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
Blue Eyes Riddle
Both of my parents have brown eyes, as do I. My brother and my wife have blue eyes. Using the simple brown-blue model (two genes; a brown gene dominates blue gene), what are the chances of my first child having blue eyes?
Hint: Given my brother's blue eyes, what are the odds on my pair of eye-color genes?
1 in 3.
Since my brother has blue eyes (bb), both of my parents carry one brown and one blue gene (Bb). The three possibilities for my genotype, equally likely, are BB, Bb, and bB. Thus, there is a 2/3 chance that I carry a blue gene.
If I carry a blue gene, there is a 50% chance I will pass it on to my first child (and, obviously, 0% if I carry two brown genes).
Since my child will certainly get a blue gene from my wife, my gene will determine the eye color.
Multiplying the probabilities of those two independent events, there is a chance of 1/2 x 2/3 = 1/3 of my passing on a blue gene. Did you answer this riddle correctly?
YES NO
Since my brother has blue eyes (bb), both of my parents carry one brown and one blue gene (Bb). The three possibilities for my genotype, equally likely, are BB, Bb, and bB. Thus, there is a 2/3 chance that I carry a blue gene.
If I carry a blue gene, there is a 50% chance I will pass it on to my first child (and, obviously, 0% if I carry two brown genes).
Since my child will certainly get a blue gene from my wife, my gene will determine the eye color.
Multiplying the probabilities of those two independent events, there is a chance of 1/2 x 2/3 = 1/3 of my passing on a blue gene. 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:
Kindness And Cruelty
Capable of Kindness and cruelty, I take victims when I sour. I can be on your side or wrong you. I bring gifts though you already have me. What am I?
Hint:
Man With A Wooden Leg
Hint:
You cant take a picture with a wooden leg. You need a camera (or iPad or cell phone) to take a picture. Did you answer this riddle correctly?
YES NO
YES NO
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:
Different Shapes And Sizes
I'm white but I'm not a sheet of paper
I come in different shapes and sizes but I'm not a snowflake
I can be broken but Im not a window
I can be brittle but Im not peanut butter
I'm sometimes humerus but Im not funny
I am?
I come in different shapes and sizes but I'm not a snowflake
I can be broken but Im not a window
I can be brittle but Im not peanut butter
I'm sometimes humerus but Im not funny
I am?
Hint:
Soccer Kick Riddle
Hint:
An Absentminded Philosopher Riddle
An absentminded philosopher forgot to wind up the only clock in his house. He had no radio, television, telephone, internet, or any other means of ascertaining the time. He therefore decided to travel by foot to his friend's house, a few miles down a straight desert road. He stayed there for the night and when he came back home the following morning, he was able to set his clock to the correct time. Assuming the philosopher always walks at the same speed, how did he know the exact time upon his return? Note: this is not a trick question. The Philosopher did not bring anything to his friend's house, nor did he bring anything back with him on his trip home.
Hint: We can assume that the journey to his friend's and back took exactly the same amount of time.
He Philosopher winds the grandfather clock to a random time right before leaving, 9:00 for example. Although this is not the right time, the clock can now be used to measure elapsed time. As soon as he arrives at his friend's house, the Philosopher looks at the time on his friend's clock. Let's say the time is 7:15. He stays overnight and then, before leaving in the morning, he looks at the clock one more time. Let's say the time is now 10:15 (15 hours later). When the Philosopher arrives home, he looks at his grandfather clock. Let's say his clock reads 12:40. By subtracting the time he set it to when he left (9:00) from the current time (12:40) he knows that he has been gone for 15 hours and 40 minutes. He knows that he spent 15 hours at his friends house, so that means he spent 40 minutes walking. Since he walked at the same speed both ways, it took him 20 minutes to walk from his friend's home back to his place. So the correct time to set the clock to in this example would therefore be 10:15 (the time he left his friend's house) + 20 minutes (the time it took him to walk home) = 10:35. Did you answer this riddle correctly?
YES NO
YES NO
The Quietest Whimper
I talk, but I do not speak my mind
I hear words, but I do not listen to thoughts
When I wake, all see me
When I sleep, all hear me
Many heads are on my shoulders
Many hands are at my feet
The strongest steel cannot break my visage
But the softest whisper can destroy me
The quietest whimper can be heard.
What am I?
I hear words, but I do not listen to thoughts
When I wake, all see me
When I sleep, all hear me
Many heads are on my shoulders
Many hands are at my feet
The strongest steel cannot break my visage
But the softest whisper can destroy me
The quietest whimper can be heard.
What am I?
Hint:
Dropping Her Bouquet Riddle
Hint:
Dont Trust The Ocean
Hint:
School Ladder Riddle
Hint:
Eskimo Toys Riddle
Hint:
Add Your Riddle Here
Have some tricky riddles of your own? Leave them below for our users to try and solve.