Found In A Kitchen Riddle
I have a handle but Im not a car door
Im found in a kitchen but Im not a cupboard door
I sometimes spread things but Im not a sneeze
Im used to cut things but Im not a pair of scissors
I have a blade but Im not grass
Im found in a kitchen but Im not a cupboard door
I sometimes spread things but Im not a sneeze
Im used to cut things but Im not a pair of scissors
I have a blade but Im not grass
Hint:
Made Of Metal Riddle
Im made of metal but Im not a saucepan
I have a lip but I dont have any teeth
I have a clapper but I dont have any hands
I can ring but Im not a phone
I can peal but Im not a banana
I can jingle but Im not music in an advertisement
I am?
I have a lip but I dont have any teeth
I have a clapper but I dont have any hands
I can ring but Im not a phone
I can peal but Im not a banana
I can jingle but Im not music in an advertisement
I am?
Hint:
Coming Down The Chimney
When used it can warm you up
But try not to burn your hand
When he comes down a chimney
This is where Santa would land?
But try not to burn your hand
When he comes down a chimney
This is where Santa would land?
Hint:
Born In Mourning
I have a name, but it isn't my name. My face shows signs of age. I always mean the same thing, no matter what I say. I'm born in mourning, and I last 'til the end of days. Men plant me, but I never grow. They run from me, but I never move. They look at me and see their future, rotting in the fields where I bloom. What am I?
Hint:
Mortal Privation Riddle
Marking mortal privation, when firmly in place.
An enduring summation, inscribed in my face.
What am I?
An enduring summation, inscribed in my face.
What am I?
Hint:
Lazily In The Sun
My head bobs lazily in the sun. You think I'm cute, For my face is yellow, My hair is white, and my body is green. What am I?
Hint:
Full Of Holes
Hint:
Roll The Dice
A gambler goes to bet. The dealer has 3 dice, which are fair, meaning that the chance that each face shows up is exactly 1/6.
The dealer says: "You can choose your bet on a number, any number from 1 to 6. Then I'll roll the 3 dice. If none show the number you bet, you'll lose $1. If one shows the number you bet, you'll win $1. If two or three dice show the number you bet, you'll win $3 or $5, respectively."
Is it a fair game?
The dealer says: "You can choose your bet on a number, any number from 1 to 6. Then I'll roll the 3 dice. If none show the number you bet, you'll lose $1. If one shows the number you bet, you'll win $1. If two or three dice show the number you bet, you'll win $3 or $5, respectively."
Is it a fair game?
Hint: What will happen if there are 6 gamblers, each of whom bet on a different number?
It's a fair game. If there are 6 gamblers, each of whom bet on a different number, the dealer will neither win nor lose on each deal.
If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1.
If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1.
If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1.
In each case, the dealer neither wins nor loses. Hence it's a fair game. Did you answer this riddle correctly?
YES NO
If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1.
If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1.
If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1.
In each case, the dealer neither wins nor loses. Hence it's a fair game. Did you answer this riddle correctly?
YES NO
Three People In A Room
Three people enter a room and have a green or blue hat placed on their head. They cannot see their own hat, but can see the other hats.
The color of each hat is purely random. They could all be green, or blue, or any combination of green and blue.
They need to guess their own hat color by writing it on a piece of paper, or they can write 'pass'.
They cannot communicate with each other in any way once the game starts. But they can have a strategy meeting before the game.
If at least one of them guesses correctly they win $50,000 each, but if anyone guess incorrectly they all get nothing.
What is the best strategy?
The color of each hat is purely random. They could all be green, or blue, or any combination of green and blue.
They need to guess their own hat color by writing it on a piece of paper, or they can write 'pass'.
They cannot communicate with each other in any way once the game starts. But they can have a strategy meeting before the game.
If at least one of them guesses correctly they win $50,000 each, but if anyone guess incorrectly they all get nothing.
What is the best strategy?
Hint:
Simple strategy: Elect one person to be the guesser, the other two pass. The guesser chooses randomly 'green' or 'blue'. This gives them a 50% chance of winning.
Better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down 'pass'.
It works like this ('-' means 'pass'):
Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose
Result: 75% chance of winning! Did you answer this riddle correctly?
YES NO
Better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down 'pass'.
It works like this ('-' means 'pass'):
Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose
Result: 75% chance of winning! Did you answer this riddle correctly?
YES NO
The Secret Santa Exchange
A group of ten friends decide to exchange gifts as secret Santas. Each person writes his or her name on a piece of paper and puts it in a hat. Then each person randomly draws a name from the hat to determine who has him as his or her secret Santa. The secret Santa then makes a gift for the person whose name he drew.
When it's time to exchange presents, each person walks over to the person he made the gift for and holds his or her left hand in his right hand.
What is the probability that the 10 friends holding hands form a single continuous circle?
When it's time to exchange presents, each person walks over to the person he made the gift for and holds his or her left hand in his right hand.
What is the probability that the 10 friends holding hands form a single continuous circle?
Hint: It's not as difficult as it seems.
It's the number of ways the friends can form a circle divided by the number of ways the names can be drawn out of the hat.
1/10
For a group of n friends, there are n! (n factorial) ways to draw the names out of the hat. Since a circle does not have a beginning and end, choose one person as the beginning and end of the circle. There are now (n-1)! ways to distribute the remaining people around the circle. Thus the probability of forming a single circle is
(n-1)! / n!
Since n! = (n-1)! * n (for n > 1), this can be rewritten as
(n-1)! / (n*(n-1)!)
Factoring out the (n-1)! from the numerator and denominator leaves
1/n
as the probability. Did you answer this riddle correctly?
YES NO
For a group of n friends, there are n! (n factorial) ways to draw the names out of the hat. Since a circle does not have a beginning and end, choose one person as the beginning and end of the circle. There are now (n-1)! ways to distribute the remaining people around the circle. Thus the probability of forming a single circle is
(n-1)! / n!
Since n! = (n-1)! * n (for n > 1), this can be rewritten as
(n-1)! / (n*(n-1)!)
Factoring out the (n-1)! from the numerator and denominator leaves
1/n
as the probability. Did you answer this riddle correctly?
YES NO
The 3 Inch Cube Riddle
A 3 inch cube is painted on all sides with RED. The cube is then cut into small cubes of dimension 1 inch. All the so cut cubes are collected and thrown on a flat surface. What is the probability that all the top facing surfaces have RED paint on them?
Hint: Visualize the core of the cube.
ZERO.
The core of the 3 inch cube when cut, has all faces that are not painted. Hence at least one cube with no painted face always occurs. Did you answer this riddle correctly?
YES NO
The core of the 3 inch cube when cut, has all faces that are not painted. Hence at least one cube with no painted face always occurs. Did you answer this riddle correctly?
YES NO
A Thought In Your Mind
I can bring a smile to your face, a tear to your eye or even a thought to your mind, but I can't be seen. What Am I?
Hint:
A Town With No Houses
Hint:
On Display In December
I have numbers on me but Im not a phone
I have several doors but Im not a car
Im on display in December but Im not a nativity scene
Im not seen after Christmas Day but Im not a cookie for Santa
I often contain chocolate but Im not an Easter egg
What am I?
I have several doors but Im not a car
Im on display in December but Im not a nativity scene
Im not seen after Christmas Day but Im not a cookie for Santa
I often contain chocolate but Im not an Easter egg
What am I?
Hint:
Prints In The Sand
Im something with five digits
But I am not a hand
When you walk along the beach
I leave prints in the sand
What could I be?
But I am not a hand
When you walk along the beach
I leave prints in the sand
What could I be?
Hint:
Add Your Riddle Here
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