Hidden Gems Riddle
Find the names of 10 gems or precious stones hidden in the following story. Each one spans at least two words.
Sir Gade rode toward a castle atop a zebra at a steady gallop, a long way from home. He arrived at the gate and the keeper knelt in shame. Thy steed must be tired. Sir Gade replied, I am on direct orders from the King. Let me through. A cougar, nettled by the noise, emerged from a shrub. You must save me! cried the gate man. Sir Gade, eyes filled with rancor, alighted and gazed toward the sky. An item fell from his cloak as drove his sword into the cougars spine. Let me pass! cried Sir Gade. The gate keeper, stunned from his saga, tentatively opened the gate, then retrieved the fallen item.
Sir Gade rode toward a castle atop a zebra at a steady gallop, a long way from home. He arrived at the gate and the keeper knelt in shame. Thy steed must be tired. Sir Gade replied, I am on direct orders from the King. Let me through. A cougar, nettled by the noise, emerged from a shrub. You must save me! cried the gate man. Sir Gade, eyes filled with rancor, alighted and gazed toward the sky. An item fell from his cloak as drove his sword into the cougars spine. Let me pass! cried Sir Gade. The gate keeper, stunned from his saga, tentatively opened the gate, then retrieved the fallen item.
Hint:
1. topaz atop a zebra
2. opal gallop, a long
3. amethyst shame. Thy steed
4. diamond replied, I am on direct
5. garnet cougar, nettled
6. ruby shrub. You
7. coral rancor, alighted
8. kyanite sky. An item
9. spinel spine. Let
10. agate saga, tentatively Did you answer this riddle correctly?
YES NO
2. opal gallop, a long
3. amethyst shame. Thy steed
4. diamond replied, I am on direct
5. garnet cougar, nettled
6. ruby shrub. You
7. coral rancor, alighted
8. kyanite sky. An item
9. spinel spine. Let
10. agate saga, tentatively Did you answer this riddle correctly?
YES NO
The Black Widow Riddle
Hint:
John's Three Daughters Riddle
John has three daughters who are all unmarried. The youngest always lies, the oldest always tells the truth, and the one in the middle either tells the truth or lies. A very rich young man comes to John's house and says he wishes to marry one of his daughters. Naturally he wants to marry the oldest or the youngest so he will always know if she is lying or telling the truth. John agrees but says he can only ask one of the girls a yes or no question to decide which one he marries. They all look the same age.
What one question does he ask one of the daughters at random to figure out which daughter is the youngest or oldest?
What one question does he ask one of the daughters at random to figure out which daughter is the youngest or oldest?
Hint:
"Is she older than her?"
Explanation: (He would ask one of the daughters if one of the other daughters is older than the last daughter). He always should pick the younger daughter based on what he knows. If he asks the older daughter and she says yes, then the youngest daughter will be known. If he asks the older daughter and she says no, then the youngest daughter is the other one. If he asks the youngest daughter and she says yes, she is lying and he will still pick the oldest. If he asks the youngest and she says no, he will just pick the other like in the first case. If he asks the middle daughter it doesn't matter because both will be acceptable choices. Did you answer this riddle correctly?
YES NO
Explanation: (He would ask one of the daughters if one of the other daughters is older than the last daughter). He always should pick the younger daughter based on what he knows. If he asks the older daughter and she says yes, then the youngest daughter will be known. If he asks the older daughter and she says no, then the youngest daughter is the other one. If he asks the youngest daughter and she says yes, she is lying and he will still pick the oldest. If he asks the youngest and she says no, he will just pick the other like in the first case. If he asks the middle daughter it doesn't matter because both will be acceptable choices. Did you answer this riddle correctly?
YES NO
Shooting Mr. Door Riddle
Mr. Box drove up to the house where Mr. Door lived carrying a box in his hand. He was smiling as he rang the doorbell. When Mr. Door opened the door, Mr. Boxs smile melted away. Mr. Box took out a gun and shot Mr. Door. Why?
Hint:
Mr. Box and Mr. Door had been in the navy together years ago. They had both been on a ship that had shipwrecked in the Pacific. A group of men had survived on a liferaft but didnt have any food. Mr. Door was a navy doctor. He suggested that he would amputate each of their left arms for food. There was, however, no other doctor available to amputate Mr. Doors arm. To make it fair, Mr. Door promised that when they were back home again, he would have his arm amputated also. Mr. Door had sent Mr. Box an arm in the mail. Mr. Box assumed it was Mr. Doors arm, but when Mr. Box visited Mr. Door and saw that he still had both of his arms, he became very angry that Mr. Door had not followed through on his side of the bargain and shot him. Did you answer this riddle correctly?
YES NO
YES NO
Easy To Flex Riddle
When I am moved a lot, liquid I spew.
When I am hit, color I change.
And color, I come in quite a range.
What I cover is very complex, and I am very easy to flex.
What am I?
Hint:
5 Men Riddle
There were five men going to church and it started to rain. The four that ran got wet and the one that stood still stayed dry. How did the one stay dry?
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
Three Rats Riddle
Three rats are sitting at the three corners of an equilateral triangle. Each rat starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the rats collide?
Hint:
So lets think this through. The rats can only avoid a collision if they all decide to move in the same direction (either clockwise or rati-clockwise). If the rats do not pick the same direction, there will definitely be a collision. Each rat has the option to either move clockwise or rati-clockwise. There is a one in two chance that an rat decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision. Did you answer this riddle correctly?
YES NO
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
Little Billy's Calculator
Little Billy has a calculator with 15 buttons. He has 10 keys for 0-9, a key for addition, multiplication, division, and subtraction. Finally, he has an = sign. However, Mark the Meanie messed up the programming on Billy's calculator. Now, whenever Billy presses any of the number keys, it comes up with a random single-digit number. The same goes for the four operations keys (+,-,x, /). So whenever Billy tries to press the + button, the calculator chooses randomly between addition, multiplication, subtraction, and division. The only key left untouched was the = sign.
Now, if Billy were to press one number key, one operation key, then another number key, then the = button, what are the chances the answer comes out to 6?
Now, if Billy were to press one number key, one operation key, then another number key, then the = button, what are the chances the answer comes out to 6?
Hint: Think about how many ways he could possibly get 6.
There is a 4% chance.
There are 16 possible ways to get 6.
0+6
1+5
2+4
3+3
6+0
5+1
4+2
9-3
8-2
7-1
6-0
1x6
2x3
6x1
3x2
6/1
There are 400 possible button combinations.
When Billy presses any number key, there are 10 possibilities; when he presses any operation key, there are 4 possibilities.
10(1st#)x4(Operation)x10(2nd#)=400
16 working combinations/400 possible combinations= .04 or 4% Did you answer this riddle correctly?
YES NO
There are 16 possible ways to get 6.
0+6
1+5
2+4
3+3
6+0
5+1
4+2
9-3
8-2
7-1
6-0
1x6
2x3
6x1
3x2
6/1
There are 400 possible button combinations.
When Billy presses any number key, there are 10 possibilities; when he presses any operation key, there are 4 possibilities.
10(1st#)x4(Operation)x10(2nd#)=400
16 working combinations/400 possible combinations= .04 or 4% 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
An Island That Has 3 Gods
There is an Island that has 3 gods. One god always tells a lie, and the other always tells the truth. The third god has a random behavior. To top it off, these three gods, being jerks, answer in their own languages such that you are unable to tell which word, between "ja" or "da", means "no" or "yes". You have 3 questions to work out the True god, the false god, and the Random god.
Hint:
Question 1: (To any of the three gods) If I were to ask you "Is that the random god," would your answer be "ja?" (This questions, no matter the answer, will enable you to tell which god is not random i.e. the god who is either False or True)
Question 2: (To either the True or False god) If I asked you "are you false," would your answer be "ja?"
Question 3: (To the same god you asked the second question) If I asked you "whether the first god I spoke to is random," would your answer be "ja?" Did you answer this riddle correctly?
YES NO
Question 2: (To either the True or False god) If I asked you "are you false," would your answer be "ja?"
Question 3: (To the same god you asked the second question) If I asked you "whether the first god I spoke to is random," would your answer be "ja?" Did you answer this riddle correctly?
YES NO
A Laundered Robbery
Hint:
Three Gods Riddle
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.
What three questions can you ask?
What three questions can you ask?
Hint:
A possible solution is:
Q1: Ask god B, "If I asked you 'Is A Random?', would you say ja?". If B answers ja, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers da, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, you know the identity of a god who is not Random.
Q2: Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you False?', would you say ja?". Since he is not Random, an answer of da indicates that he is True and an answer of ja indicates that he is False.
Q3: Ask the same god the question: "If I asked you 'Is B Random?', would you say ja?". If the answer is ja, B is Random; if the answer is da, the god you have not yet spoken to is Random. The remaining god can be identified by elimination. Did you answer this riddle correctly?
YES NO
Q1: Ask god B, "If I asked you 'Is A Random?', would you say ja?". If B answers ja, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers da, either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, you know the identity of a god who is not Random.
Q2: Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you False?', would you say ja?". Since he is not Random, an answer of da indicates that he is True and an answer of ja indicates that he is False.
Q3: Ask the same god the question: "If I asked you 'Is B Random?', would you say ja?". If the answer is ja, B is Random; if the answer is da, the god you have not yet spoken to is Random. The remaining god can be identified by elimination. Did you answer this riddle correctly?
YES NO
Add Your Riddle Here
Have some tricky riddles of your own? Leave them below for our users to try and solve.