The Blind Mammals Riddle
The fact this mammal has webbed wings
Makes it a one of a kind
And contrary to the saying
None of these creatures are blind
What are these mammals?
Makes it a one of a kind
And contrary to the saying
None of these creatures are blind
What are these mammals?
Hint:
Two Tablets Of Stone
He led Israelites out of Egypt
And went up Mount Sinai alone
He came back down with ten commandments
Written on two tablets of stone
Who is this man?
And went up Mount Sinai alone
He came back down with ten commandments
Written on two tablets of stone
Who is this man?
Hint:
Under The Cup Riddle
You decide to play a game with your friend where your friend places a coin under one of three cups. Your friend would then switch the positions of two of the cups several times so that the coin under one of the cups moves with the cup it is under. You would then select the cup that you think the coin is under. If you won, you would receive the coin, but if you lost, you would have to pay.
As the game starts, you realise that you are really tired, and you don't focus very well on the moving of the cups. When your friend stops moving the cups and asks you where the coin is, you only remember a few things:
He put the coin in the rightmost cup at the start.
He switched two of the cups 3 times.
The first time he switched two of the cups, the rightmost one was switched with another.
The second time he switched two of the cups, the rightmost one was not touched.
The third and last time he switched two of the cups, the rightmost one was switched with another.
You don't want to end up paying your friend, so, using your head, you try to work out which cup is most likely to hold the coin, using the information you remember.
Which cup is most likely to hold the coin?
As the game starts, you realise that you are really tired, and you don't focus very well on the moving of the cups. When your friend stops moving the cups and asks you where the coin is, you only remember a few things:
He put the coin in the rightmost cup at the start.
He switched two of the cups 3 times.
The first time he switched two of the cups, the rightmost one was switched with another.
The second time he switched two of the cups, the rightmost one was not touched.
The third and last time he switched two of the cups, the rightmost one was switched with another.
You don't want to end up paying your friend, so, using your head, you try to work out which cup is most likely to hold the coin, using the information you remember.
Which cup is most likely to hold the coin?
Hint: Write down the possibilities. Remember that there are only three cups, so if the rightmost cup wasn't touched...
The rightmost cup.
The rightmost cup has a half chance of holding the coin, and the other cups have a quarter chance.
Pretend that Os represent cups, and Q represents the cup with the coin.
The game starts like this:
OOQ
Then your friend switches the rightmost cup with another, giving two possibilities, with equal chance:
OQO
QOO
Your friend then moves the cups again, but doesn't touch the rightmost cup. The only switch possible is with the leftmost cup and the middle cup. This gives two possibilities with equal chance:
QOO
OQO
Lastly, your friend switches the rightmost cup with another cup. If the first possibility shown above was true, there would be two possibilities, with equal chance:
OOQ
QOO
If the second possibility shown above (In the second switch) was true, there would be two possibilities with equal chance:
OOQ
OQO
This means there are four possibilities altogether, with equal chance:
OOQ
QOO
OOQ
OQO
This means each possibility equals to a quarter chance, and because there are two possibilities with the rightmost cup having the coin, there is a half chance that the coin is there. Did you answer this riddle correctly?
YES NO
The rightmost cup has a half chance of holding the coin, and the other cups have a quarter chance.
Pretend that Os represent cups, and Q represents the cup with the coin.
The game starts like this:
OOQ
Then your friend switches the rightmost cup with another, giving two possibilities, with equal chance:
OQO
QOO
Your friend then moves the cups again, but doesn't touch the rightmost cup. The only switch possible is with the leftmost cup and the middle cup. This gives two possibilities with equal chance:
QOO
OQO
Lastly, your friend switches the rightmost cup with another cup. If the first possibility shown above was true, there would be two possibilities, with equal chance:
OOQ
QOO
If the second possibility shown above (In the second switch) was true, there would be two possibilities with equal chance:
OOQ
OQO
This means there are four possibilities altogether, with equal chance:
OOQ
QOO
OOQ
OQO
This means each possibility equals to a quarter chance, and because there are two possibilities with the rightmost cup having the coin, there is a half chance that the coin is there. Did you answer this riddle correctly?
YES NO
Fighting In A Truel
Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. All three competitors know one another's shooting odds. If you are Mr. Black, where should you shoot first for the highest chance of survival?
Hint: Think from the points of view of Mr. Gray and Mr. White, not just Mr. Black.
He should shoot at the ground.
If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before. Did you answer this riddle correctly?
YES NO
If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before. 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
Four Balls In A Bowl
This is a famous paradox probability riddle which has caused a great deal of argument and disbelief from many who cannot accept the correct answer.
Four balls are placed in a bowl. One is Green, one is Black and the other two are Yellow. The bowl is shaken and someone draws two balls from the bowl. He looks at the two balls and announces that at least one of them is Yellow. What are the chances that the other ball he has drawn out is also Yellow?
Four balls are placed in a bowl. One is Green, one is Black and the other two are Yellow. The bowl is shaken and someone draws two balls from the bowl. He looks at the two balls and announces that at least one of them is Yellow. What are the chances that the other ball he has drawn out is also Yellow?
Hint:
1/5
There are six possible pairings of the two balls withdrawn,
Yellow+Yellow
Yellow+Green
Green+Yellow
Yellow+Black
Black+Yellow
Green+Black.
We know the Green + Black combination has not been drawn.
This leaves five possible combinations remaining. Therefore the chances tbowl the Yellow + Yellow pairing has been drawn are 1 in 5.
Many people cannot accept tbowl the solution is not 1 in 3, and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as Yellow before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one. Did you answer this riddle correctly?
YES NO
There are six possible pairings of the two balls withdrawn,
Yellow+Yellow
Yellow+Green
Green+Yellow
Yellow+Black
Black+Yellow
Green+Black.
We know the Green + Black combination has not been drawn.
This leaves five possible combinations remaining. Therefore the chances tbowl the Yellow + Yellow pairing has been drawn are 1 in 5.
Many people cannot accept tbowl the solution is not 1 in 3, and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as Yellow before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one. 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
Chances Of A 2nd Girl Riddle
Tipli and Pikli are a married couple (dont ask me who he is and who she is)
They have two children, one of the child is a boy. Assume safely that the probability of each gender is 1/2.
What is the probability that the other child is also a boy?
They have two children, one of the child is a boy. Assume safely that the probability of each gender is 1/2.
What is the probability that the other child is also a boy?
Hint: It is not 1/2 as you would first think.
1/3
This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.
The following are possible combinations of two children that form a sample space in any earthly family:
Boy - Girl
Girl - Boy
Boy - Boy
Girl - Girl
Since we know one of the children is a boy, we will drop the girl-girl possibility from the sample space.
This leaves only three possibilities, one of which is two boys. Hence the probability is 1/3 Did you answer this riddle correctly?
YES NO
This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.
The following are possible combinations of two children that form a sample space in any earthly family:
Boy - Girl
Girl - Boy
Boy - Boy
Girl - Girl
Since we know one of the children is a boy, we will drop the girl-girl possibility from the sample space.
This leaves only three possibilities, one of which is two boys. Hence the probability is 1/3 Did you answer this riddle correctly?
YES NO
The Blue And Red Dice Riddle
Timothy and Urban play a game with two dice. But they do not use the numbers. Some of the faces are painted red and the others blue. Each player throws the dice in turn. Timothy wins when the two top faces are the same color. Urban wins when the colors are different. Their chances are even.
The first die has 5 red faces and 1 blue face. How many red and how many blue are there on the second die?
The first die has 5 red faces and 1 blue face. How many red and how many blue are there on the second die?
Hint:
Each die has 6 faces. When two dice are thrown, there are 36 equally possible results. For chances to be even, there must be 18 ways of getting the same color on top. Let X be the number of red faces on the second die. We have: 18 = 5X + 1(6 - X)
X = 3
The second die must have 3 red faces and 3 blue faces. Did you answer this riddle correctly?
YES NO
X = 3
The second die must have 3 red faces and 3 blue faces. 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
Yahtzee Riddle
The game of Yahtzee is played with five dice. On the first turn, a player rolls all five dice, and then may decide to keep any, all, or none of the dice aside before rolling again. Each player has a maximum of three rolls to try to get a favorable combination of dice "kept" on the side.
If a player rolls two 2s and two 4s on his/her first roll, and keeps all four of these dice aside, what is the probability of getting a full house (three of one value and two of another) in one of his/her next two rolls? (ie what is the probability of getting either a 2 or a 4 in one of the next two rolls?)
If a player rolls two 2s and two 4s on his/her first roll, and keeps all four of these dice aside, what is the probability of getting a full house (three of one value and two of another) in one of his/her next two rolls? (ie what is the probability of getting either a 2 or a 4 in one of the next two rolls?)
Hint: Think of the probability of NOT getting a full house.
5/9
The answer is NOT 2/3 because you cannot add probabilities. On each roll, the probability of getting a 2 or a 4 is 1/3, so therefore, the probability of not getting a 2 or a 4 is 2/3. Since the die is being rolled twice, square 2/3 to get a 4/9 probability of NOT getting a full house in two rolls. The probability of getting a full house is therefore 1 - 4/9, or 5/9. Did you answer this riddle correctly?
YES NO
The answer is NOT 2/3 because you cannot add probabilities. On each roll, the probability of getting a 2 or a 4 is 1/3, so therefore, the probability of not getting a 2 or a 4 is 2/3. Since the die is being rolled twice, square 2/3 to get a 4/9 probability of NOT getting a full house in two rolls. The probability of getting a full house is therefore 1 - 4/9, or 5/9. 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
Russian Roulette Riddle
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?
Hint:
Russian Roulette
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Puzzle ID: #17681
Fun: *** (2.59)
Difficulty: ** (2.07)
Category: Probability
Submitted By: JMCLEOD****
Corrected By: cnmne
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?
Answer
Label the chambers 1 through 6. Chambers 1 through 3 have bullets and chambers 4 through 6 are empty. After you spin the cylinder there are six possible outcomes:
1. Chamber 1 is fired first: Player 1 loses
2. Chamber 2 is fired first: Player 1 loses
3. Chamber 3 is fired first: Player 1 loses
4. Chamber 4 is fired first: Player 2 loses (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)
5. Chamber 5 is fired first: Player 1 loses (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)
6. Chamber 6 is fired first: Player 2 loses (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)
Therefore player 2 has an 4/6 or 2/3 chance of winning. Did you answer this riddle correctly?
YES NO
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Puzzle ID: #17681
Fun: *** (2.59)
Difficulty: ** (2.07)
Category: Probability
Submitted By: JMCLEOD****
Corrected By: cnmne
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?
Answer
Label the chambers 1 through 6. Chambers 1 through 3 have bullets and chambers 4 through 6 are empty. After you spin the cylinder there are six possible outcomes:
1. Chamber 1 is fired first: Player 1 loses
2. Chamber 2 is fired first: Player 1 loses
3. Chamber 3 is fired first: Player 1 loses
4. Chamber 4 is fired first: Player 2 loses (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)
5. Chamber 5 is fired first: Player 1 loses (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)
6. Chamber 6 is fired first: Player 2 loses (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)
Therefore player 2 has an 4/6 or 2/3 chance of winning. Did you answer this riddle correctly?
YES NO
The Traffic Light Riddle
There is a traffic light at the top of a hill. Cars can't see the light until they are 200 feet from the light.
The cycle of the traffic light is 30 seconds green, 5 seconds yellow and 20 seconds red.
A car is traveling 45 miles per hour up the hill.
What is the probability that the light will be yellow when the driver first crests the hill and that if the driver continues through the intersection at her present speed that she will run a red light?
The cycle of the traffic light is 30 seconds green, 5 seconds yellow and 20 seconds red.
A car is traveling 45 miles per hour up the hill.
What is the probability that the light will be yellow when the driver first crests the hill and that if the driver continues through the intersection at her present speed that she will run a red light?
Hint:
The probability of the driver encountering a yellow light and the light turning red before the car enters the intersection is about 5.5%.
At 45 mph the car is traveling at 66 feet/second and will take just over 3 seconds (3.03) to travel the 200 feet to the intersection. Any yellow light that is in the last 3.03 seconds of the light will cause the driver to run a red light.
The entire cycle of the light is 55 seconds. 3.03/55 = 5.5%. Did you answer this riddle correctly?
YES NO
At 45 mph the car is traveling at 66 feet/second and will take just over 3 seconds (3.03) to travel the 200 feet to the intersection. Any yellow light that is in the last 3.03 seconds of the light will cause the driver to run a red light.
The entire cycle of the light is 55 seconds. 3.03/55 = 5.5%. 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.