Using A Contact Lens
These come in different colours
But they are not felt tip pens
To help these things work better
You might use a contact lens
What could I be?
But they are not felt tip pens
To help these things work better
You might use a contact lens
What could I be?
Hint:
The Murder Of Ray Whitcombe
Ray Whitcombe is found dead in his office at his desk. The police have narrowed the suspects down to three people: Mrs. Barbara Whitcombe, Ray's wife; Mr. Jason McCubbins, Ray's business partner; and Mr. Harold Nichols, Ray's best friend. All three visited Mr. Whitcombe the day of his murder, but all three provide the police with stories of explanation as to the reason for their visit. Police found Mr. Whitcombe with his wrist watch still on his right arm, a torn up picture of his wife laying on the floor beside the trash can, and an ink pen in his right hand. On the desk, the police found a name plate, a telephone that was off the hook, and a personal calendar turned to the July 5th page with 7B91011 written on it. After examining this evidence, the police knew their suspect. Who was it?
Hint:
Jason McCubbins, Ray's business partner. The calendar is the clue to solving this murder. The police realized that since Mr. Whitcombe was wearing his watch on his right arm, he must be left handed. But the pen was found in his right hand. Realizing that the number on the calendar was written in a hurry and with his opposite hand, police matched the written number with the months of the year. So the B was an 8, thereby giving us 7-8-9-10-11: July, August, September, October, November. Use the first letter of each month and it spells J-A-S-O-N. Did you answer this riddle correctly?
YES NO
YES NO
99 Points Riddle
While out bowling with his friends, a man managed to throw eight strikes (all ten pins knocked down in a single throw) and not a single gutter ball during the entire game. To his amazement, his final score was only 99 points! Assuming there were no penalties or fouls, can you come up with a ten frame scorecard with eight strikes and a final score of only 99 points?
Hint: If you knock down a single pin, for example at the far left of the back row, then repeat the same identical shot on your second throw, you'll score 0 points for your second throw (because there's no pin there anymore), but it's not a gutter ball as the s
Just to reiterate the hint, if you knock down a single pin, for example at the far left of the back row, then repeat the same identical shot on your second throw, you'll score 0 points for your second throw (because there's no pin there anymore), but it's not a gutter ball as the shot did not enter the gutter. Did you answer this riddle correctly?
YES NO
YES NO
Found In Tall Buildings
I have buttons but Im not a shirt
I have doors but Im not a house
I go up and down but Im not an umbrella
I need at least two stories but Im not a book of fairytales
Im found in tall buildings but Im not a penthouse
What am I?
I have doors but Im not a house
I go up and down but Im not an umbrella
I need at least two stories but Im not a book of fairytales
Im found in tall buildings but Im not a penthouse
What am I?
Hint:
Found In A Cave Riddle
Im black but Im not a marker pen
Im a mammal but Im not a whale
I can fly but Im not a plane
I go before man to make a superhero but Im not a spider
I might be found in a cave but Im not a stalagmite
I'm a?
Im a mammal but Im not a whale
I can fly but Im not a plane
I go before man to make a superhero but Im not a spider
I might be found in a cave but Im not a stalagmite
I'm a?
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
7B91011 Riddle
Ray Whitcombe was found dead in his office at his desk. The police have narrowed the suspects down to three people:
Mrs. Barbara Whitcombe, Ray's wife
Mr. Jason McCubbins, Ray's business partner
Mr. Harold Nichols, Ray's best friend
All three visited Mr. Whitcombe the day of his murder, but all three provided the police with stories of explanation as to the reason for their visit.
Police found Mr. Whitcombe with his wrist watch still on his right arm, a torn up picture of his wife laying on the floor beside the trash can, and an ink pen in his right hand. On the desk, the police found a name plate, a telephone that was off the hook, and a personal calendar turned to the July 5th page with 7B91011 written on it. After examining this evidence, the police knew their suspect.
Who was it?
Mrs. Barbara Whitcombe, Ray's wife
Mr. Jason McCubbins, Ray's business partner
Mr. Harold Nichols, Ray's best friend
All three visited Mr. Whitcombe the day of his murder, but all three provided the police with stories of explanation as to the reason for their visit.
Police found Mr. Whitcombe with his wrist watch still on his right arm, a torn up picture of his wife laying on the floor beside the trash can, and an ink pen in his right hand. On the desk, the police found a name plate, a telephone that was off the hook, and a personal calendar turned to the July 5th page with 7B91011 written on it. After examining this evidence, the police knew their suspect.
Who was it?
Hint:
Jason McCubbins, Ray's business partner.
The calendar was the clue to solving this murder. The police realized that since Mr. Whitcombe was wearing his watch on his right arm, he must have been left-handed. Realizing that the number on the calendar was written in a hurry and with his opposite hand, police matched the written number with the months of the year. So the B was an 8, thereby giving us 7-8-9-10-11: July, August, September, October, November. Use the first letter of each month and it spells J-A-S-O-N. Did you answer this riddle correctly?
YES NO
The calendar was the clue to solving this murder. The police realized that since Mr. Whitcombe was wearing his watch on his right arm, he must have been left-handed. Realizing that the number on the calendar was written in a hurry and with his opposite hand, police matched the written number with the months of the year. So the B was an 8, thereby giving us 7-8-9-10-11: July, August, September, October, November. Use the first letter of each month and it spells J-A-S-O-N. Did you answer this riddle correctly?
YES NO
Mike Mom Has Three Sons Riddle
Hint:
The correct answer is mike.
4 sons name:
1. Penny
2. Nickel
3. Dime
4. Mike Did you answer this riddle correctly?
YES NO
4 sons name:
1. Penny
2. Nickel
3. Dime
4. Mike Did you answer this riddle correctly?
YES NO
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