

also I've never heard of "group theory" hahaha

[Self_Indulging_Prick]

i've got a question can they all check a box and then all check another box that way when they all check two boxes they should get rougly 2 our of the hundred without losing anybody and contionue on so forth...

[Self_Indulging_Prick]

damn im tired i mean roughly 4 i think you would have 4900 chances with the number of chests gradually lowering with a one in a hudnred chance at first...hmm

[FalcoLombardi]

this warden seems a little troubled to me

reminds me of Belick from prison break

[Self_Indulging_Prick]

4950 i suppose.. and its 2 boxes...

[Self_Indulging_Prick]

by the time you get half way through you should with the luck of prbability have the odds go up 50% so it would take much less tries

[Self_Indulging_Prick]

i forget how to calculate the exponential gain here but at first you'll slowly gain your ratio and loose less turns and get more chests right by the end you shouldnt lose as many and i think the odds actually arent too bad considering doing it this way if its allowed by the rules would make it so no one gets eliminated till the end... i wish kit was still online... i shoulda typed this stuff i woke up a little anyway ima whip that white **** that accumalates on the corner of your lips when your thirsty....

[Kit-]

Maybe I should rephrase.
Each inmate gets to choose fifty boxes. Once the first inmate goes in, as far as any one inmate is concerned, every other inmate has vanished off the face of the earth. [hint size="heug"]Thus, it doesn't matter if they all go at the same time, as long as they all get 100 separate, identical rooms.[/hint] How should each prisoner choose his/her fifty boxes?
So no more guesses about how they can pass each other information.

[Self_Indulging_Prick]

ok, i suppose i would chose a few boxes in varied locations and take the data in on what numbers and where and match it and make an algarythym or whatever taht word is or should be cuz with probability you can figure out by where it isnt where it should be specially by the kinda numbers you find else where kinda doubt this what you mean either though... but im bored

[Jammer325]

so, does every inmate see 100 boxes in the room even if the guy before was correct?

Edit: Have everyone pick a different combination of boxes so no 50 picked between all 100 people are the same. Don't know if this would get up to 30%, but it would increase the odds.
Edit2: **** group theory. I'm agreeing with shash on this one. I'm guessing less than 10 people on these forums even know what group theory is.

[Kit-]

Oh whatever.
Strategy for each inmate:
First box: the box number corresponding to his/her ID number.
Each box after that: the box number corresponding to the ID number found in the previous box.
When does this succeed? Only when there are no cycles of order 51 or higher, which is roughly 30.6% of the time. If there is a cycle of order 51 or higher, each person whose ID is in this cycle fails. The first case obviously has a probability of 50%, with the chance gradually decreasing from there until it reaches 30.6% at 50 people, because there is guaranteed to be at least one person in a cycle of 51 if there is one present. The cool thing is that even if you have 5,000 people sharing 100 ID numbers, the chance doesn't go down.

Someone else post one.

[JTOTHEACKIE2]

well if the warden gets shot they can all be freed!!

[Rolenquin]

I think you guys will like this one I got it off the http://www.ocf.berkeley.edu/ website it's easier than the last too

10 straight-jacketed prisoners are on death row. Tomorrow they will be arranged in single file, all facing one direction. The guy in the front of the line (he can't see anything in front of him) will be called the 1st guy, and the guy in the back of the line (he can see the heads of the other nine people) will be called the 10th guy. An executioner will then put a hat on everyone's head; the hat will either be black or white, totally random. Prisoners cannot see the color of their own hat. The executioner then goes to the 10th guy and asks him what color hat he is wearing; the prisoner can respond with either "black" or "white". If what he says matches the color of the hat he's wearing, he will live. Else, he dies. The executioner then proceeds to the 9th guy, and asks the same question, then asks the 8th guy ... this continues until all of the prisoners have been queried.

This is the night before the execution. The prisoners are allowed to get together to discuss a plan for maximizing the number of lives saved tomorrow. What is the optimal plan?

[BLaCkHawKDowN615]

have the 10th guy tell everyone else their colors and let him guess based on which colors have been picked the most. this would definately work if the masks were 5 white 5 black but idk if that is the case.

Edit: or the guy in the back could just rub his hat on the guy in front of him and hope some fibers come off

[~HentaiXXX~]

i'm sad...cause i still haven't figured that out...='(

[Doug31]

Assuming there will always be 5 of each of black hats and white hats, the guy in the back says the color that there are only 4 of in front of him, then the next guy says the color that there are only 4 of still in front of him and counting what the previous guy said, and so on, until they get to the front. Or, can they only see the hat of the person in front of them?