Best riddles

short

I'm in the colosseum

I'm in you, But not in him, I go up, But not down, I'm in the colosseum, But not a tower, I'm in a puzzle, But not a riddle.
The letter U.
88.50 %
36 votes

short

Never even share a brief touch

There is one thing that goes round the house and also inside the house but never even share a brief touch. What is it?
Sun.
88.50 %
36 votes

Anagram of wolf

What emotion is an anagram of a homonym of an antonym of a homonym of an anagram of wolf?
Fear. The anagram of wolf is fowl is foul. The antonym of fowl is fair. The homonym of fair is fare. The anagram of fareis fear, which is the emotion.
88.50 %
36 votes

Light as a feather

This is as light as a feather, yet no man can hold it for long. What am I?
Your Breath.
88.50 %
36 votes

logicshort

Akbar and Birbal

One day, Emperor Akbar posed a question to Birbal. He asked him what Birbal would choose if he offered either justice or a gold coin. "The gold coin," said Birbal without hesitation. On hearing this, Akbar was taken aback. "You would prefer a gold coin to justice?" he asked, not believing his own ears. "Yes," said Birbal. The other courtiers were amazed by Birbal's display of idiocy. They were full of glee that Birbal had finally managed himself to do what these courtiers had not been able to do for a long time - discredit Birbal in the emperor's eyes! "I would have been disappointed if this was the choice made even by my lowliest of servants," continued the emperor. "But coming from you it's not only disappointing, but shocking and sad. I did not know you were so debased!" How did Birbal justify his answer to the enraged and hurt Emperor?
"One asks for what one does not have, Your Majesty." said Birbal, smiling gently and in quiet tones. "Under Your Majesty´s rule, justice is available to everybody. But I am a spendthrift and always short of money and therefore I said I would choose the gold coin." The answer immensely pleased the emperor and respect for Birbal was once again restored in the emperor's eyes.
88.42 %
49 votes

logic

Twelve balls, one different

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.
88.42 %
49 votes

cleanfunny

An orange

What kind of flower lives between your mouth amd chin? Two-lips.
It concentrates.
88.42 %
49 votes