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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?

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just because i don't like your signature, im gonna go ahead and just answer this one early.

An ump and a catcher

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just because i don't like your signature, im gonna go ahead and just answer this one early.

An ump and a catcher

Correct sir.

PS. I have nothing against Parise. I've been a Devils fan for a long time, and I'm sick and tired of our franchise losing top class players for nothing (through UFA).

It's a long time since that we got some good assets in return instead of losing them for nothing.

Riddle # 2

There are three light switches in a three storey house marked A-B-C, two are up, one is down.

On the top floor there are three light bulbs in rooms 1-2-3.

Two are on, one is off.

You can only go upstairs once, to see what bulbs are on.

You have only one attempt at one switch (up or down) to figure out what switches control which light bulb(s).

Once you've switched a switch, you go back upstairs to figure out which switch controls which light bulb...

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Correct sir.

PS. I have nothing against Parise. I've been a Devils fan for a long time, and I'm sick and tired of our franchise losing top class players for nothing (through UFA).

It's a long time since that we got some good assets in return instead of losing them for nothing.

Riddle # 2

There are three light switches in a three storey house marked A-B-C, two are up, one is down.

On the top floor there are three light bulbs in rooms 1-2-3.

Two are on, one is off.

You can only go upstairs once, to see what bulbs are on.

You have only one attempt at one switch (up or down) to figure out what switches control which light bulb(s).

Once you've switched a switch, you go back upstairs to figure out which switch controls which light bulb...

Turn one off and immediately go upstairs. Whichever switch is on obviously controls the light that is on. Then you feel the bulbs, whichever one still feels warm is the one you switched off.

I got one...

You are on a game show. There are three closed doors. Behind one of them is a car. You choose a door and then the host of the show opens one of the other doors, one they know doesn't have the car. He then gives you the option of switching doors. Why should you switch?

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Turn one off and immediately go upstairs. Whichever switch is on obviously controls the light that is on. Then you feel the bulbs, whichever one still feels warm* is the one you switched off.

I got one...

You are on a game show. There are three closed doors. Behind one of them is a car. You choose a door and then the host of the show opens one of the other doors, one they know doesn't have the car. He then gives you the option of switching doors. Why should you switch?

Correct sir.

*careful-you'll still burn your fingers ;-)

You switch because they know you had the wrong choice in the first place.(your first choice was wrong).

Riddle #3

A building in the desert. 300 miles from anywhere (in all directions)

This building is 50 feet high, 50 feet deep and 50 feet wide.

Inside the building is a metal beam 45 feet off the ground spanning wall to wall horizontally. ( five feet off the ceiling)

Hanging from the beam is a six feet piece of rope attached to it is a 36"x36"x36" box (inside is gold)

The box stands 36' off the ground.

There is only one door that is a standard 78"x30" to get inside the building (there are no windows)

The roof is as flat as the walls forming a prefect 50'x50' cube.

Outside is a flatbed 18 wheeler truck~there is nothing on the bed of the truck.

There are no ladders~only one man put the box where it is.

All that remains below the box on the ground is a pool of water about 2 inches deep.

How did one man get that box 36' above his head and tie it to the beam with no ladders and no help...

Reason for edit~~~bad math.

Edited by Neutral Zone Trap

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Correct sir.

*careful-you'll still burn your fingers ;-)

You switch because they know you had the wrong choice in the first place.(your first choice was wrong).

I don't think I understand your answer, or maybe I didn't explain the question well. First you pick a door that may or may not have the car. Then, whether you picked the right door or one of the wrong ones, the host opens one of the doors that does not have the car. So at this point you still might have chosen correctly the first time, either your door or the third door has the car. So why should you switch?

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Aw, man, they just did the door one on Mythbusters, but I haven't seen that one yet. :doh1:

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I don't think I understand your answer, or maybe I didn't explain the question well. First you pick a door that may or may not have the car. Then, whether you picked the right door or one of the wrong ones, the host opens one of the doors that does not have the car. So at this point you still might have chosen correctly the first time, either your door or the third door has the car. So why should you switch?

There is a 33% chance your 1st pick is right and a 66% chance one of the other 2 doors have the car. As they eliminate one door for you, there is a 66% chance the remaining door holds the car..

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Riddle #3

A building in the desert. 300 miles from anywhere (in all directions)

This building is 50 feet high, 50 feet deep and 50 feet wide.

Inside the building is a metal beam 45 feet off the ground spanning wall to wall horizontally. ( five feet off the ceiling)

Hanging from the beam is a six feet piece of rope attached to it is a 36"x36"x36" box (inside is gold)

The box stands 36' off the ground.

There is only one door that is a standard 78"x30" to get inside the building (there are no windows)

The roof is as flat as the walls forming a prefect 50'x50' cube.

Outside is a flatbed 18 wheeler truck~there is nothing on the bed of the truck.

There are no ladders~only one man put the box where it is.

All that remains below the box on the ground is a pool of water about 2 inches deep.

How did one man get that box 36' above his head and tie it to the beam with no ladders and no help...

Reason for edit~~~bad math.

I'm going to guess it has something to do with filling the room with water so he could float on it until it was high enough for him to reach the beam. Not sure how the truck factors in or how he gets the water into the room though.

There is a 33% chance your 1st pick is right and a 66% chance one of the other 2 doors have the car. As they eliminate one door for you, there is a 66% chance the remaining door holds the car..

Correct.

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There is a 33% chance your 1st pick is right and a 66% chance one of the other 2 doors have the car. As they eliminate one door for you, there is a 66% chance the remaining door holds the car..

But that wouldnt matter because then the odds of your door having the car would increase

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But that wouldnt matter because then the odds of your door having the car would increase

Nope, the odds that you pick the right door in the beginning are always 33%. It took me a while to understand this too. There are three possible outcomes:

(1) You pick the right door in the beginning. You are then shown one of the wrong doors. If you switch you lose.

(2) You first choose wrong door A. You are then shown wrong door B. If you switch you get the car.

(3) You first choose wrong door B. You are then shown wrong door A. If you switch you get the car.

If you switch, there are two outcomes that result in you winning the car. The only way you lose if you switch is if you chose the correct door in the first place, but the chance of that happening is only 33%, so if you switch you have a 66% chance of winning.

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There is a 33% chance your 1st pick is right and a 66% chance one of the other 2 doors have the car. As they eliminate one door for you, there is a 66% chance the remaining door holds the car..

I understand this but logically it still doesnt make sense, each door has 33% regardless of when it is opened to me...when they eliminate then it recalcs to 50-50

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Nope, the odds that you pick the right door in the beginning are always 33%. It took me a while to understand this too. There are three possible outcomes:

(1) You pick the right door A in the beginning. You are then shown one of the wrong doors. If you switch you lose.

(2) You first choose wrong door A. You are then shown wrong door B. If you switch you get the car.

(3) You first choose wrong door B. You are then shown wrong door A. If you switch you get the car.

(4) You first choose right door B in the beginning. You are then shown one of the wrong doors. If you switch you lose.

If you switch, there are two outcomes that result in you winning the car. The only way you lose if you switch is if you chose the correct door in the first place, but the chance of that happening is only 33%, so if you switch you have a 66% chance of winning.

I corrected above to why i think it is still a 50-50

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i dont like the game show one. no matter what, those two remaining doors are going to have the same exact chance of having the car. the 3rd door's percentage should just be distributed equally to the two remaining doors

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i dont like the game show one. no matter what, those two remaining doors are going to have the same exact chance of having the car. the 3rd door's percentage should just be distributed equally to the two remaining doors

haha thats what Im saying! anyone have another riddle?

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Correct sir.

Riddle #3

A building in the desert. 300 miles from anywhere (in all directions)

This building is 50 feet high, 50 feet deep and 50 feet wide.

Inside the building is a metal beam 45 feet off the ground spanning wall to wall horizontally. ( five feet off the ceiling)

Hanging from the beam is a six feet piece of rope attached to it is a 36"x36"x36" box (inside is gold)

The box stands 36' off the ground.

There is only one door that is a standard 78"x30" to get inside the building (there are no windows)

The roof is as flat as the walls forming a prefect 50'x50' cube.

Outside is a flatbed 18 wheeler truck~there is nothing on the bed of the truck.

There are no ladders~only one man put the box where it is.

All that remains below the box on the ground is a pool of water about 2 inches deep.

How did one man get that box 36' above his head and tie it to the beam with no ladders and no help...

Reason for edit~~~bad math.

I am really curious to the answer for this. I find it interesting that they use 50' deep. also the fact that its in the desert and there is a pool of water...

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The 18 wheeler flat bed truck is VERY important...

Edited by Neutral Zone Trap

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Nope, the odds that you pick the right door in the beginning are always 33%. It took me a while to understand this too. There are three possible outcomes:

(1) You pick the right door in the beginning. You are then shown one of the wrong doors. If you switch you lose.

(2) You first choose wrong door A. You are then shown wrong door B. If you switch you get the car.

(3) You first choose wrong door B. You are then shown wrong door A. If you switch you get the car.

If you switch, there are two outcomes that result in you winning the car. The only way you lose if you switch is if you chose the correct door in the first place, but the chance of that happening is only 33%, so if you switch you have a 66% chance of winning.

I became obsessed with this after watching the movie "21" it took me a while to understand it....

its called the monty hall problem : http://en.wikipedia.org/wiki/Monty_Hall_problem

here is a simulator http://people.hofstra.edu/steven_r_costenoble/MontyHall/MontyHallSim.html

i did 50 trials of each method.

when staying with my original door i had a success rate of 0.4 (20 out of 50)

when taking the switch i had a success rate of 0.68 (34 out of 50)

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I became obsessed with this after watching the movie "21" it took me a while to understand it....

its called the monty hall problem : http://en.wikipedia.org/wiki/Monty_Hall_problem

here is a simulator http://people.hofstra.edu/steven_r_costenoble/MontyHall/MontyHallSim.html

i did 50 trials of each method.

when staying with my original door i had a success rate of 0.4 (20 out of 50)

when taking the switch i had a success rate of 0.68 (34 out of 50)

Bingo. The Wikipedia page does a good job explaining it.

A simple way to demonstrate that a switching strategy really does win two out of three times on the average is to simulate the game with playing cards (Gardner 1959b; vos Savant 1996:8). Three cards from an ordinary deck are used to represent the three doors; one 'special' card such as the Ace of Spades should represent the door with the car, and ordinary cards, such as the two red twos, represent the goat doors.

The simulation, using the following procedure, can be repeated several times to simulate multiple rounds of the game. One card is dealt face-down at random to the 'player', to represent the door the player picks initially. Then, looking at the remaining two cards, at least one of which must be a red two, the 'host' discards a red two. If the card remaining in the host's hand is the Ace of Spades, this is recorded as a round where the player would have won by switching; if the host is holding a red two, the round is recorded as one where staying would have won.

This experiment is likely to approximate the probability of winning, and running the experiment over enough rounds should not only verify that the player does win by switching two times out of three, but show why. After one card has been dealt to the player, it is already determined whether switching will win the round for the player; and two times out of three the Ace of Spades is in the host's hand.

If this is not convincing, the simulation can be done with the entire deck, dealing one card to the player and keeping the other 51 (Gardner 1959b; Adams 1990). In this variant the Ace of Spades goes to the host 51 times out of 52, and stays with the host no matter how many non-Ace cards are discarded.

That switching has a probability of 2/3 of winning the car runs counter to many people's intuition. If there are two doors left, then why isn't each door 1/2? It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. His initial probability of winning is 1 out of 1,000,000. The game host goes down the line of doors, opening each one to show 999,998 goats in total, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat—the chance that the player's door is correct hasn't changed. A rational player should switch.

Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. It's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens 999,998 for you, leaving, deliberately, the one with the prize. Clearly, one would choose to open the other 999,999 doors rather than keep the one.

Edited by devilsfan26

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The 18 wheeler flat bed truck is VERY important...

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.

Edited by devilsfan26

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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.

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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

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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....

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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.

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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

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