Logic Riddles - Page 5

Hey, sat for maybe 10 minutes. Skipped one I didn't get lol. Not read any answers or googled anything.

1. I take of all my clothes, and take shit, then lift all 3 over

2. Maybe he grabs a cup of coffe at the 18th floor, or has a crush at someone he wants to go past. Or maybe he wants to get some exercise in. And at the end of the day, he's much more fatigued, and cba to hassle of walking, so he just elevator's down.
Maybe his kid died in the 18th floor, and he has made this ritual. Maybe he gets his work assignment on the 18th floor, then heads on up to his office.

3. On top of his chest (not around him) is a loaded gun, harmonica and a note saying "I lost my kid, cba to live, so I'm just gonna puff my kids favourite song then end it" and his brains is blown out. (haha this one was easy)

4. I'm greedy I want 3:1, and if you keep pushing it I won't like you as much.

5. *Cries*

6. I'll turn "ON" to switches to the left, Nr,1,2. And I'll have them on for a few minutes before entering. Then I'll turn OFF nr2, and only have on nr,1, then enter the room. Now I'f the light is on, it's Nr1, if it's not on, and the bulb is warm, it's nr2, if it's not warm it's nr3

7. Shoot at the guy who never misses, still feels like it doesn't matter much lol

On October 23 2012 21:49 HotChip wrote: Another classic: A father and a son were in a car. Out of nowhere a truck hit the car. Two ambulances came and the father and son were bought to different hospitals. The doctor was sent in for an operation for the boy and says "I can't operate on this boy this is my son!" How was that possible?

The Grandfather of son, so the father would be his son?

On October 25 2012 20:46 NewbSaibot wrote: What weighs more a pound of quarters or a pound of feathers lololol

same

+ Show Spoiler +

On October 28 2012 21:26 Art.Cascade wrote: Show nested quote + On October 28 2012 19:55 jeffv8x_-_16 wrote: 1) A girl sees with her own eyes how a man murders two people and then burns down their house. Why doesn't she call the police? 2) The boss of a softwarecompany is murdered. Four employees have a clear motive. While being interrogated they are recorded saying; Bart: It was Tom. Tom: It was Erik. Stefan: It definitely wasn't me. Erik: Tom is lying if he claims it was me. Only one of these men is speaking the truth. Who killed the boss? 2) + Show Spoiler + Both Tom and Erik cannot be lying, because if Tom lies, then Erik doesn't. Which means that both Bart and Stefan ARE lying. Which means that it was Stefan. Conistency check: Bart; lies Tom: lies Stefan: lies Erik: truth

That is correct

On October 28 2012 23:38 Rapoza wrote: Show nested quote + On October 28 2012 19:55 jeffv8x_-_16 wrote: 1) A girl sees with her own eyes how a man murders two people and then burns down their house. Why doesn't she call the police? + Show Spoiler + She is deaf-mute OR + Show Spoiler + Her name is Debra Morgan lol

Not the answer I'm looking for, although yours is probably better then mine

On October 24 2012 10:38 HotChip wrote: Show nested quote + On October 24 2012 09:19 player999 wrote: doctor is his mom duh winner

a woman doctor? preposterous!

On October 28 2012 21:32 Art.Cascade wrote: And another of my favourites, from TL actually. 100 damn smart and organised prisoners. The guards set up a room, with 100 boxes, and place the (unique) name of all prisoner in the boxes. The guard collect one prisoner at a time from his cell to the room with boxes, and let the prisoner open 50 of the 100 boxes, one after the other. If any prisoner fails to find his own name in any of the 50 boxes he opens, then all the prisoners are killed. If all prisoners manage to find their name in the 50 tries, then they all go free. They have time to prepare a strategy before the box-opening starts, but once the first prisoner is in the room opening boxes, no more communication between the prisoners is possible. The prisoners do not know in which order they will be called in the room, and the guards set up the room in the exact same way for each prisoner. How can the prisoners maximise their chances to get free? hint: they can get well above 10%.

If this is some kind of trick question then please pm me or something. This one is driving me crazy. I can't figure out how they could ever get anywhere near 10%. But unless it's a trick question I don't wanna give up just yet...

On October 30 2012 17:12 Almebeast wrote: Show nested quote + On October 28 2012 21:32 Art.Cascade wrote: And another of my favourites, from TL actually. 100 damn smart and organised prisoners. The guards set up a room, with 100 boxes, and place the (unique) name of all prisoner in the boxes. The guard collect one prisoner at a time from his cell to the room with boxes, and let the prisoner open 50 of the 100 boxes, one after the other. If any prisoner fails to find his own name in any of the 50 boxes he opens, then all the prisoners are killed. If all prisoners manage to find their name in the 50 tries, then they all go free. They have time to prepare a strategy before the box-opening starts, but once the first prisoner is in the room opening boxes, no more communication between the prisoners is possible. The prisoners do not know in which order they will be called in the room, and the guards set up the room in the exact same way for each prisoner. How can the prisoners maximise their chances to get free? hint: they can get well above 10%. If this is some kind of trick question then please pm me or something. This one is driving me crazy. I can't figure out how they could ever get anywhere near 10%. But unless it's a trick question I don't wanna give up just yet...

No trick question would make it to my favourites.

When I first saw it I had the exact same reaction. At first glance, it seems that each prisoner has 50% to get their name, which would mean a 1/2^100 (about 10^(-30)) probability for everyone to hit. I don't think I solved it until the day after or something. + Show Spoiler + [small hint]

There is a flaw in the calculation giving 1/2^100, and you have to exploit the shit out of that flaw, and you are fine. The bigger hint is what this flaw is.

Wow, LP does not like several spoilers in the same post! :o Trying in separate post instead... yes, that worked.

+ Show Spoiler + [bigger hint]

--- Nuked ---

Wasn't aware this was so unclear.

I'll give you an example. You are prisoner Adam.

1) The night before you talk to the other prisoners and come up with something smart.
2) You get locked into your cell all alone, no more communication with anyone.
3) You sit and wait, knowing that any second one of the other prisoners can fail to find their name and get everyone killed. You have no clue in what order people are being let in the room though, but you know it is unlikely that you will go first.
4) Let say that you are lucky, and don't die before you get into the room. At some point the guards will come and pick you up.
5) You get into the room with 100 boxes. You do not know how many other prisoners, if any, have been in here. You only know that if any others have been there, they have found their name, as you are still alive.
6) You chose to start by opening, for example, box number 72. It has "Bart" on it. Not your name. Damn.
7) You then chose to process to, for example, box 10, which has "Carl". Damn.
8) And so on. If you don't find your name in 50 tries, you and ll other prisoenrs die.
9) Let's assume you find your name (for example in the 46:th box you open). If you were the last prisoner to go into the room, you win and all are free! If not, you are led back to your cell to wait. The boxes are put in order for next guy.
10) You now sit in your cell, waiting for the other prisoners to go into the room, hoping that they will find their name as well.

I'm sure you can invent silly tricks like "I beat up all the guards", "i tilt the boxes with Carls name just a little to the left" etc. And there are riddles for that, but here you can get well above 10% with just a smart way to select what boxes to open.

On October 30 2012 17:12 Almebeast wrote: Show nested quote + On October 28 2012 21:32 Art.Cascade wrote: And another of my favourites, from TL actually. 100 damn smart and organised prisoners. The guards set up a room, with 100 boxes, and place the (unique) name of all prisoner in the boxes. The guard collect one prisoner at a time from his cell to the room with boxes, and let the prisoner open 50 of the 100 boxes, one after the other. If any prisoner fails to find his own name in any of the 50 boxes he opens, then all the prisoners are killed. If all prisoners manage to find their name in the 50 tries, then they all go free. They have time to prepare a strategy before the box-opening starts, but once the first prisoner is in the room opening boxes, no more communication between the prisoners is possible. The prisoners do not know in which order they will be called in the room, and the guards set up the room in the exact same way for each prisoner. How can the prisoners maximise their chances to get free? hint: they can get well above 10%. If this is some kind of trick question then please pm me or something. This one is driving me crazy. I can't figure out how they could ever get anywhere near 10%. But unless it's a trick question I don't wanna give up just yet...

i dont think that part is right. i dont get how it can be possible if they dont know what order they are going into the room

On October 30 2012 17:55 Art.Cascade wrote: Show nested quote + On October 30 2012 17:12 Almebeast wrote:   On October 28 2012 21:32 Art.Cascade wrote: And another of my favourites, from TL actually. 100 damn smart and organised prisoners. The guards set up a room, with 100 boxes, and place the (unique) name of all prisoner in the boxes. The guard collect one prisoner at a time from his cell to the room with boxes, and let the prisoner open 50 of the 100 boxes, one after the other. If any prisoner fails to find his own name in any of the 50 boxes he opens, then all the prisoners are killed. If all prisoners manage to find their name in the 50 tries, then they all go free. They have time to prepare a strategy before the box-opening starts, but once the first prisoner is in the room opening boxes, no more communication between the prisoners is possible. The prisoners do not know in which order they will be called in the room, and the guards set up the room in the exact same way for each prisoner. How can the prisoners maximise their chances to get free? hint: they can get well above 10%. If this is some kind of trick question then please pm me or something. This one is driving me crazy. I can't figure out how they could ever get anywhere near 10%. But unless it's a trick question I don't wanna give up just yet... No trick question would make it to my favourites. When I first saw it I had the exact same reaction. At first glance, it seems that each prisoner has 50% to get their name, which would mean a 1/2^100 (about 10^(-30)) probability for everyone to hit. I don't think I solved it until the day after or something. + Show Spoiler + There is a flaw in the calculation giving 1/2^100, and you have to exploit the shit out of that flaw, and you are fine. The bigger hint is what this flaw is.

After a prisoner finds his own name, is the box put back or does he keep it?
So the boxes are put back in the same way/location (and same condition) for every prisoner and the prisoners are called in at random? But are they are aware they are prisoner number x going or is the cell an isolation chamber?

+ Show Spoiler +

On October 31 2012 00:16 Art.Cascade wrote: 3) You sit and wait, knowing that any second one of the other prisoners can fail to find their name and get everyone killed. You have no clue in what order people are being let in the room though, but you know it is unlikely that you will go first.

I think this assumption is wrong, I think they know beforehand in what order they are let in.

Than my solution would be:
+ Show Spoiler +

On October 28 2012 21:32 Art.Cascade wrote: And another of my favourites, from TL actually. 100 damn smart and organised prisoners. The guards set up a room, with 100 boxes, and place the (unique) name of all prisoner in the boxes. The guard collect one prisoner at a time from his cell to the room with boxes, and let the prisoner open 50 of the 100 boxes, one after the other. If any prisoner fails to find his own name in any of the 50 boxes he opens, then all the prisoners are killed. If all prisoners manage to find their name in the 50 tries, then they all go free. They have time to prepare a strategy before the box-opening starts, but once the first prisoner is in the room opening boxes, no more communication between the prisoners is possible. The prisoners do not know in which order they will be called in the room, and the guards set up the room in the exact same way for each prisoner. How can the prisoners maximise their chances to get free? hint: they can get well above 10%.

Don't close the boxes back... as long as the first guy gets it its pretty much a lock from there.

--- Nuked ---

On October 31 2012 09:10 HotChip wrote: Show nested quote + On October 31 2012 00:16 Art.Cascade wrote: 3) You sit and wait, knowing that any second one of the other prisoners can fail to find their name and get everyone killed. You have no clue in what order people are being let in the room though, but you know it is unlikely that you will go first. I think this assumption is wrong, I think they know beforehand in what order they are let in. Than my solution would be: + Show Spoiler + The prisoner get each assigned a number from 1-100 and that number represents their number in line to take a look at the boxes. Each prisoner has to remember all the numbers of all prisoners. The first one starts with box number one, if it's not his name in the box he will next take a look at the box with the same number as the prisoner in box 1 was assigned. He will continue to do this until he finds his name. Prisoner 2 comes in and does exactly the same thing except he starts with box number 2, since he KNOWS prisoner 1 found his box starting with box number 1. This goes on and on and you can see how that after 50 prisoners, there is no chance of failure since half of the prisoners have found their box using this loop strategy. So the chance of success are: 1 - (1/51 + 1/52 + ... + 1/100) = 0.312 = 31,2%

This is the correct solution, well done!

And it is not needed to let the prisoners in the room in any specific order. Whenever you are let in the room, you can assume that all the other prisoners have (or will) manage(d) to find their name, because if they don't it doesn't matter if you find yours anyways.

7) Shoot yourself (if you gonna aim at somebody)

Best at missing, others will go for better shooters, u get second chance, moore chance of u hitting

for the prisoner box problem, why not placed "used" boxes in one of 26 columns with corresponding rows by alphabetical order?

is there math that proves that this choice is not optimal?

On November 02 2012 21:47 cnew27 wrote: for the prisoner box problem, why not placed "used" boxes in one of 26 columns with corresponding rows by alphabetical order? is there math that proves that this choice is not optimal?

The guards set the room and boxes in order for each prisoner. There is no way to communicate with the other prisoners once the box-opening has started, which includes leaving messages inside the room.

If you want to find "tricks" like that you can go on for infinity (which can be fun), but that is not the aim of this riddle as there is a nice mathematical solution, so just assume that the guards are putting a stop to any "outside the box" (lulz...) tricks for now.

so the prisoners know the order or not?