Logic Riddles

Played poker with a guy named Barry tonight. He owns a pretty big trading firm in Chicago that screens thousands of college grads each year to see if they are a good fit to be a trader. Apparently the interview process includes these odd logic puzzles to see how you think on your feet. He gave me a few tonight that were interesting to think about: (Try to answer each question in 5 minutes time, goal is to create an answer that makes some sense)

1. There are three 1 pound balls that need to be carried over a bridge. The bridge can only hold a total weight of you + 2 balls. Your task is to transport all 3 balls across the bridge in one trip. How do you do it?

2. A man works on the 50th floor. When he starts work, he takes the elevator to the 18th floor, then uses the stairs to walk up the remaining 32 floors. On the way back home, he takes the elevator all the way down. Why does he do this?

3. You find a man lying dead in the desert. There is nothing around him. You know what killed him, and you know exactly what he was doing before he died. What was he doing before he died and what killed him?

4. If I told you that I could flip a coin and pick heads or tails correctly 2 out of every 3 coin flips, what odds would you lay me?

5. Using only one mathematical symbol, and all the numbers between 0 and 9 (once and without repeating) make the equation equal 99.

And now for the harder ones that you can take your time with.

6. There are 3 light switches outside of a room. One of the switches is connected to a light bulb inside the room. Each of the three switches can be either 'ON' or 'OFF'. You are allowed to set each switch the way you want it before entering the room. Once you enter the room the door will lock you inside until you can determine which switch controls the bulb. How do you figure out which switch controls the bulb?

7. Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. If you are Mr. Black, where should you shoot first for the highest chance of survival?


Post others if you guys find more interesting ones.

5.
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7.
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2.
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wild guess on 1
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Read in Scottish accent (more fun lol)
Four old Vikings have only one lit torch. They are standing besides a narrow bridge in the middle of the night after a long day of battle, the bridge only holds two big Vikings at a time so to get home they must cross the bridge two at a time.
Each of them are wounded differently so their crossing time over the bridge is not equally long:

Viking A is rather healthy, it takes him 5 minutes to cross the bridge
Viking B is wounded, it takes him 10 minutes to cross the bridge
Viking C is wounded, it takes him 20 minutes to cross the bridge
Viking D is badly wounded, it takes him 25 minutes to cross the bridge

It is so dark that they cannot see unless one Viking carries the lit torch that they have over the bridge.
e.g. if A and C travel together over, their time would be 20 minutes and someone would have to bring the torch back to B and D waiting on the other side.

What is the shortest time for all of them to cross the bridge and get home safe and sound? And in what order should they go?

4.
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Odds of guessing 3 out of 3 is: 1/8

Eight different results:

Odds of guessing 2 out of 3 is: Right Wrong Right
Wrong Right Right
Right Right Wrong = 3/8 = 37,5%
To be +EV you should lay someone lower odds than 3/8

Right or wrong ?

On October 18 2012 09:25 TeSpartan wrote: Read in Scottish accent (more fun lol) Four old Vikings have only one lit torch. They are standing besides a narrow bridge in the middle of the night after a long day of battle, the bridge only holds two big Vikings at a time so to get home they must cross the bridge two at a time. Each of them are wounded differently so their crossing time over the bridge is not equally long: Viking A is rather healthy, it takes him 5 minutes to cross the bridge Viking B is wounded, it takes him 10 minutes to cross the bridge Viking C is wounded, it takes him 20 minutes to cross the bridge Viking D is badly wounded, it takes him 25 minutes to cross the bridge It is so dark that they cannot see unless one Viking carries the lit torch that they have over the bridge. e.g. if A and C travel together over, their time would be 20 minutes and someone would have to bring the torch back to B and D waiting on the other side. What is the shortest time for all of them to cross the bridge and get home safe and sound? And in what order should they go?

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2 is easy. Player999 got it.

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ok went through them and wrote down thought process


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question 4 he means what odds to give him that he gets one single flip right, answers above are off
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1. Build a raft.

2. The elevator broke on his way up.

3. If I already know, doesn't that defeat the purpose of the question?

4. I wouldn't lay you any because the coin is probably rigged.

5. im bad at math!!! ...

ill post the others after thinking

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Popularish but one of my favs:
You are at a cross road, and you see two kids playing. One of them always tells the truth, the other always lies. You don't know which one is which.

You get to ask one question to one of them find out the correct direction.

On October 18 2012 13:47 Moksha wrote: 1) + Show Spoiler + a) bowling-roll 1 ball to otherside b) walk across with the 2 remaining balls c) ??? d) PROFIT!!!1 Popularish but one of my favs: You are at a cross road, and you see two kids playing. One of them always tells the truth, the other always lies. You don't know which one is which. You get to ask one question to one of them find out the correct direction.

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On October 18 2012 13:47 Moksha wrote: 1) + Show Spoiler + a) bowling-roll 1 ball to otherside b) walk across with the 2 remaining balls c) ??? d) PROFIT!!!1 Popularish but one of my favs: You are at a cross road, and you see two kids playing. One of them always tells the truth, the other always lies. You don't know which one is which. You get to ask one question to one of them find out the correct direction.

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1. How big is the bridge?

2. Exercise

3. He was doing the thing that killed him, and he died because he was doing that.

4. I would lay you 1:100. No, 1:1000. Fuck it, 1:1000000000. I will lay you the lowest possible odds I can convince you to take.

5. 12345678 = 99. I used the numbers between 0 and 9, and I made them equal 99. The equal symbol itself is a mathematical symbol.

6. Turn 2 switches on. Then randomly guess which of the 2 is the light. It's worth the 50/50 gamble plus you might get lucky and the light will still be off when you enter. There's no guarantee the bulb will be accessible to touch, and some bulbs dont produce a perceptible level of heat.

7. Forfeit your turn and shoot into the air. You cant risk #3 living and killing you, you need to let #2 have his chance first. Upon seeing this they are likely to duel each other giving you more time. Then just roll the dice and try to kill whoever is left.


At first these answers might seem silly, but the goal is to be pragmatic. Usually these kinds of questions are meant to assess your line of thinking, not get a correct answer, since obviously there can be no correct answer most of the time. I've heard of interviewers asking questions like "how many golf balls could fill this room?" and "explain a database to your 5 year old nephew in 3 sentences". It's a personality test.

I had a few round of interviewing with a prop trading firm, and these questions are pretty much exactly the type of questions I was asked in the first round

On October 18 2012 20:49 mnj wrote: I had a few round of interviewing with a prop trading firm, and these questions are pretty much exactly the type of questions I was asked in the first round

Tell us your answers so we know what not to say

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Let's say that you were to shrink to the size of a couple of inches and were put into a blender. How do you get out?

rofl missread nvm

''If you are Mr. Black, where should you shoot first for the highest chance of survival?''

On October 18 2012 23:41 HeRoS)eNGagE wrote: ''If you are Mr. Black, where should you shoot first for the highest chance of survival?''

Yeah I noticed that too but decided it cosmo probably misworded it.

7.

He should miss on purpose because Gray will most likely shoot White and White will shoot Gray.

If Gray kills White, Black will win 1/3 of the time.

If Gray does not kill White, White will kill Gray 100% of the time.

If Black miss on purpose, he will always have 1/3 of chance to win regardless of who's left because he will have the first shot.

Sorry for no-spoiler for got how to use it.

i think its retarded to use such logic stories at a job interview, i heard 60% of these from a logic story book so if u know they test u like this, u can prepare really easily
they should invent new ones for every applicant but then again the comparison wouldnt be fair since everyone gets totally diferent tasks

the probability ones are the only valid ones imho

I think it is dumb to ask riddles, such as 1, 2, 3 or 6 as either you already know the answer or it is just a guess.
I read some asked by google for recruitment interviews, (or at least, it was said that they were) and they were way more original. But it was more about logic or pragmatism with quantification assessments for example.
Here the results don't really matter but the reasoning does.

Found some :

5 Google Interview Questions
1. What's the next number in this sequence: 10, 9, 60, 90, 70, 66 … ?

2. You're in a car with a helium balloon on a string that is tied to the floor. The windows are closed. When you step on the gas pedal, what happens to the balloon—does it move forward, move backward, or stay put?

3. Using only a four-minute hourglass and a seven-minute hourglass, measure exactly nine minutes—without the process taking longer than nine minutes.

4. A book has N pages, numbered the usual way, from 1 to N. The total number of digits in the page numbers is 1,095. How many pages does the book have?

5. A man pushed his car to a hotel and lost his fortune. What happened?

for more :
http://www.businessinsider.com/15-goo...ill-make-you-feel-stupid-2009-11?op=1

1. It was said that the balls weigh 1 pound each which is the approximate weight of your clothes. Undress and take all of them with you

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On October 18 2012 22:21 Gnarly wrote: Let's say that you were to shrink to the size of a couple of inches and were put into a blender. How do you get out?

C'mon guys, it's easy. This was one of the questions for a current higher up in Google for their job interview.

^^jump out? =)

Any physics people know if juggling is even a legit solution to the first riddle? Sounds like troll physics

On October 20 2012 03:58 blackjacki2 wrote: Any physics people know if juggling is even a legit solution to the first riddle? Sounds like troll physics

Actually if you only consider it as a static problem, it sounds legit and it is asked this way. But as a dynamic problem, I'm not sure.
I like way more the answer with taking off the clothes.

On October 19 2012 16:09 Gnarly wrote: Show nested quote + On October 18 2012 22:21 Gnarly wrote: Let's say that you were to shrink to the size of a couple of inches and were put into a blender. How do you get out? C'mon guys, it's easy. This was one of the questions for a current higher up in Google for their job interview.

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Yeah, juggling doesn't really work because whenever you catch or throw a ball it's effectively going to "weigh" more, since you are accelerating it. So for instance at the moment when you are holding one ball and catching another, you are going to exceed the maximum weight.

On October 19 2012 17:37 nlloser60 wrote: ^^jump out? =)

This guy probably youtubed it.

The idea is that if you were to shrink, the universal laws we know of don't apply. If they did, you wouldn't be able to shrink! Shoot, you could even gain super strength and punch the glass to break it and waltz your way out.

1. I would remove clothing and make myself throw up until i had lost a pound
2. he likes to excercise in the morning to wake himself up
3. he was died of starvation and he was trying to find water before he died
4. since hes so sure i would give him odds that heavily favor me
5. 98 + 1
6. stand outside the room and check each light individually
7. mr gray because theres a 66 chance i miss and then it makes the most sense for
him to shoot mr white with a higher succes rate tahn me and then i get to shoot at him again

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On October 20 2012 12:42 Gnarly wrote: Show nested quote + On October 19 2012 17:37 nlloser60 wrote: ^^jump out? =) This guy probably youtubed it.

What do you mean I youtubed it.

On October 20 2012 21:52 nlloser60 wrote: Show nested quote + On October 20 2012 12:42 Gnarly wrote:   On October 19 2012 17:37 nlloser60 wrote: ^^jump out? =) This guy probably youtubed it. What do you mean I youtubed it.

*Probably*

Really, it just means that you were right.

"Heard on the Street" is one of the well known brainteaser books used to prepare for these interviews.

On October 18 2012 10:45 Daut wrote: + Show Spoiler + if you shoot at mr gray (1/3) you die immediately, (2/3) you miss, then (2/3)*(2/3) you are HU with gray when he shoots white. this leads to an endless string of 1/3*2/3 type things since you can both miss forever by running bad and it has a formula that i dont remember about cantor sets but we can estimate. your odds of winning are (4/9)*(1/3+2/27+4/243....this seems like it approaches 4/9 would be my guess). (2/3)*(1/3) its HU vs mr white (who kills mr gray after he misses) and (2/3)*(1/3)*(1/3) you WIN (2/3)*(1/3)*(2/3) you die. so your odds of winning are.....(4/9)*(4/9)+ 2/27. so you win 16/81+6/81=22/81 if you shoot at mr white first (2/3) you miss. 2/9 you HU with mr white. (4/27) you win, 8/27 you die. then 4/9 you HU with gray. (16/81) you win, (20/81) you die. (1/3) you hit white. then odds of winning are (1/27+2/243+4/2187....about 1/21?) 16/81+1/21 is smaller than 22/81 shoot at gray first.

If you shoot at white first your odds of winning are 19.857/81 compared to shooting gray at 22/81 thus our best hope for survival lies in the higher percentage because we are calculating the times we win. 27.161% for Gray and 24.514% when shooting white. Am i in the wrong here? Can you elaborate in laymans terms why shooting gray is superior to shooting white because logic tells me we are missing some ev on killing white if we don't go for him as aften as possible. I'd prefer the loop over a certain K.O

we also have to calculate our chances if we miss on purpose

How can it ever be shoot gray first, the only benefit of shooting gray first is you most often get to shoot gray twice to his once but there is no actual value in our first shot so its pointless, miss on purpose.

On October 23 2012 10:57 EvilSky wrote: How can it ever be shoot gray first, the only benefit of shooting gray first is you most often get to shoot gray twice to his once but there is no actual value in our first shot so its pointless, miss on purpose.

true, either miss or shoot white, do the math, tho I'm guessing missing is best

1.) D e f e c a t e

wait im not sure thats right meh i dont have time for this shit -.-;

no wait u have to shoot the guy that is 3/3 because if u hit him and kill him ur not AUTO-dead. u still have 1/3 chance of surviving and then either way they wont be shooting you if miss so u get to go again.

missing is most +Ev tho

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?

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?

It's possible because you did not state that the son and father are related, just simply stated that there is a son and a father.

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?

previous answer fits better, but i like this photograph.
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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?

Because they were both on the same car i'm gonna assume they were a grandfather, who is also a father, and a son, whose dad is a doctor.

doctor is his mom duh

On October 24 2012 09:19 player999 wrote: doctor is his mom duh

winner

Or gay father

Oh, I like these! Let me add two of my favourites.

8. Many will have heard this, but we can't miss out on the Monty Hall problem:
You are in a game show, and can choose between 3 closed boxes. One of them contains a car, the other two are empty. You pick box number 1. The show leader then opens box number 2, and it is empty. The show leader then presents you with the option to switch to box number 3. Should you switch?

9. An hour glass with all the sand in the bottom is places on a very accurate scale. The hour glass is turned and the sand start flowing down. Does the scale show more, less or same as before?

I'm a bit curious what the right answer, if there is any, is to the bridge and the three balls. Essentially the options are
a) to reinforce the bridge somehow.
b) loose the weight of the last ball on yourself somehow (clothes, etc)
c) transport the ball across in a way so that it doesn't put it's weight on the bridge at the same time you do, or not at all. Like pulling with a rope after you, rolling across first, throwing, raft, etc.

And neither of those really sound like an answer that I'd expect from riddle like these. Either I missed something, or this is an "impossible" question to just test the inventiveness of people.

On October 22 2012 18:50 player999 wrote: we also have to calculate our chances if we miss on purpose

Gray will obv aim for white. If he hits, it will be showdown between you and gray guy, and you go first. Call this scenario A. If he missed, you will have one chance to take down white, then you die. This is scenario B.

You can win in scenario A on shot 1 (ie, gray hits white, you hit gray), on shot 2 (gray hits white, you miss gray, gray miss you, you hit gray), 3, etc. Let's call these scenario A1, A2, A3 etc. The probabilities for these are

A1: 2/3 * 1/3
A2: 2/3 * (2/3 * 1/3) * 1/3
A3: 2/3 * (2/3 * 1/3) * (2/3 * 1/3) *1/3
...

Ie, each extra iteration gives an extra factor 2/3*1/3 = 2/9. The sum of all these are

2/3 * Sum_{i from 0 to infinity} (2/9^i) * 1/3.

The geometric series Sum_i (x^i) is 1/(1-x), so the probability to win through scenario A is 2/3 * 1/(1-2/9) * 1/3 = 2/9 * 9/(9-2) = 2/7.

The probability win through scenario B is simply 1/3*1/3 = 1/9 as gray have to miss, and you have to hit.

So total chance to win if we miss on purpose is 2/7 + 1/9 = 18/63 + 7/63 = 25/63 which is just below 40%. Not too bad imo, considering how bad of a shot mr black is.

^Just want to say +1 to the above. I did the calculations a few days ago and got the same result, but was too lazy to type it on the computer.

I can add that if you happen to dislike geometric series for some reason, you can find probability to win in scenario A in a different way, which is kind of neat imo =). Let Pbg denote the probability of black winning when black is HU with gray, and black is about to shoot. Pgb is the prob of black winning when gray is about to shoot.

Pbg = 1/3 + 2/3*Pgb
Pgb = 1/3*Pbg

Solving the above 2x2 system of equations yields Pbg = 3/7, Pgb = 1/7.

On October 25 2012 01:20 Art.Cascade wrote: Oh, I like these! Let me add two of my favourites. 8. Many will have heard this, but we can't miss out on the Monty Hall problem: You are in a game show, and can choose between 3 closed boxes. One of them contains a car, the other two are empty. You pick box number 1. The show leader then opens box number 2, and it is empty. The show leader then presents you with the option to switch to box number 3. Should you switch?

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9. An hour glass with all the sand in the bottom is places on a very accurate scale. The hour glass is turned and the sand start flowing down. Does the scale show more, less or same as before?

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What weighs more a pound of quarters or a pound of feathers lololol

lololo

Acting on an anonymous phone call, the police raid a house to arrest a suspected murderer. They don't know what he looks like but they know his name is John and that he is inside the house. The police bust in on a carpenter, a lorry driver, a mechanic and a fireman all playing poker. Without hesitation or communication of any kind, they immediately arrest the fireman. How do they know they've got their man?

On October 27 2012 15:39 dnagardi wrote: Acting on an anonymous phone call, the police raid a house to arrest a suspected murderer. They don't know what he looks like but they know his name is John and that he is inside the house. The police bust in on a carpenter, a lorry driver, a mechanic and a fireman all playing poker. Without hesitation or communication of any kind, they immediately arrest the fireman. How do they know they've got their man?

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On October 27 2012 17:03 Spicy wrote: Show nested quote + On October 27 2012 15:39 dnagardi wrote: Acting on an anonymous phone call, the police raid a house to arrest a suspected murderer. They don't know what he looks like but they know his name is John and that he is inside the house. The police bust in on a carpenter, a lorry driver, a mechanic and a fireman all playing poker. Without hesitation or communication of any kind, they immediately arrest the fireman. How do they know they've got their man? + Show Spoiler + is it because firemen have name tags?

no, they dont

On October 27 2012 15:39 dnagardi wrote: Acting on an anonymous phone call, the police raid a house to arrest a suspected murderer. They don't know what he looks like but they know his name is John and that he is inside the house. The police bust in on a carpenter, a lorry driver, a mechanic and a fireman all playing poker. Without hesitation or communication of any kind, they immediately arrest the fireman. How do they know they've got their man?

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On October 28 2012 08:37 kingpowa wrote: Show nested quote + On October 27 2012 15:39 dnagardi wrote: Acting on an anonymous phone call, the police raid a house to arrest a suspected murderer. They don't know what he looks like but they know his name is John and that he is inside the house. The police bust in on a carpenter, a lorry driver, a mechanic and a fireman all playing poker. Without hesitation or communication of any kind, they immediately arrest the fireman. How do they know they've got their man? + Show Spoiler + 3 women, one man ?

correct

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quarters i think

unless there's a trick

On October 28 2012 16:38 bigredhoss wrote: quarters i think

Not sure if serious

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?

On October 28 2012 19:47 SolarM wrote: Show nested quote + On October 28 2012 16:38 bigredhoss wrote: quarters i think Not sure if serious

whatever dude i gave it a shot

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 +

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%.

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On October 28 2012 23:17 Rapoza 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 prisoners are allowed to go into the room, one at a time, and 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. How can the prisoners maximise their chances to get free? hint: they can get well above 10%. + Show Spoiler + 50% (first time cap) 1- They run a list with their names (alphabetic order); 2- Everybody memorize the next name; 3- They enter using alphabetic order; 4- Guy enters and piles opened boxes in a alphabetic order; 5- Trow box with his name away from pile; 6- Trow box with next name away from pile in the other side (if he finds it); 7- Next guy enters and searches for the box in the other side away from the pile. If there inst a box, search in the remaining not piled boxes; 8- Repeat step 4 -> 8 until its done.

I knew I should have been clearer!
The guards prepare the room in the exact same way for each prisoner. There is no way for them to communicate once they started opening boxes. And the prisoners don't get to choose in what order they enter the room.

I'll edit to be more clear, sorry.

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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?

On November 05 2012 04:30 gymnast wrote: so the prisoners know the order or not?

On October 28 2012 21:32 Art.Cascade wrote: 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%.

I finally gave up and looked at the solution. It's absolutely brilliant. Kind of simple (as most brilliant things are), but still very hard to come up with. No silly tricks or anything though. Art.Cascade has explained it really well.

It's kind of funny how people find different things difficult. The hints didn't help me at all because those were the things that I realized almost instantly. It's also fairly obvious that the order in which the prisoners are let in doesn't matter, since they have to assume that the other guys have found/will find their names, or else they're all screwed anyway. Still, I couldn't come up with the solution. I guess I lack the creativity or something ^^

You should definitely not feel bad for not solving the prisoner problem. I don't think many percent manage to find the solution to this riddle themselves. Actually, one of the reasons that I like this problem is that I was the only one (or at least first, don't know what's happened since) solving it in the thread on teamliquid, and it took me over 24 hours. Most of the other puzzles posted at that time (it was a bit of a riddle frenzy at TL for some time) were solved within minutes.

More problems!!!
A) Santas raindeers and helpers
1) Santa has 100 raindeers. Or any large number, maybe he has infinite number of raindeers, who knows? Anyway, not important. A lot of raindeers.
2) They are in booths in the stable. Let's say for simplicity that the booths are numbered 1, 2, ...
3) The booths start all closed.
4) The first of santas little helper (helper number 1) opens all the booths.
5) The second helper (helper number 2) closes all even numbered booths.
6) Helper 3 goes to every booth with a number divisible with 3 (ie, booth 3, 6, 9, ...) and change the state of the booth. Ie, if it's open, he closes it, and if it's closed, he opens it. So booth 3 will be open, so he closes it. Booth 6 is closed, so he opens it. Booth 9 is open, so he closes it. etc.
7) And it continues like this. Helper number n will go to every booth divisible with n, and change the state of the booth. There are as many helpers as booths, so the last helper will just open/close the very last booth.
8) When all the helpers are done, which booths are open, and which are closed?

B) Eye colours at island (hard)
1) An island with 100 inhabitants, all of them very intelligent, logical, rational, and aware that all the other inhabitants are the same.
2) All the inhabitants have either brown or blue eyes, and they are all aware of that.
3) All inhabitants knows the eye colour of all other inhabitants, but no inhabitant knows their own eye colour.
4) If an inhabitant somehow figures out his own eye colour by sunset, he/she will leave the island during the night. The inhabitants see each other daily, so all other will notice before sunset if someone leaves the island.
5) 50 of the inhabitants have blue eyes, 50 have brown. The inhabitants don't know these numbers though. (So for example a inhabitant with brown eyes will see 50 with blue eyes, and 49 with brown eyes, but doesn't know if he is the 51:st blue, or 50:th brown.)
6) The inhabitants never talk about eye colour. Seeing how they are all very intelligent etc, and everyone likes the island, they make absolutely sure to not communicate in any way whatsoever about even the faintest clue regarding eye colours.
7) One day (let's call it day 0 for simplicity) a stranger arrives at the island, gathers all the inhabitants, and loudly declares (everyone hears, everyone knows that everyone else hears, everyone knows that everyone else also knows that everyone hears, etc) that there is at least one person has brown eyes. All the inhabitants trust him (everyone knows that he cannot lie or something).
8) The stranger leaves. Assuming that nothing else disturbs the inhabitants, what will happen when?

I think I got the questions right, but please correct me or ask if anything is unclear.

On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 8) When all the helpers are done, which booths are open, and which are closed?

+ Show Spoiler +

I don't understand something in the second one.
You say in 5) that there are 50 brown eyes ones, and 50 blue eyes ones, and in 3) that all inhabitants know the eye colour of the other inhabitants.
So that should imply, that every inhabitant, knows his own eye colour : I see 49 brown eyes inhabitants, so I have blue eyes, and thus I leave.

A classical one that I will try to put in other words, so that it is not easily found with google.

There are 5 kids who just got 12 candies for halloween and they have to share it. Youngest one has to propose a way to distribute the candies among the kids. Then, they proceed to vote, and if the strict majority rejects the proposition, the youngest kid is evicted and leaves with nothing. And the 2nd younger kid will then have to propose a distribution, and they will again vote for it, and may or not accept or evict this kid. And this goes on, until the distribution is accepted.

Assuming that there are all loving candies and are not friends (they don't matter if a kid is evicted), what is the distribution the youngest have to make in order to get some candies ?

On November 06 2012 06:42 kingpowa wrote: Show nested quote + On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 8) When all the helpers are done, which booths are open, and which are closed? + Show Spoiler + The closed ones are all x² with x²<n and n.

nope

+ Show Spoiler +

On November 06 2012 06:47 kingpowa wrote: I don't understand something in the second one. You say in 5) that there are 50 brown eyes ones, and 50 blue eyes ones, and in 3) that all inhabitants know the eye colour of the other inhabitants. So that should imply, that every inhabitant, knows his own eye colour : I see 49 brown eyes inhabitants, so I have blue eyes, and thus I leave.

I've heard this one before so I won't answer, but the point is that they don't know there are 50 people with brown eyes and 50 with blue eyes.

spoiler because maybe hint:
+ Show Spoiler +

On November 06 2012 08:21 KoeBawlt wrote: Show nested quote + On November 06 2012 06:42 kingpowa wrote:   On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 8) When all the helpers are done, which booths are open, and which are closed? + Show Spoiler + The closed ones are all x² with x²<n and n. nope + Show Spoiler + door n is closed if it is divisible by an even number of numbers between 1 and n.

Your answer is not wrong, but I prefer mine.
go further.

On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 1) Santa has 100 raindeers. Or any large number, maybe he has infinite number of raindeers, who knows? Anyway, not important. A lot of raindeers. 2) They are in booths in the stable. Let's say for simplicity that the booths are numbered 1, 2, ... 3) The booths start all closed. 4) The first of santas little helper (helper number 1) opens all the booths. 5) The second helper (helper number 2) closes all even numbered booths. 6) Helper 3 goes to every booth with a number divisible with 3 (ie, booth 3, 6, 9, ...) and change the state of the booth. Ie, if it's open, he closes it, and if it's closed, he opens it. So booth 3 will be open, so he closes it. Booth 6 is closed, so he opens it. Booth 9 is open, so he closes it. etc. 7) And it continues like this. Helper number n will go to every booth divisible with n, and change the state of the booth. There are as many helpers as booths, so the last helper will just open/close the very last booth. 8) When all the helpers are done, which booths are open, and which are closed? .

+ Show Spoiler +

On November 06 2012 08:21 KoeBawlt wrote: Show nested quote + On November 06 2012 06:42 kingpowa wrote:   On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 8) When all the helpers are done, which booths are open, and which are closed? + Show Spoiler + The closed ones are all x² with x²<n and n. nope + Show Spoiler + door n is closed if it is divisible by an even number of numbers between 1 and n.

You are both right.

Show nested quote + On November 06 2012 06:47 kingpowa wrote: I don't understand something in the second one. You say in 5) that there are 50 brown eyes ones, and 50 blue eyes ones, and in 3) that all inhabitants know the eye colour of the other inhabitants. So that should imply, that every inhabitant, knows his own eye colour : I see 49 brown eyes inhabitants, so I have blue eyes, and thus I leave. I've heard this one before so I won't answer, but the point is that they don't know there are 50 people with brown eyes and 50 with blue eyes. spoiler because maybe hint: + Show Spoiler + So when a brown eyed person sees 50 blue eyed people and 49 brown eyed people she doesn't know whether there are 51 blue eyed people and 49 brown eyed people or 50 of each.

Exactly, thanks. The inhabitants don't know they are 50-50, that's just a piece of information I give to you as the problem solver. I'll edit my post to clarify.

On November 06 2012 15:27 HotChip wrote: Show nested quote + On November 06 2012 05:33 Art.Cascade wrote: More problems!!! A) Santas raindeers and helpers 1) Santa has 100 raindeers. Or any large number, maybe he has infinite number of raindeers, who knows? Anyway, not important. A lot of raindeers. 2) They are in booths in the stable. Let's say for simplicity that the booths are numbered 1, 2, ... 3) The booths start all closed. 4) The first of santas little helper (helper number 1) opens all the booths. 5) The second helper (helper number 2) closes all even numbered booths. 6) Helper 3 goes to every booth with a number divisible with 3 (ie, booth 3, 6, 9, ...) and change the state of the booth. Ie, if it's open, he closes it, and if it's closed, he opens it. So booth 3 will be open, so he closes it. Booth 6 is closed, so he opens it. Booth 9 is open, so he closes it. etc. 7) And it continues like this. Helper number n will go to every booth divisible with n, and change the state of the booth. There are as many helpers as booths, so the last helper will just open/close the very last booth. 8) When all the helpers are done, which booths are open, and which are closed? . + Show Spoiler + It's a sequence wich goes like this: open - 2xClosed - open - 4xClosed - open - 6xClosed - open - 8xClosed - open 10xClosed...

close, but you must have made a mistake somewhere. Go through the first few booths manually and you will see that your solution isn't accurate, and I think you will be able to fix it. The principle seems right though, as the answer is almost right.

see post 17 and 18. already asked.

thank you for pointing this out, didn't notice that ^^

On November 06 2012 07:03 kingpowa wrote: A classical one that I will try to put in other words, so that it is not easily found with google. There are 5 kids who just got 12 candies for halloween and they have to share it. Youngest one has to propose a way to distribute the candies among the kids. Then, they proceed to vote, and if the strict majority rejects the proposition, the youngest kid is evicted and leaves with nothing. And the 2nd younger kid will then have to propose a distribution, and they will again vote for it, and may or not accept or evict this kid. And this goes on, until the distribution is accepted. Assuming that there are all loving candies and are not friends (they don't matter if a kid is evicted), what is the distribution the youngest have to make in order to get some candies ?

+ Show Spoiler +

For the reindeer,

+ Show Spoiler +

I hope they are not really question to hire brokers cuz damn those are retarded.

Mixing exact and precise question with "creative" bullshit is absurd.


What odds i lay on the flip? 9999999999:1... cuz why not? its more profitable and it doesnt say you will reject Ev- offers and we are already guessing the height of the mother fucking elevator guy, those questions are are useful to determine somebodys intelligence as much as a coloring book, shame on any company who uses these.



rant off... carry on

Why absurd? It is incredibly important to be able to identify a problem that is impossible to solve with a direct mathematical approach, that has to be worked around. Many very intelligent people have wasted huge amount of time trying to solve impossible problems.

In real life you will not know if the problem you have in front of you is solvable with mathematics, or if you need "creative bullshit".

On November 10 2012 11:16 waga wrote: Show nested quote + On November 06 2012 07:03 kingpowa wrote: A classical one that I will try to put in other words, so that it is not easily found with google. There are 5 kids who just got 12 candies for halloween and they have to share it. Youngest one has to propose a way to distribute the candies among the kids. Then, they proceed to vote, and if the strict majority rejects the proposition, the youngest kid is evicted and leaves with nothing. And the 2nd younger kid will then have to propose a distribution, and they will again vote for it, and may or not accept or evict this kid. And this goes on, until the distribution is accepted. Assuming that there are all loving candies and are not friends (they don't matter if a kid is evicted), what is the distribution the youngest have to make in order to get some candies ? + Show Spoiler + 2 + 4 x 2.5 = 12

nop. Can't break the candy.
When a kid proposes a sharing, the other ones when voting, will think "If I evict him, will I get more candies ?"
And it is actually about optimal choices.

On November 11 2012 08:19 kingpowa wrote: Show nested quote + On November 10 2012 11:16 waga wrote:   On November 06 2012 07:03 kingpowa wrote: A classical one that I will try to put in other words, so that it is not easily found with google. There are 5 kids who just got 12 candies for halloween and they have to share it. Youngest one has to propose a way to distribute the candies among the kids. Then, they proceed to vote, and if the strict majority rejects the proposition, the youngest kid is evicted and leaves with nothing. And the 2nd younger kid will then have to propose a distribution, and they will again vote for it, and may or not accept or evict this kid. And this goes on, until the distribution is accepted. Assuming that there are all loving candies and are not friends (they don't matter if a kid is evicted), what is the distribution the youngest have to make in order to get some candies ? + Show Spoiler + 2 + 4 x 2.5 = 12 nop. Can't break the candy. When a kid proposes a sharing, the other ones when voting, will think "If I evict him, will I get more candies ?" And it is actually about optimal choices.

+ Show Spoiler +

On November 11 2012 15:08 bober1 wrote: Show nested quote + On November 11 2012 08:19 kingpowa wrote:   On November 10 2012 11:16 waga wrote:   On November 06 2012 07:03 kingpowa wrote: A classical one that I will try to put in other words, so that it is not easily found with google. There are 5 kids who just got 12 candies for halloween and they have to share it. Youngest one has to propose a way to distribute the candies among the kids. Then, they proceed to vote, and if the strict majority rejects the proposition, the youngest kid is evicted and leaves with nothing. And the 2nd younger kid will then have to propose a distribution, and they will again vote for it, and may or not accept or evict this kid. And this goes on, until the distribution is accepted. Assuming that there are all loving candies and are not friends (they don't matter if a kid is evicted), what is the distribution the youngest have to make in order to get some candies ? + Show Spoiler + 2 + 4 x 2.5 = 12 nop. Can't break the candy. When a kid proposes a sharing, the other ones when voting, will think "If I evict him, will I get more candies ?" And it is actually about optimal choices. + Show Spoiler + start with 2 kids. Youngest takes all the candies because the strict majority can't reject it. So its 12-0. If you have 3 kids, the 2nd youngest knows knows if he votes the youngest off he gets all the candy so he always votes the youngest off. The oldest knows if he votes the youngest off he gets none. So you give the oldest 1 candy. 11-0-1 If you have 4 kids, the 2nd youngest knows he can get 11 out of 12 candies if he votes the youngest off so hes always going to vote the youngest off unless he gives him 11 candies. Youngest can just give one candy to the 2nd oldest now because if it gets to the case with 3 kids he gets none. 11-0-1-0 So with 5 kids, 2nd youngest always will vote youngest off unless he gets 11, so youngest should give 1 candy to the oldest and 1 candy to the middle position because they get none if youngest get voted off. 10-0-1-0-1

gg

On November 11 2012 06:35 Art.Cascade wrote: Why absurd? It is incredibly important to be able to identify a problem that is impossible to solve with a direct mathematical approach, that has to be worked around. Many very intelligent people have wasted huge amount of time trying to solve impossible problems. In real life you will not know if the problem you have in front of you is solvable with mathematics, or if you need "creative bullshit".

that has to be the worst rationalization ive heard.

Many intelligent people would have saved so much time if they only knew smartass responses to ridiculous newspaper fun trivia questions.

I gave a perfect example why its ridiculous, ill give 99999999 : 1 odds question, see?