Best long riddles

logicclean

You have twelve balls, identical in every way except that one of them weighs slightly less or more than the balls. You have a balance scale, and are allowed to do 3 weighings to determine which ball has the different weight, and whether the ball weighs more or less than the other balls. What process would you use to weigh the balls in order to figure out which ball weighs a different amount, and whether it weighs more or less than the other balls?
Take eight balls, and put four on one side of the scale, and four on the other. If the scale is balanced, that means the odd ball out is in the other 4 balls. Let's call these 4 balls O1, O2, O3, and O4. Take O1, O2, and O3 and put them on one side of the scale, and take 3 balls from the 8 "normal" balls that you originally weighed, and put them on the other side of the scale. If the O1, O2, and O3 balls are heavier, that means the odd ball out is among these, and is heavier. Weigh O1 and O2 against each other. If one of them is heavier than the other, this is the odd ball out, and it is heavier. Otherwise, O3 is the odd ball out, and it is heavier. If the O1, O2, and O3 balls are lighter, that means the odd ball out is among these, and is lighter. Weigh O1 and O2 against each other. If one of them is lighter than the other, this is the odd ball out, and it is lighter. Otherwise, O3 is the odd ball out, and it is lighter. If these two sets of 3 balls weigh the same amount, then O4 is the odd ball out. Weight it against one of the "normal" balls from the first weighing. If O4 is heavier, then it is heavier, if it's lighter, then it's lighter. If the scale isn't balanced, then the odd ball out is among these 8 balls. Let's call the four balls on the side of the scale that was heavier H1, H2, H3, and H4 ("H" for "maybe heavier"). Let's call the four balls on the side of the scale that was lighter L1, L2, L3, and L4 ("L" for "maybe lighter"). Let's also call each ball from the 4 in the original weighing that we know aren't the odd balls out "Normal" balls. So now weigh [H1, H2, L1] against [H3, L2, Normal]. -If the [H1, H2, L1] side is heavier (and thus the [H3, L2, Normal] side is lighter), then this means that either H1 or H2 is the odd ball out and is heavier, or L2 is the odd ball out and is lighter. -So measure [H1, L2] against 2 of the "Normal" balls. -If [H1, L2] are heavier, then H1 is the odd ball out, and is heavier. -If [H1, L2] are lighter, then L2 is the odd ball out, and is lighter. -If the scale is balanced, then H2 is the odd ball out, and is heavier. -If the [H1, H2, L1] side is lighter (and thus the [H3, L2, Normal] side is heavier), then this means that either L1 is the odd ball out, and is lighter, or H3 is the odd ball out, and is heavier. -So measure L1 and H3 against two "normal" balls. -If the [L1, H3] side is lighter, then L1 is the odd ball out, and is lighter. -Otherwise, if the [L1, H3] side is heavier, then H3 is the odd ball out, and is heavier. If the [H1, H2, L1] side and the [H3, L2, Normal] side weigh the same, then we know that either H4 is the odd ball out, and is heavier, or one of L3 or L4 is the odd ball out, and is lighter. So weight [H4, L3] against two of the "Normal" balls. If the [H4, L3] side is heavier, then H4 is the odd ball out, and is heavier. If the [H4, L3] side is lighter, then L3 is the odd ball out, and is lighter. If the [H4, L3] side weighs the same as the [Normal, Normal] side, then L4 is the odd ball out, and is lighter.
72.91 %
88 votes
cleanpoemssimple

Pronounced as one letter, And written with three, Two letters there are, And two only in me. I’m double, I’m single, I’m black blue and gray, I’m read from both ends, And the same either way.
Eye.
72.74 %
117 votes
cleanpoemssimplelogic

The front of me is the source of a song Or to kiss with a fervor of love lifelong. My back is a plant fit for a queen, Crafted by needle, chemical, or machine.
Necklace
72.70 %
83 votes
logicstorydetective

Emperor Akbar once ruled over India. He was a wise and intelligent ruler; and he had in his court the Nine Gems, his nine advisors, who were each known for a particular skill. One of these Gems was Birbal, known for his wit and wisdom. The story below is one of the examples of his wit. Do you have it in you to find the answer? One day the Emperor Akbar stumbled on a small rock in the royal gardens and momentarily went off balance. He was in a bad mood that day and the incident only served to make him more angry. Finding a target for his mood of the day, he ordered the gardener's arrest and execution. Birbal heard of this and visited the gardener in the cell where he was being held awaiting execution. Birbal had known the gardener for many years and also knew of the gardener's immense respect and sense of loyalty for the king. He decided to help the gardener escape the death sentence and explained his plan to the gardener, who reluctantly agreed to go along. The next day the gardener was asked what his last wish was before he was hanged, as was custom. The gardener requested an audience with the emperor. This wish was granted, but when the man neared the throne he tried to attack the emperor. The emperor was shocked and demanded an explanation. The gardener looked at Birbal, who stepped forward and explained why the gardener had attacked the emperor. The emperor immediately realised how unjust he had been and ordered the release of the gardener. How did Birbal manage this?
"Your Majesty," said Birbal, "there is probably no person more loyal to you than this unfortunate gardener. Fearing that people would say you hanged him for a silly reason and question your sense of justice, he went out of his way to give you a genuine reason for hanging him."
72.49 %
95 votes
cleanlogicmath

Mick and John were in a 100 meter race. When Mick crossed the finish line, John was only at the 90 meter mark. Mick suggested they run another race. This time, Mick would start ten meters behind the starting line. All other things being equal, will John win, lose, or will it be a tie in the second race?
John will lose again. In the second race, Mick started ten meters back. By the time John reaches the 90 meter mark, Mick will have caught up him. Therefore, the final ten meters will belong to the faster of the two. Since Mick is faster than John, he will win the final 10 meters and of course the race.
72.45 %
65 votes
logiccleverstory

A king decided to let a prisoner try to escape the prison with his life. The king placed 2 marbles in a jar that was glued to a table. One of the marbles was supposed to be black, and one was supposed to be blue. If the prisoner could pick the blue marble, he would escape the prison with his life. If he picked the black marble, he would be executed. However, the king was very mean, and he wickedly placed 2 black marbles in the jars and no blue marbles. The prisoner witnessed the king only putting 2 black marbles in the jars. If the jar was not see-through and the jar was glued to the table and that the prisoner was mute so he could not say anything, how did he escape with his life?
The prisoner grabbed one of the marbles from the jar and concealed it in his hand. He then swallowed it, and picked up the other marble and showed everyone. The marble was black, and since the other marble was swallowed, it was assumed to be the blue one. So the mean king had to set him free.
72.34 %
56 votes
cleanlogicwhat am I

I'm a fruit. If you take away my first letter, I'm a crime. If you take away my first two letters, I'm an animal. If you take away my first and last letters, I'm a form of music. What am I?
Grape.
72.33 %
69 votes
logicmysteryscarydetective

On the first day of the school a young girl was found murdered. Police suspect four male teachers and question them. They were asked what they were doing at 8:00 am. Mr. Walter: I was driving to school and I was late. Mr. Thomas: I was checking English exam papers. Mr. Benjamin: I was reading the newspaper. Mr. Calvin: I was with my wife in my office. The police arrested the killer. How did the police find the murderer?
Mr.Thomas as he cannot be checking exam papers on the first day of school.
72.30 %
2289 votes
logicmathclean

What is the least number of people that need to be in a room such that there is greater than a 50% chance that at least two of the people have the same birthday?
Only 23 people need to be in the room. Our first observation in solving this problem is the following: (the probability that at least 2 people have the same birthday + the probability that nobody has the same birthday) = 1.0 What this means is that there is a 100% chance that EITHER everybody in the room has a different birthday, OR at least two people in the room have the same birthday (and these probabilities don't add up to more than 1.0 because they cover mutually exclusive situations). With some simple re-arranging of the formula, we get: the probability that at least 2 people have the same birthday = (1.0 - the probability that nobody has the same birthday) So now if we can find the probability that nobody in the room has the same birthday, we just subtract this value from 1.0 and we'll have our answer. The probability that nobody in the room has the same birthday is fairly straightforward to calculate. We can think of this as a "selection without replacement" problem, where each person "selects" a birthday at random, and we then have to figure out the probability that no two people select the same birthday. The first selection has a 365/365 chance of being different than the other birthdays (since none have been selected yet). The next selection has a 364/365 chance of being different than the 1 birthday that has been selected so far. The next selection has a 363/365 chance of being different than the 2 birthdays that have been selected so far. These probabilities are multiplied together since each is conditional on the previous. So for example, the probability that nobody in a room of 3 people have the same birthday is (365/365 * 364/365 * 363/365) =~ 0.9918 More generally, if there are n people in a room, then the probability that nobody has the same birthday is (365/365 * 364/365 * ... * (365-n+2)/365 * (365-n+1)/365) We can plug in values for n. For n=22, we get that the probability that nobody has the same birthday is 0.524, and thus the probabilty that at least two people have the same birthday is (1.0 - 0.524) = 0.476 = 47.6%. Then for n=23, we get that the probability that nobody has the same birthday is 0.493, and thus the probabilty that at least two people have the same birthday is 1.0 - 0.493) = 0.507 = 50.7%. Thus, once we get to 23 people we have reached the 50% threshold.
72.19 %
135 votes