38 replies to this topic

[Neutral Zone Trap]

Alright here's my rethought answer...

The guy loads up the truck with enough blocks of ice that are small enough to fit through the door. He drives through the desert to the building at night during the winter so the ice doesn't melt (though I suppose this could depend on where the desert is located). He then stacks the blocks of ice on top of each other until he can climb them like a staircase to reach the beam and tie the box to it. As the temperature rises during the day, the ice melts leaving nothing inside except the box tied to the beam and the pool of water.

Well played Df26, you win.

[Devils Dose]

The mental hurdle that people must clear with the game show one is that the opening of the bad door by the show's host is meaningless, it does not re-inform the original probabilities.

Here's my alternative explanation of the problem, an equivalent version of the game:
You are presented with 3 doors, one of which conceals a car. You are told to designate one door as The 1, and the remaining two are called The Pair. You are asked to choose The 1 or The Pair, and if your selection contains the car, you win. What would you do? Obviously you would choose The Pair, for better odds. Game Over

"But what about the part where they reveal an empty door?" you're probably wondering. In the above, that would be the equivalent of the host turning to you and saying, "But you do realize that The Pair contains at least one empty door, don't you?" To which you say, "Thanks, Sherlock. I think a kindergartener could have told me that."


If this still doesn't work for you, imagine that there are 1 million doors, and that they open 999,998 of them and then ask you if you want to switch.

Edited by Devils Dose, 14 December 2011 - 09:31 PM.

[UnderDogX]

now I have to get into this because I still haven't cleared that mental hurdle

If one of the three doors is going to be opened to reveal an empty door regardless of the one you picked then the percentage is only perceived as being evenly distributed when it actually isn't. We see it as 33% for each door because we perceive it to be even. But it isn't. One door will be opened and empty regardless so the percentage is distributed as 50% your choice is correct and 50% the other two have the winning door. In a sense the third door is just meaningless you are never really choosing from 3 doors you are only choosing from the right one and the wrong one....

[devilsfan26]

now I have to get into this because I still haven't cleared that mental hurdle

If one of the three doors is going to be opened to reveal an empty door regardless of the one you picked then the percentage is only perceived as being evenly distributed when it actually isn't. We see it as 33% for each door because we perceive it to be even. But it isn't. One door will be opened and empty regardless so the percentage is distributed as 50% your choice is correct and 50% the other two have the winning door. In a sense the third door is just meaningless you are never really choosing from 3 doors you are only choosing from the right one and the wrong one....

I'm not sure if this will settle the confusion, but the host knows that the door he opens is empty. He doesn't choose which door to open until after you make your pick. No matter which door you pick, there is always at least one empty door for the host to open. If you choose an empty one (which you have a 66% chance of doing), he opens the other empty one, so the door that you would switch to is the one with the car. So since you have a 66% chance of picking an empty door in the beginning, you have a 66% chance of winning the car if you switch.

[devilsfan26]

Well played Df26, you win.

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?

Edited by devilsfan26, 15 December 2011 - 01:05 AM.

[mvparise]

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.

[gogonjdevil]

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, 15 December 2011 - 04:13 PM.

[Devils731]

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.

[devilsfan26]

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.

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.

[gogonjdevil]

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

[devilsfan26]

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.

[RowdyFan42]

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.

[lucifer91]

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.

[devilsfan26]

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, 16 December 2011 - 02:32 PM.

[Devils731]

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)

[devilsfan26]

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

[RowdyFan42]

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

[Devils Dose]

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.

[devilsfan26]

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.