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Neutral Zone Trap

Riddle...

39 posts in this topic

A man turns left three times to get home,

When he gets home there are two men waiting for him,

Both of the men wear masks...........

Who are the men?

Wait wait wait..........does your little thingy say trade parise now?.......please do everyone a favor and stop calling yourself a devs fan and grab a rangers jersey. That just goes to show how clueless some devils fans are.

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Ha nice. Great thread by the way, I feel like it's been a while since we had fun off-topic threads like this.

Here's another one...

The warden meets with 50 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a room with a switch turned off. The switch is not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room for an hour. Nobody may bring anything into the switch room. After the hour he is returned to his cell.

"No one else will enter the switch room until I lead the next prisoner there. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back.

"But, given enough time, everyone will eventually visit the switch room. At any time any one of you may declare to me, 'We have all visited the switch room.' and be 100% sure.

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."

What is the strategy they come up with so that they can be free?

I've heard this with 2 switches, but I think the solution is pretty similar.

The inmates all nominate one of them to be a leader. Then the first time only that each inmate gets taken into the room, he flips the switch up. The only person who can flip the switch down is the leader. He keeps a running total and when he has flipped the switch down 49 times he knows that all of the inmates have gone.

Edited by gogonjdevil

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I know it's breaking the riddle but I'd ask each prisoner to leave their shirt behind when they're in the room. Then when you reach 50 shirts, you're finished. :lol:

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I've heard this with 2 switches, but I think the solution is pretty similar.

The inmates all nominate one of them to be a leader. Then the first time only that each inmate gets taken into the room, he flips the switch up. The only person who can flip the switch down is the leader. He keeps a running total and when he has flipped the switch down 49 times he knows that all of the inmates have gone.

Yep you got it. I was going to ask the one with two switches next but I guess that is pointless now haha.

I know it's breaking the riddle but I'd ask each prisoner to leave their shirt behind when they're in the room. Then when you reach 50 shirts, you're finished. :lol:

Yea I guess when asking it you can make a rule saying the prisoners can't take anything with them or leave anything behind. There are other variations too, I thought of one where the room has no switch and is completely empty except for a rock. Then instead of flipping a switch, they just put the rock against the wall or something and then the leader resets it by putting it in the middle of the room.

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so do I get to post the next riddle now? Well...I'm going to:

Daniel, my son, is exactly one fifth of my age. In 21 years time, I will be exactly twice his age. My wife is exactly seven times older than my daughter, Jessica. In 8 years time, my wife will be three times older then Jessica. How old are Daniel and Jessica now?

A nice little math puzzle :)

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so do I get to post the next riddle now? Well...I'm going to:

Daniel, my son, is exactly one fifth of my age. In 21 years time, I will be exactly twice his age. My wife is exactly seven times older than my daughter, Jessica. In 8 years time, my wife will be three times older then Jessica. How old are Daniel and Jessica now?

A nice little math puzzle :)

Daniel is 7 and Jessica is 4.

There are three boxes labeled “apples,” “oranges,” and “apples and oranges.” All three labels are wrong. You can take one fruit from one box. Figure out which labels go on which boxes.

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So how does the switch thing work? Specifically, how does the leader know when everyone else has been through? I thought one of the rules was that the guard can bring in any prisoner multiple times without having to have already brought everyone in at least once.

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So how does the switch thing work? Specifically, how does the leader know when everyone else has been through? I thought one of the rules was that the guard can bring in any prisoner multiple times without having to have already brought everyone in at least once.

Because each prisoner can only flip it once...so when the leader goes back into the room and sees the switch is down, he knows it was a repeat and doesnt count that as 1. The number of prisoners is known.

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So how does the switch thing work? Specifically, how does the leader know when everyone else has been through? I thought one of the rules was that the guard can bring in any prisoner multiple times without having to have already brought everyone in at least once.

If you're not the leader, you flip the switch on when it's your first time in the room. If you've been in there already, you don't touch the switch. The leader just counts how many time he turns it back off.

Edited by devilsfan26

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Daniel is 7 and Jessica is 4.

There are three boxes labeled “apples,” “oranges,” and “apples and oranges.” All three labels are wrong. You can take one fruit from one box. Figure out which labels go on which boxes.

You take from the apple and oranges box.

Whatever fruit is in there gets that label. (said both, now let's say it was an apple, so label apple)

Whatever box you took that label from is the other fruit. (said apple, so now label it orange)

The remaining box is apples and oranges. (said orange, so now label as both)

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You take from the apple and oranges box.

Whatever fruit is in there gets that label. (said both, now let's say it was an apple, so label apple)

Whatever box you took that label from is the other fruit. (said apple, so now label it orange)

The remaining box is apples and oranges. (said orange, so now label as both)

Yep

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Okay, I get the switch thing now. I think I misread the explanation the first time, which led to the confusion. :doh1:

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Yep you got it. I was going to ask the one with two switches next but I guess that is pointless now haha.

Yea I guess when asking it you can make a rule saying the prisoners can't take anything with them or leave anything behind. There are other variations too, I thought of one where the room has no switch and is completely empty except for a rock. Then instead of flipping a switch, they just put the rock against the wall or something and then the leader resets it by putting it in the middle of the room.

An elegant solution, but that's going to take a long time to complete. You say that each prisoner is put in the switch room for an hour, right? Then the quickest the prisoners can be freed is after 98 hrs. I will win many, many lottery jackpots before that happens. I have calculated the expected time to completion of this process (until the leader asks the guards to set everyone free) and it's 2,674 hrs. That's 111 days and 10 hrs.

If the prisoners were willing to agree to even a minimal amount of risk, they could be free much earlier. After 341 hrs (about 2 weeks), there's a 95% chance that they have all visited the switch room. Not worth risking death? I hear yah. After 422 hrs, it's a 99% chance. Gator-phobic? After 536 hrs., the odds are 1 in 1,000 that not everyone has been to the room.

650 hrs. = 1 in 10,000

764 hrs. = 1 in 100,000

878 hrs. = 1 in 1,000,000

1,000 hrs (still less than half of the expected time of the process) = 1 in 11,883,772

**One assumption that I made above that is not outlined in the original wording of the riddle is that as soon as one prisoner leaves the room, the next one's hour immediately begins. This is crucial to my "play the odds" strategy because the prisoners need to have some idea of how many of them have gone to the room in order to know when the roll of the dice will give them the odds that they desire. However, the more infrequent the visits, the longer it takes for that sure-thing process to work, and if just one prisoner (or the leader) gets transferred, paroled, or dies before changing the switch in the room, then they will all be in jail for life.

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An elegant solution, but that's going to take a long time to complete. You say that each prisoner is put in the switch room for an hour, right? Then the quickest the prisoners can be freed is after 98 hrs. I will win many, many lottery jackpots before that happens. I have calculated the expected time to completion of this process (until the leader asks the guards to set everyone free) and it's 2,674 hrs. That's 111 days and 10 hrs.

If the prisoners were willing to agree to even a minimal amount of risk, they could be free much earlier. After 341 hrs (about 2 weeks), there's a 95% chance that they have all visited the switch room. Not worth risking death? I hear yah. After 422 hrs, it's a 99% chance. Gator-phobic? After 536 hrs., the odds are 1 in 1,000 that not everyone has been to the room.

650 hrs. = 1 in 10,000

764 hrs. = 1 in 100,000

878 hrs. = 1 in 1,000,000

1,000 hrs (still less than half of the expected time of the process) = 1 in 11,883,772

**One assumption that I made above that is not outlined in the original wording of the riddle is that as soon as one prisoner leaves the room, the next one's hour immediately begins. This is crucial to my "play the odds" strategy because the prisoners need to have some idea of how many of them have gone to the room in order to know when the roll of the dice will give them the odds that they desire. However, the more infrequent the visits, the longer it takes for that sure-thing process to work, and if just one prisoner (or the leader) gets transferred, paroled, or dies before changing the switch in the room, then they will all be in jail for life.

Well yea that's true if you think about it more practically, but the point of the riddle is to figure out the process by which they can determine when everyone has been there. Suppose instead of breaking out of prison they were trying to win enough money for each person that they could all live like kings for the rest of their lives. Perhaps then they would be more likely to wait it out and keep living their normal lives until they were 100% sure.

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