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:
Two In A Row Riddle
A certain mathematician, his wife, and their teenage son all play a fair game of chess. One day when the son asked his father for 10 dollars for a Saturday night date, his father puffed his pipe for a moment and replied, "Let's do it this way. Today is Wednesday. You will play a game of chess tonight, tomorrow, and a third on Friday. If you win two games in a row, you get the money."
"Whom do I play first, you or mom?"
"You may have your choice," said the mathematician, his eyes twinkling.
The son knew that his father played a stronger game than his mother. To maximize his chance of winning two games in succession, should he play father-mother-father or mother-father-mother?
"Whom do I play first, you or mom?"
"You may have your choice," said the mathematician, his eyes twinkling.
The son knew that his father played a stronger game than his mother. To maximize his chance of winning two games in succession, should he play father-mother-father or mother-father-mother?
Hint: Who does he need to beat to win?
Father-mother-father
To beat two games in a row, it is necessary to win the second game. This means that it would be to his advantage to play the second game against the weaker player. Though he plays his father twice, he has a higher chance of winning by playing his mother second. Did you answer this riddle correctly?
YES NO
To beat two games in a row, it is necessary to win the second game. This means that it would be to his advantage to play the second game against the weaker player. Though he plays his father twice, he has a higher chance of winning by playing his mother second. 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
The Serial Killer Husband
A man kills his wife. Many people watch him doing so. Yet no one will ever be able to accuse him of murder. Why?
Hint:
A Fathers Murder
A man goes to his mother funeral, there, he meets a woman. They go out and the part there separate ways. The man forgets to get the woman's phone number. Three days later he kills his Father...Why?
Hint:
So the woman would go to his father's funeral and he can get her number this time....98% of people who got this right turned out to be serial killers... Did you answer this riddle correctly?
YES NO
YES NO
Crossing Safety Riddle
Two boys and a man need to cross a river. They can only use the canoe. It will hold only the man OR the two boys' weight. How can they all get across safely?
Answer:
Answer:
Hint:
The two boys go across. One of them get out. The other one goes back. He gets out and the man gets in. He goes across. Then the man gets out and the other boy gets in and goes across. Then the boy that was left gets in and now they both go across together. Did you answer this riddle correctly?
YES NO
YES NO
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:
Laying An Egg Riddle
Hint:
It didn't roll since when did roosters start laying eggs? Did you answer this riddle correctly?
YES NO
YES NO
Widows Sister Riddle
Hint:
An Absentminded Philosopher Riddle
An absentminded philosopher forgot to wind up the only clock in his house. He had no radio, television, telephone, internet, or any other means of ascertaining the time. He therefore decided to travel by foot to his friend's house, a few miles down a straight desert road. He stayed there for the night and when he came back home the following morning, he was able to set his clock to the correct time. Assuming the philosopher always walks at the same speed, how did he know the exact time upon his return? Note: this is not a trick question. The Philosopher did not bring anything to his friend's house, nor did he bring anything back with him on his trip home.
Hint: We can assume that the journey to his friend's and back took exactly the same amount of time.
He Philosopher winds the grandfather clock to a random time right before leaving, 9:00 for example. Although this is not the right time, the clock can now be used to measure elapsed time. As soon as he arrives at his friend's house, the Philosopher looks at the time on his friend's clock. Let's say the time is 7:15. He stays overnight and then, before leaving in the morning, he looks at the clock one more time. Let's say the time is now 10:15 (15 hours later). When the Philosopher arrives home, he looks at his grandfather clock. Let's say his clock reads 12:40. By subtracting the time he set it to when he left (9:00) from the current time (12:40) he knows that he has been gone for 15 hours and 40 minutes. He knows that he spent 15 hours at his friends house, so that means he spent 40 minutes walking. Since he walked at the same speed both ways, it took him 20 minutes to walk from his friend's home back to his place. So the correct time to set the clock to in this example would therefore be 10:15 (the time he left his friend's house) + 20 minutes (the time it took him to walk home) = 10:35. Did you answer this riddle correctly?
YES NO
YES NO
Going To St. Ives
As I was going to St. Ives,
I met a man with seven wives,
Each wife had seven sacks,
Each sack had seven cats,
Each cat had seven kits:
Kits, cats, sacks, and wives,
How many were there going to St. Ives?
I met a man with seven wives,
Each wife had seven sacks,
Each sack had seven cats,
Each cat had seven kits:
Kits, cats, sacks, and wives,
How many were there going to St. Ives?
Hint:
One. As John McClane learns, this is a classic trick question. If the narrator meets the group on the way to St. Ives, then they must be going in the opposite direction and the math calculations are simply a bit of trickery meant to misdirect. Did you answer this riddle correctly?
YES NO
YES NO
Red Everything Riddle
Hint:
Money In The Freezer Riddle
Hint:
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