The worlds first riddle!

Do you two live next door to each other? You are way too good for me. Try this;

Four men are buried up to their necks in the ground. They cannot move so can only look forward. Between A & B is a brick wall, which cannot be seen through. They know that between them are 4 hats, 2 are black and 2 are white, but they do not know which colour they are wearing. In order to avoid being shot one of them must call out to the executioner the colour of the hat they are wearing. If they get it wrong, everyone will be shot. They are not allowed to talk to each other and have 10 minutes to work it out.

Note A & C have black hats.

1 Which one of them calls out

2 Why is he 100% certain of the colour of his hat

This is not a trick question. .
There are no outside influences, nor any other ways of communicating. They cannot move and are buried in a straight line. A & B can only see their respective sides of the brick wall, C can see B and D can see B & C.

On a certain house, the two halves of the roof are unequally pitched; one-half slopes downward at an angle of 62 degrees and the other half at an angle of 77 degrees. Suppose a rooster lays an egg right on the peak. On which side of the roof would the egg fall

I'm not sure how they are lined?
A, (brick wall), B, C, D - so that A and B can't see anything, C can see B, and D can see C and B?

If that's it then it's really easy (but I think I haven't understood completely) - because if D would see two hats of same colour in front of him, he would know colour of his hat - in addition, C knows that D does not see two hats of same colour, so he figures that he has to have different colour then B.
So he calls out - saying different colour then one on B's head...

However, I don't think I understood it correctly, especially since I think that very very similar hat question was already on this topic.

and no, Magnum and I are not living next door to each other, but colours on our flags are in same order, so maybe that's the clue?

A graphic came with the question above, but it would not copy. However, you are right it was too easy for you. This one caused a few problems.


A student fails his Mathematics exam and goes postal. He kidnaps three of his teachers, all well-known mathematicians, and keeps them in captivity. He puts a hat on the head of each one with a number on it.

He tells his captives, "All of the numbers are positive and none of them is zero. They are whole numbers, and one of them is the sum of the other two. All you have to do is figure out what number your hat has on it . . . or else! Mwuhahahah"

The three numbers are 5, 3, 2. (To simplify explanations we will call the one with '5' A, with '3' B; and with '2' C) None of them knows what is on his own hat but he can see the other two. No talking allowed.

Can they do it or are they doomed

•100.010 You are trapped in a small phone booth shaped room. In the middle of each side of the room there is a hole. In each hole there is a light switch. You can't see in the holes but you can reach your hands in them and flip the switches. You may stick your hands in any two holes at the same time and flip none, either, or both of the switches as you please. Nothing will happen until you remove both hands from the holes.

You succeed if you get all the switches into the same position, after which time you will immediately be released from the room. Unless you escape, after removing your hands the room will spin around, disorienting you so you can't tell which side is which. How can you escape

well I'm not saying it's generally very easy task, just that this one was easiest in "hats series"

But this with numbers is still bugging me...c ya later if I come up with something

okay, I actually got it - call it a sudden enlightening will send you by pm though, don't want to spoil everything to others, let them think sometime too...it's a very good one, I will call it a lucky shot, it came to me while I was thinking that I should quit for today (over midnight in Croatia)

pm sent

MOU got it in one.

Three players enter a room and a red or blue hat is placed on each person's head. The color of each hat is determined by a coin toss, with the outcome of one coin toss having no effect on the others. Each person can see the other players' hats but not his own.

No communication of any sort is allowed, except for an initial strategy session before the game begins. Once they have had a chance to look at the other hats, the players must simultaneously guess the color of their own hats or pass. The group shares a hypothetical $3 million prize if at least one player guesses correctly and no players guess incorrectly.

The same game can be played with any number of players. The general problem is to find a strategy for the group that maximizes its chances of winning the prize.

One obvious strategy for the players, for instance, would be for one player to always guess "red" while the other players pass. This would give the group a 50 percent chance of winning the prize.

Question: Can the group do better than 50%


Four men and four women are shipwrecked on a deserted island. Eventually each person falls in love with one person and is loved by one person. You are given the following information:
o Chad loves the girl who is in love with David.
o Arthur loves the girl who loves the man who loves Ellen.
o Bruce loves the girl who loves the man who loves Mary.
o Gloria does not love Bruce.
o Helen loves a man who does not love Gloria.
o There is no mutual love interest (nobody loves the person who loves them back)
o Nobody is homosexual or narcissist.

Who loves who

what do you mean by simultaneously? they have to say it in the same moment?
if not, then they can agree before the game that one that see two hats of same colour is the one that will say "pass" first (and one has to see two hats of same colour in every single combination - if all hats are R or B then it's obvius, while if there is 2B and 1R, or 2R and 1B, then one with single colour will see two B or two R), and then the other too both know colour of their hats....

"what do you mean by simultaneously? they have to say it in the same moment?"

I took it to mean, ?'at the same time' more or less.

The ?'official' answer is;

a) If the strategy is "honest" the chance of guessing correctly if one names a color and two say "pass" is 50% no matter what.
If two guess and one passes the chance of both being correct is 25%
If all three guess the chance of all three being correct is 12.5%.
This will be true no matter what the strategy.

b) If the strategy is a "Cheat", e.g. they agree that only 'C' makes a guess while 'A' and 'B' both say "Pass" and at the same time both 'A' and 'B' look at 'C' if he has a red hat and at one another if he has a blue hat, the chance of them winning is 100%.

c) Here is how it works out with all the possible combinations;

Person A - R R R R B B B B
Person B - R B R B R B R B
Person C - R R B B R R B B

In the case of the first and last combination (all red or all blue) all three will simultaneously pass -- and they will lose.

In all the other cases -- one person will see two of one color -- and will guess the opposite color -- and they will win.

Therefore, the odds are 6 in 8 -- or 3 in 4. 75%

I hope that is clear. Damn hats.

my brain hurts but i believe that:

Bruce – Ellen
Arthur – Helen
Chad – Mary
David – Gloria

Gloria – Arthur
Ellen – Chad
Mary – David
Helen – Bruce

assuming that all the teeth are the same size, that makes R the largest one. i got the conclusion: 15015

of course, in lateral thinking, you could say that the largest one has to move 11, though the word will read a bit crooked. :wink:

i am a huge moron. after reading ONLY THE FIRST PAGE, i sent a reply not realizing there are 21 REPLY PAGES!!!

though i take comfort in knowing that i was somewhat right with 15015, i just forgot to divide by 143 again to get the number of revolutions, being 105. and i will ALWAYS stand by this answer because the letter Y is never written with all three lines of the letter being equidistant and equilateral.

writing the letter Y like a child in order to get a cooler, smaller answer is pretty unfair, you lateral thinking jerks. :wink:

don't worry, one of my first posts was reply on conversation about european baseball and last post was (I noticed that later) around six months old

yeah, it would be too simple
however, here's small help: without any doubt you were thinking in right way about this task, but you stopped too fast

"Ellen loves Chad who loves Mary who loves David who loves Gloria who loves Arthur who loves Helen who loves Bruce who loves Ellen."

MOU in a short space of time you have reached ?'Genius' status.

Welcome Ivan. Quote, "writing the letter Y like a child in order to get a cooler, smaller answer is pretty unfair, you lateral thinking jerks."

May I say, fairness has nothing to do with it. I am only six years old. :wink:

Now for something a bit different. This question was sent to me with the answer, but no matter how hard I try, I cannot understand the logic, or make it work if the booth spins in a random way and you can never know how many times you have been there before. Can anyone give a simple answer?



You are trapped in a small phone booth shaped room. In the middle of each side of the room there is a hole. In each hole there is a light switch. You can't see in the holes but you can reach your hands in them and flip the switches. You may stick your hands in any two holes at the same time and flip none, either, or both of the switches as you please. Nothing will happen until you remove both hands from the holes. You succeed if you get all the switches into the same position, after which time you will immediately be released from the room. Unless you escape, after removing your hands the room will spin around, disorienting you so you can't tell which side is which. How can you escape

Here is the solution. At the end of any step you may win, otherwise proceed to the next step. D=Off. U=On.

1. Pick two adjacent holes and turn both switches to on.
2. Pick two opposite holes and turn both switches to on. Assuming I didn't win then I must have a DUUU configuration.
3. Pick two opposite holes. If one switch is down turn up and win. Otherwise turn one down for a DDUU configuration.
4. Pick two adjacent holes. If the switches are in the same position, then switch both (and win). If different, switch them for a DUDU configuration.
5. Pick opposite holes and switch both switches (and win)



Tricky, but no math knowledge required.

Four mathematicians have the following conversation:
Alice: I am insane.
Bob: I am pure.
Charlie: I am applied.
Dorothy: I am sane.
Alice: Charlie is pure.
Bob: Dorothy is insane.
Charlie: Bob is applied.
Dorothy: Charlie is sane.

You are also given that:
o Pure mathematicians tell the truth about their beliefs.
o Applied mathematicians lie about their beliefs.
o Sane mathematicians beliefs are correct.
o Insane mathematician's beliefs are incorrect.

There are four possible kinds of mathematicians:
1. Pure and sane
2. Pure and insane
3. Applied and sane
4. Applied and insane

Describe the four mathematicians

i was a bit late i suppose but i've found the same answer:

Arthur-->Helen-->Bruce-->Ellen-->Chad-->Mary-->David-->Gloria-->Arthur

I assume that each letter is different number and that it is between 0/9??

Quote, "I assume that each letter is different number and that it is between 0/9??"

You assume correctly. However, before you use up too much time, I warn you it is a very difficult question to answer. Good luck

If you still want to give it a try, M = 1

while O is 0; S is 9, N is E+1, and I leave you there

try: pm sent