Meeting Your Fate
If you ever happen to see one of these etched
With your name, birth year and a future date
You might want to be worried that there is someone
Who knows when it is youll meet your fate
What is this?
With your name, birth year and a future date
You might want to be worried that there is someone
Who knows when it is youll meet your fate
What is this?
Hint:
Winged Creatures Riddle
These winged creatures often scare people
If you touch one you are brave
Theyre often found just hanging out
Upside down inside a cave
What are these creatures?
If you touch one you are brave
Theyre often found just hanging out
Upside down inside a cave
What are these creatures?
Hint:
Bruce Wayne Is This Type Of Man
Its black, furry and flies about
Try not to be scared if you can
It likes to hang out upside down
And Bruce Wayne is this type of man
Who am I?
Try not to be scared if you can
It likes to hang out upside down
And Bruce Wayne is this type of man
Who am I?
Hint:
The Loaded Revolver Riddle
Henry has been caught stealing cattle, and is brought into town for justice. The judge is his ex-wife Gretchen, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Henry from close range. To make things a little better for Henry, Gretchen tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot. If Henry is still alive, she will then either take another shot, or spin the chamber again before shooting.
Henry is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower. He steels himself as Gretchen loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Gretchen asks, "Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?"
What should Henry choose?
Henry is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower. He steels himself as Gretchen loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Gretchen asks, "Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?"
What should Henry choose?
Hint:
Henry should have Gretchen pull the trigger again without spinning.
We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.
If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position. Did you answer this riddle correctly?
YES NO
We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.
If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position. 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
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 Cheap Mp3 Player
My MP3 player is cheap 'n' nasty and has now broken: it is stuck on 'Shuffle'. In this mode it starts with whatever track you put it on, but then plays tracks in a random order. The only restriction is it never plays a song that's already been played that day.
I purchased my favourite murder mystery book in audio format, and put the first 6 chapters on my MP3 player. (Each chapter is exactly 1 track.) There's nothing else on my player at the moment. What is the probability that I will hear the 6 chapters in order as I listen today, without having to change tracks at all? (Obviously, I will ensure it plays chapter 1 first.)
The next day I empty the player before putting on the next 6 chapters. This time I also transfer a CD of mine with 11 songs on. I don't mind songs coming in between the chapters of my book, as long as the chapters are in order. What's the probability of that happening now?
I purchased my favourite murder mystery book in audio format, and put the first 6 chapters on my MP3 player. (Each chapter is exactly 1 track.) There's nothing else on my player at the moment. What is the probability that I will hear the 6 chapters in order as I listen today, without having to change tracks at all? (Obviously, I will ensure it plays chapter 1 first.)
The next day I empty the player before putting on the next 6 chapters. This time I also transfer a CD of mine with 11 songs on. I don't mind songs coming in between the chapters of my book, as long as the chapters are in order. What's the probability of that happening now?
Hint:
With only 6 tracks on the player:
The first chapter has been set to play first. The probability of the next 5 chapters playing in order is 1/5! = 1/120.
With the music on the player as well:
Seeing as I don't care about when the music plays, it doesn't change anything. The answer is still 1/120. Did you answer this riddle correctly?
YES NO
The first chapter has been set to play first. The probability of the next 5 chapters playing in order is 1/5! = 1/120.
With the music on the player as well:
Seeing as I don't care about when the music plays, it doesn't change anything. The answer is still 1/120. Did you answer this riddle correctly?
YES NO
The Prime Number Riddle
Two hundred people in an auditorium are asked to think of a single digit number from 1 to 9 inclusive and write it down. All those who wrote down a prime number are now asked to leave. Ninety people remain behind in the hall. How many of these are expected to have written down an odd number?
Hint: Remember that 1 is not a prime number.
Those that remain behind must have written {1,4,6,8,9} and from this only {1,9} are odd. The probability of an odd number is thus 2/5.
Expected number of odds is 2/5 * 90 = 36 Did you answer this riddle correctly?
YES NO
Expected number of odds is 2/5 * 90 = 36 Did you answer this riddle correctly?
YES NO
I Have A Tail Riddle
Hint:
Standing My Test
I'm seen to fly, described as hard. I can be your currency and heal all wounds, but not many things can stand my test. I am...
Hint:
Snowman Baby Crib Riddle
Hint:
The Safest Room
Hint:
They'd both be safe because the lions that havent eaten in 3 months would be dead. Did you answer this riddle correctly?
YES NO
YES NO
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:
Getting Put Into Shoes
This is a body part
That can be found in twos
It is a type of length
And gets put into shoes
They are?
That can be found in twos
It is a type of length
And gets put into shoes
They are?
Hint:
I Come As A Pair
I come as a pair but Im not jeans
I have several digits but Im not a cellphone
I have an arch but Im not a bridge
I have nails but I dont have a hammer
Im part of the body but Im not a hand
What part of the body am I?
I have several digits but Im not a cellphone
I have an arch but Im not a bridge
I have nails but I dont have a hammer
Im part of the body but Im not a hand
What part of the body am I?
Hint:
Add Your Riddle Here
Have some tricky riddles of your own? Leave them below for our users to try and solve.