Miss Millie's Parrot
When Miss Millie purchased her new parrot, the salesman assured her that it would repeat any word it heard. About a week later, Miss Millie returned the parrot complaining it hadn't uttered a single word. Given that the salesman had spoken the truth about the parrot's abilities, why wouldn't the bird talk?
Hint:
The Parrot Doors Riddle
There are two doors. One door lead to Heaven, while the other leads to Hell. A parrot stands in front of each door. One parrot always tells a lie, while the other always tells the truth. You do not know which parrot or door is which. You are allowed to only ask one question. So, what one question must you ask to determine which door is which, so you can finally go to Heaven? (Hint: The question involves what one parrot would say about the doors.)
Hint:
It doesn't matter which parrot you ask the question to, but the question would be, "What door would the other parrot say is Heaven?". Then you would choose the other door. Did you answer this riddle correctly?
YES NO
YES NO
Sitting On A Pirates Shoulder
This is a certain type of bird
But it is not a hawk
It sits on a pirate's shoulder
While letting out a squawk
But it is not a hawk
It sits on a pirate's shoulder
While letting out a squawk
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
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
The Gardners Riddle
Gretchen and Henry were discussing their new neighbors, the Gardners. Gretchen mentioned that she met two of the daughters, and they each had blond hair.
"I have met all of the sisters," replied Henry, "and the probability that both of the girls you met would have had blond hair, assuming you were equally likely to meet any of the sisters, is exactly 50%. Do you know how many children there are?"
After thinking for a minute, Gretchen asks if the family is abnormally large. When Henry replies that it is not, Gretchen tells him how many girls are in the family. What number did she say?
"I have met all of the sisters," replied Henry, "and the probability that both of the girls you met would have had blond hair, assuming you were equally likely to meet any of the sisters, is exactly 50%. Do you know how many children there are?"
After thinking for a minute, Gretchen asks if the family is abnormally large. When Henry replies that it is not, Gretchen tells him how many girls are in the family. What number did she say?
Hint:
Gretchen said that there were 4 girls in the family, three of whom were blond.
This would make the probability that she saw two blonds (3/4) * (2/3), which equals 1/2.
Other numbers would work, but the next pair would lead to a rather large family. Did you answer this riddle correctly?
YES NO
This would make the probability that she saw two blonds (3/4) * (2/3), which equals 1/2.
Other numbers would work, but the next pair would lead to a rather large family. 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
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
Safe And Secure Riddle
As a whole, I am both safe and secure.
Behead me, and I become a place of meeting.
Behead me again, and I am the partner of ready.
Restore me, and I become the domain of beasts.
What am I?
Behead me, and I become a place of meeting.
Behead me again, and I am the partner of ready.
Restore me, and I become the domain of beasts.
What am I?
Hint:
Part Of Your Body
I can hold things but Im not a bag
Im used to write things down but Im not a pen
I have digits but Im not a cellphone
I have nails but Im not a hook
Im part of your body but Im not a foot
I am a?
Im used to write things down but Im not a pen
I have digits but Im not a cellphone
I have nails but Im not a hook
Im part of your body but Im not a foot
I am a?
Hint:
I Make You Gay
When young, I am sweet in the sun. When middle-aged, I make you gay. When old, I am valued more than ever. What am I?
Hint:
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
Unwilling To Kiss
First think of the person who lives in disguise,
Who deals in secrets and tells naught but lies.
Next, tell me whats always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard
During the search for a hard-to-find word.
Now string them together, and answer me this,
Which creature would you be unwilling to kiss?
Who deals in secrets and tells naught but lies.
Next, tell me whats always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard
During the search for a hard-to-find word.
Now string them together, and answer me this,
Which creature would you be unwilling to kiss?
Hint:
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