WHAT HAS FACE WITH TWO HANDS BUT N RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS

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Riddles and Answers © 2024

Born In Mourning

Hint:
A tombstone
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Found In A Graveyard Riddle

Hint:
Tombstone
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Mortal Privation Riddle

Hint:
A tombstone
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Circus Performers Riddle

Hint:
Bats
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Psychopath Test Riddle

Hint:
She reasoned that if the guy appeared at her mother's funeral, then he might appear another family funeral.

If you answered this correctly, you think like a psychopath. This was a test by a famous American psychologist used to test if one has the same mentality as a killer. Many arrested serial killers took part in this test and answered correctly.
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A Character In Thomas And Friends

Hint:
Im a daisy
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Lazily In The Sun

Hint:
A daisy
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Swallowed Up By A Whale

Hint:
Jonah
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Two In A Row Riddle

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.
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Pearl Problems Riddle

Hint: If you took out 2 pearls, you would have about a 50% chance of getting 2 gold bars. However, you can take even more pearls and still retain the 50% chance.
Take out 5000 pearls. If the remaining pearl is white, then you've won 5000 gold bars!
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Knights Of The Round Table Riddle

Hint: Does it matter if they are sitting clockwise or counterclockwise? Or where the oldest sits?
The odds are 11:1. (The probability is 1/12.)

Imagine they sat down in age order, with each person randomly picking a seat. The first person is guaranteed to pick a seat that "works". The second oldest can sit to his right or left, since these five can sit either clockwise or counterclockwise. The probability of picking a seat that works is thus 2/4, or 1/2. The third oldest now has three chairs to choose from, one of which continues the progression in the order determined by the second person, for a probability of 1/3. This leaves two seats for the fourth oldest, or a 1/2 chance. The youngest would thus be guaranteed to sit in the right seat, since there is only one seat left. This gives 1 * 1/2 * 1/3 * 1/2 * 1 = 1/12, or 11:1 odds against.
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Hint: Their dad is a very smart person.
Believe it or not, both Mike and James have a 1/2 chance of winning.

James wins if:
-he calls both coin flips right = 1/2 x 1/2 = 1/4
OR
-he does not call both coin flips right, Mike does not call the die roll correctly, and he guesses the number on the spinner right = 3/4 x 5/6 x 2/5 = 30/120 = 1/4

1/4 + 1/4 = 1/2

Mike wins if:
-James does not call both coin flips right and he calls the die roll correctly = 3/4 x 1/6 = 3/24 = 1/8
OR
-James does not call both coin flips right, he does not call the die roll correctly, and Mike does not guess the number on the spinner right = 3/4 x 5/6 x 3/5 = 45/120 = 3/8

1/8 + 3/8 = 1/2

Of course, dad could have just flipped a coin
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The Gardners Riddle

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.
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Little Billy's Calculator

Hint: Think about how many ways he could possibly get 6.
There is a 4% chance.

There are 16 possible ways to get 6.

0+6
1+5
2+4
3+3
6+0
5+1
4+2
9-3
8-2
7-1
6-0
1x6
2x3
6x1
3x2
6/1

There are 400 possible button combinations.

When Billy presses any number key, there are 10 possibilities; when he presses any operation key, there are 4 possibilities.

10(1st#)x4(Operation)x10(2nd#)=400

16 working combinations/400 possible combinations= .04 or 4%
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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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