No Farting In Public
Hint:
Throwing A Basketball Riddle
A man takes a basketball and throws it as hard as he can. There is nothing in front, behind, or on either side of him, and yet, the ball comes back and hits him square in the face. How can this be?
Hint:
The Parrot Doors Riddle
There are two doors. One door lead to Heaven, while the other leads to Hell. A parrot stands in front of each door. One parrot always tells a lie, while the other always tells the truth. You do not know which parrot or door is which. You are allowed to only ask one question. So, what one question must you ask to determine which door is which, so you can finally go to Heaven? (Hint: The question involves what one parrot would say about the doors.)
Hint:
It doesn't matter which parrot you ask the question to, but the question would be, "What door would the other parrot say is Heaven?". Then you would choose the other door. Did you answer this riddle correctly?
YES NO
YES NO
I Can Fly Riddle
Im colorful but Im not a rainbow
Im sometimes found on a shoulder but Im not a bag strap
I have two legs but Im not an ostrich
I can talk but Im not a person
I can fly but Im not an airplane
I'm a...
Im sometimes found on a shoulder but Im not a bag strap
I have two legs but Im not an ostrich
I can talk but Im not a person
I can fly but Im not an airplane
I'm a...
Hint:
Hearing Santas Sleigh
If you hear these jingling around
The night just before Christmas Day
You had better try to get to sleep
As you are hearing Santas sleigh
Whats jingling?
The night just before Christmas Day
You had better try to get to sleep
As you are hearing Santas sleigh
Whats jingling?
Hint:
Elevator Accident Riddle
Im in an elevator with two other people. When it reaches the first floor, one person gets out and six get in. When it reaches the second floor, three people get out and twelve get in. At the third floor, five leave and nine enter. It rises to the fourth floor, one person gets on and the doors close. Suddenly, the elevator cable snaps and the car smashes to the ground. No one survives the fall, yet Im alive and know exactly how many people go on and off the elevator at every floor. How is this possible?
Hint:
I got off at the first floor. Im a security guard and knew how many people got on and off the elevator by watching the surveillance footage. Did you answer this riddle correctly?
YES NO
YES NO
Put It In A Glove
This is something you might hold
Of a person that you love
When it's really cold outside
You might put it in a glove
Of a person that you love
When it's really cold outside
You might put it in a glove
Hint:
2000 Degrees Celsius
Hint:
Swallowed Up By A Whale
Instead of going to Nineveh
This person decided to bail
Thrown overboard while out to sea
He was swallowed up by a whale
He is...
This person decided to bail
Thrown overboard while out to sea
He was swallowed up by a whale
He is...
Hint:
A Pebble And A Sling
Its amazing what this person did
With a pebble and a sling
He got rid of a giant
And eventually became king
Who was this person?
With a pebble and a sling
He got rid of a giant
And eventually became king
Who was this person?
Hint:
Good Friday Riddle
You would have seen three of these
On the Calvary hillside
One of which carried Jesus
On Good Friday when he died
What was it?
On the Calvary hillside
One of which carried Jesus
On Good Friday when he died
What was it?
Hint:
The Miracle Mountain Riddle
A hiker climbs all day up a steep mountain path and arrives at the mountain top where he camps overnight. The next day he begins the descent down the same trail to the bottom of the mountain when suddenly he looks at his watch and exclaims, "That is amazing! I was at this very same spot at exactly the same time of day yesterday on my way up."
What is the probability that a hiker will be at exactly the same spot on the mountain at the same time of day on his return trip, as he was on the previous day's hike up the mountain?
Is the probability closest to (A) 99% or (B) 50% or (C) 0.1% ?
What is the probability that a hiker will be at exactly the same spot on the mountain at the same time of day on his return trip, as he was on the previous day's hike up the mountain?
Is the probability closest to (A) 99% or (B) 50% or (C) 0.1% ?
Hint: This is not a trick. His watch works perfectly well. He does not sit in the same spot all day or any other such device, although it would not change the answer if he did!
The answer is (A). Since it must happen, the probability is actually 1 (100%).
Explanation: Firstly, consider 2 men, one starting from the top of the mountain and hiking down while the other starts at the bottom and hikes up. At some time in the day, they will cross over. In other words they will be at the same place at the same time of day.
Now consider our man who has walked up on one day and begins the descent the next day. Imagine there is someone (a second person) shadowing his exact movements from the day before. When he meets his shadower (it must happen) it will be the exact place that he was the day before, and of course they are both at this spot at the same time.
Contrary to our common sense, which seems to say that this is an extremely unlikely event, it is a certainty.
NOTE: There is one unlikely event here, and that is that he will notice the time when he is at the correct location on both days, but that was not what the question asked. Did you answer this riddle correctly?
YES NO
Explanation: Firstly, consider 2 men, one starting from the top of the mountain and hiking down while the other starts at the bottom and hikes up. At some time in the day, they will cross over. In other words they will be at the same place at the same time of day.
Now consider our man who has walked up on one day and begins the descent the next day. Imagine there is someone (a second person) shadowing his exact movements from the day before. When he meets his shadower (it must happen) it will be the exact place that he was the day before, and of course they are both at this spot at the same time.
Contrary to our common sense, which seems to say that this is an extremely unlikely event, it is a certainty.
NOTE: There is one unlikely event here, and that is that he will notice the time when he is at the correct location on both days, but that was not what the question asked. Did you answer this riddle correctly?
YES NO
The Last Cookie Riddle
Mike and James are arguing over who gets the last cookie in the jar, so their dad decides to create a game to settle their dispute. First, Mike flips a coin twice, and each time James calls heads or tails in the air. If James gets both calls right, he gets the last cookie. If not, Mike picks a number between one and six and then rolls a die. If he gets the number right, he gets the last cookie. If not, James picks two numbers between one and five, then spins a spinner with numbers one through five on it. If the spinner lands on one of James' two numbers, he gets the last cookie. If not, Mike does.
Who is more likely to win the last cookie, Mike or James? And what is the probability that person wins it?
Who is more likely to win the last cookie, Mike or James? And what is the probability that person wins it?
Hint: Their dad is a very smart person.
Believe it or not, both Mike and James have a 1/2 chance of winning.
James wins if:
-he calls both coin flips right = 1/2 x 1/2 = 1/4
OR
-he does not call both coin flips right, Mike does not call the die roll correctly, and he guesses the number on the spinner right = 3/4 x 5/6 x 2/5 = 30/120 = 1/4
1/4 + 1/4 = 1/2
Mike wins if:
-James does not call both coin flips right and he calls the die roll correctly = 3/4 x 1/6 = 3/24 = 1/8
OR
-James does not call both coin flips right, he does not call the die roll correctly, and Mike does not guess the number on the spinner right = 3/4 x 5/6 x 3/5 = 45/120 = 3/8
1/8 + 3/8 = 1/2
Of course, dad could have just flipped a coin Did you answer this riddle correctly?
YES NO
James wins if:
-he calls both coin flips right = 1/2 x 1/2 = 1/4
OR
-he does not call both coin flips right, Mike does not call the die roll correctly, and he guesses the number on the spinner right = 3/4 x 5/6 x 2/5 = 30/120 = 1/4
1/4 + 1/4 = 1/2
Mike wins if:
-James does not call both coin flips right and he calls the die roll correctly = 3/4 x 1/6 = 3/24 = 1/8
OR
-James does not call both coin flips right, he does not call the die roll correctly, and Mike does not guess the number on the spinner right = 3/4 x 5/6 x 3/5 = 45/120 = 3/8
1/8 + 3/8 = 1/2
Of course, dad could have just flipped a coin Did you answer this riddle correctly?
YES NO
100 Blank Cards Riddle
Someone offers you the following deal:
There is a deck of 100 initially blank cards. The dealer is allowed to write ANY positive integer, one per card, leaving none blank. You are then asked to turn over as many cards as you wish. If the last card you turn over is the highest in the deck, you win; otherwise, you lose.
Winning grants you $50, and losing costs you only the $10 you paid to play.
Would you accept this challenge?
There is a deck of 100 initially blank cards. The dealer is allowed to write ANY positive integer, one per card, leaving none blank. You are then asked to turn over as many cards as you wish. If the last card you turn over is the highest in the deck, you win; otherwise, you lose.
Winning grants you $50, and losing costs you only the $10 you paid to play.
Would you accept this challenge?
Hint: Perhaps thinking in terms of one deck is the wrong approach.
Yes!
A sample strategy:
Divide the deck in half and turn over all lower 50 cards, setting aside the highest number you find. Then turn over the other 50 cards, one by one, until you reach a number that is higher than the card you set aside: this is your chosen "high card."
Now, there is a 50% chance that the highest card is contained in the top 50 cards (it is or it isn't), and a 50% chance that the second-highest card is contained in the lower 50. Combining the probabilities, you have a 25% chance of constructing the above situation (in which you win every time).
This means that you'll lose three out of four games, but for every four games played, you pay $40 while you win one game and $50. Your net profit every four games is $10.
Obviously, you have to have at least $40 to start in order to apply this strategy effectively. Did you answer this riddle correctly?
YES NO
A sample strategy:
Divide the deck in half and turn over all lower 50 cards, setting aside the highest number you find. Then turn over the other 50 cards, one by one, until you reach a number that is higher than the card you set aside: this is your chosen "high card."
Now, there is a 50% chance that the highest card is contained in the top 50 cards (it is or it isn't), and a 50% chance that the second-highest card is contained in the lower 50. Combining the probabilities, you have a 25% chance of constructing the above situation (in which you win every time).
This means that you'll lose three out of four games, but for every four games played, you pay $40 while you win one game and $50. Your net profit every four games is $10.
Obviously, you have to have at least $40 to start in order to apply this strategy effectively. 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
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