Random Riddles To Solve
Solving Random Riddles
Here we've provide a compiled a list of the best random puzzles and riddles to solve we could find.Our team works hard to help you piece fun ideas together to develop riddles based on different topics. Whether it's a class activity for school, event, scavenger hunt, puzzle assignment, your personal project or just fun in general our database serve as a tool to help you get started.
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In A Bus There Is A Pregnant Lady Riddle
In a bus, there is, A pregnant lady 24 years old, A security guard 37 years old, A random woman 56 years old, Driver 68 years old, Who is the youngest?
Hint:
The baby in the pregnant woman's womb is the youngest in the Bus. Did you answer this riddle correctly?
YES NO
YES NO
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
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
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 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
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
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
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
Random Slamming Doors
This place has hardly any lights
But a lot of creaking floors
There are all kinds of strange noises
And some random slamming doors
Where is this place?
But a lot of creaking floors
There are all kinds of strange noises
And some random slamming doors
Where is this place?
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
3 Gods Riddle
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
Accepting The Bet Riddle
There is a box in which distinct numbered balls have been kept. You have to pick two balls randomly from the lot.
If someone is offering you a 2 to 1 odds that the numbers will be relatively prime, for example
If the balls you picked had the numbers 6 and 13, you lose $1.
If the balls you picked had the numbers 5 and 25, you win $2.
Will you accept that bet?
If someone is offering you a 2 to 1 odds that the numbers will be relatively prime, for example
If the balls you picked had the numbers 6 and 13, you lose $1.
If the balls you picked had the numbers 5 and 25, you win $2.
Will you accept that bet?
Hint:
Yes, you should accept the bet. Simply because the odds of picking two relatively prime numbers are 60%. It is a win-win situation for you if you keep playing. Did you answer this riddle correctly?
YES NO
YES NO
Boxes Of Balls Riddle
The first box has two white balls. The second box has two black balls. The third box has a white and a black ball.
Boxes are labeled but all labels are wrong!
You are allowed to open one box, pick one ball at random, see its color and put it back into the box, without seeing the color of the other ball.
How many such operations are necessary to correctly label the boxes?
Boxes are labeled but all labels are wrong!
You are allowed to open one box, pick one ball at random, see its color and put it back into the box, without seeing the color of the other ball.
How many such operations are necessary to correctly label the boxes?
Hint:
Just One!
Because we know all labels are wrong.
So the BW box must be either BB or WW. Selecting one ball from BW will let you know which.
And the other two boxes can then be worked out logically. Did you answer this riddle correctly?
YES NO
Because we know all labels are wrong.
So the BW box must be either BB or WW. Selecting one ball from BW will let you know which.
And the other two boxes can then be worked out logically. Did you answer this riddle correctly?
YES NO
Flip The Switch Riddle
There is a prison with 100 prisoners, each in separate cells with no form of contact. There is an area in the prison with a single light bulb in it. Each day, the warden picks one of the prisoners at random, even if they have been picked before, and takes them out to the lobby. The prisoner will have the choice to flip the switch if they want. The light bulb starts off.
When a prisoner is taken into the area with the light bulb, he can also say "Every prisoner has been brought to the light bulb." If this is true all prisoners will go free. However, if a prisoner chooses to say this and it's wrong, all the prisoners will be executed. So a prisoner should only say this if he knows it is true for sure.
Before the first day of this process begins, all the prisoners are allowed to get together to discuss a strategy to eventually save themselves.
What strategy could they use to ensure they will go free?
When a prisoner is taken into the area with the light bulb, he can also say "Every prisoner has been brought to the light bulb." If this is true all prisoners will go free. However, if a prisoner chooses to say this and it's wrong, all the prisoners will be executed. So a prisoner should only say this if he knows it is true for sure.
Before the first day of this process begins, all the prisoners are allowed to get together to discuss a strategy to eventually save themselves.
What strategy could they use to ensure they will go free?
Hint:
Only allow one prisoner to turn the light bulb off and all of the others turn it on if they have never turned it on before. If they have turned it on before they do nothing. The prisoner that can turn it off then knows they have all been there and saves them all when he has turned it off 99 times. Did you answer this riddle correctly?
YES NO
YES NO
The 100 Seat Airplane
People are waiting in line to board a 100-seat airplane. Steve is the first person in the line. He gets on the plane but suddenly can't remember what his seat number is, so he picks a seat at random. After that, each person who gets on the plane sits in their assigned seat if it's available, otherwise they will choose an open seat at random to sit in.
The flight is full and you are last in line. What is the probability that you get to sit in your assigned seat?
The flight is full and you are last in line. What is the probability that you get to sit in your assigned seat?
Hint: You don't need to use complex math to solve this riddle. Consider these two questions:
What happens if somebody sits in your seat?
What happens if somebody sits in Steve's assigned seat?
The correct answer is 1/2.
The chase that the first person in line takes your seat is equal to the chance that he takes his own seat. If he takes his own seat initially then you have a 100% chance of sitting in your seat, if he takes your seat you have a 0 percent chance. Now after the first person has picked a seat, the second person will enter the plan and, if the first person has sat in his seat, he will pick randomly, and again, the chance that he picks your seat is equal to the chance he picks someone your seat. The motion will continue until someone sits in the first persons seat, at this point the remaining people standing in line which each be able to sit in their own seats. Well how does that probability look in equation form? (2/100) * 50% + (98/100) * ( (2/98) * 50% + (96/98) * ( (2/96) * (50%) +... (2/2) * (50%) ) ) This expansion reduces to 1/2.
An easy way to see this is trying the problem with a 3 or 4 person scenario (pretend its a car). Both scenarios have probabilities of 1/2. Did you answer this riddle correctly?
YES NO
The chase that the first person in line takes your seat is equal to the chance that he takes his own seat. If he takes his own seat initially then you have a 100% chance of sitting in your seat, if he takes your seat you have a 0 percent chance. Now after the first person has picked a seat, the second person will enter the plan and, if the first person has sat in his seat, he will pick randomly, and again, the chance that he picks your seat is equal to the chance he picks someone your seat. The motion will continue until someone sits in the first persons seat, at this point the remaining people standing in line which each be able to sit in their own seats. Well how does that probability look in equation form? (2/100) * 50% + (98/100) * ( (2/98) * 50% + (96/98) * ( (2/96) * (50%) +... (2/2) * (50%) ) ) This expansion reduces to 1/2.
An easy way to see this is trying the problem with a 3 or 4 person scenario (pretend its a car). Both scenarios have probabilities of 1/2. Did you answer this riddle correctly?
YES NO
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