Dancing Feet Riddle
Oh how I love my dancing feet! They stay together - oh so neat. And when I want to walk a line, They all stay together and do double time. I count them up, ten times or more, And race on-off, across the floor. What am I?
Hint:
Fired Missiles
Two ships are in the water. They each fire one missile at the other. Neither missile hit its target. But there were no counter measures launched. And nothing entered the airspace during the time. and radar showed the misses were destroyed. How is that so?
Hint:
Two Have Ten
This is what you use to write
But it is not a pen
One of these has five fingers
And two of them have ten
But it is not a pen
One of these has five fingers
And two of them have ten
Hint:
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
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
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:
Dressed In All Black
A man dressed in all black is walking down a country lane. Suddenly, a large black car without any lights on comes round the corner and screeches to a halt. How did the car know he was there?
Hint:
Four Times To Infinity
Hint:
The Female Head Riddle
There are four in a deck of cards
And it is a size of bed
In countries with a monarchy
This is the female head
What is this?
And it is a size of bed
In countries with a monarchy
This is the female head
What is this?
Hint:
Prisoner Hat Riddle
Four inmates are cleaning up a littered beach as part of a prisoner work program. The warden, who happens to be overseeing the work, decides to play a little game with the prisoners. He tells them that if they win the game he will let them go free! He then proceeds to bury each prisoner up to his neck in sand as shown.
There is a wall between prisoners C and D (which cannot be seen through or around). Prisoner A can see prisoners B and C (by moving his head to the side). Prisoner B can see prisoner C. Prisoners C and D see only the wall.
The prisoners are immobilized in the ground and can't twist their body to see the person behind them. The warden shows them two black hats and two white hats and then puts the hats in a bag to conceal them. He then stands behind each prisoner, chooses a hat from the bag, and puts it on their head. The color of each prisoner's hat is shown in the image above.
The rules are simple. If any prisoner can figure out the color of the hat on his head, all four prisoners will be set free. But they must be sure, if one of them simply guesses and is wrong, they will all be shot dead! The prisoners are not allowed to talk to each other and they have 10 seconds.
The warden counts down "ten, nine, eight, seven". All four prisoners are silent. The warden smiles, knowing that he put the hats on in such a way that no prisoner could possibly know the color of the hat they had on. He continues "six, five, four, thr.."
"I know the color of my hat!" one of the prisoners finally blurts out.
Which prisoner called out and why is he 100% certain of the color of his hat?
There is a wall between prisoners C and D (which cannot be seen through or around). Prisoner A can see prisoners B and C (by moving his head to the side). Prisoner B can see prisoner C. Prisoners C and D see only the wall.
The prisoners are immobilized in the ground and can't twist their body to see the person behind them. The warden shows them two black hats and two white hats and then puts the hats in a bag to conceal them. He then stands behind each prisoner, chooses a hat from the bag, and puts it on their head. The color of each prisoner's hat is shown in the image above.
The rules are simple. If any prisoner can figure out the color of the hat on his head, all four prisoners will be set free. But they must be sure, if one of them simply guesses and is wrong, they will all be shot dead! The prisoners are not allowed to talk to each other and they have 10 seconds.
The warden counts down "ten, nine, eight, seven". All four prisoners are silent. The warden smiles, knowing that he put the hats on in such a way that no prisoner could possibly know the color of the hat they had on. He continues "six, five, four, thr.."
"I know the color of my hat!" one of the prisoners finally blurts out.
Which prisoner called out and why is he 100% certain of the color of his hat?
Hint:
Prisoner B.
If prisoners B and C had the same color hat on, prisoner A would have know immediately that his hat was the other color (there are only two hats of each color). Since prisoner A was silent, prisoners B and C must have different colored hats. Prisoner B realized this and knew that his hat was not the same color as prisoner C, therefore his hat must be black! Did you answer this riddle correctly?
YES NO
If prisoners B and C had the same color hat on, prisoner A would have know immediately that his hat was the other color (there are only two hats of each color). Since prisoner A was silent, prisoners B and C must have different colored hats. Prisoner B realized this and knew that his hat was not the same color as prisoner C, therefore his hat must be black! Did you answer this riddle correctly?
YES NO
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