I AM IN FOOTBALL BUT NOT CRICKET I AM PRESENT IN A DOG BUT NOT IN A CAT I AM ALSO IN A HOUSE BUT NOT BANK WHAT AM I RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS

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The Secret Santa Exchange

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.
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Yahtzee Riddle

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.
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The Traffic Light Riddle

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%.
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Reindeer Virus Riddle

Hint:
Reindeer flew
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An Absentminded Philosopher Riddle

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.
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Born In London Riddle

Hint:
Because he is still alive!
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Going To London Riddle

Hint:
1
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My Fathers Name Ridlde

Hint:
What

It stated 'WHAT' was the father's name.
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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.
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Transforming Nature Riddle

Hint:
A butterfly
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Keeping A Secret Riddle

Hint:
It has too many tellers
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Running For Its Life

Hint:
Because it was being chased by a dog-erpillar
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The Queen And The Duke

Hint:
The Queen said that: "Until you see me again you shall not reveal to anyone what happen to me." The face of the Queen is on the money.
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12 Clowns Riddle

Hint:
1. The others were coming from the fair.
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Age Of Three Daughters Riddles

Hint:
3, 3, and 8. The only groups of 3 factors of 72 to have non-unique sums are 2 6 6 and 3 3 8 (with a sum of 14). The rest have unique sums:

2 + 2 + 18 = 22
2 + 3 + 12 = 18
2 + 4 + 9 = 15
3 + 4 + 6 = 13

The house number alone would have identified any of these groups. Since more information was required, we know the sum left the answer unknown. The presence of a single oldest child eliminates 2 6 6, leaving 3 3 8 as the only possible answer.
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