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Riddles and Answers © 2024

Be Careful At Night

Hint:
Bats
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Bat School Riddle

Hint:
The alphabat.
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The Youngest Level Of Girl Scouts

Hint:
A daisy
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Swallowed Up By A Whale

Hint:
Jonah
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Displayed In December

Hint:
A nativity scene
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Dimples And A Nose

Hint:
A new baby!
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Never Eaten Riddle

Hint:
A pack of cards
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Fastening Two People

Hint:
A wedding ring
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Under The Cup Riddle

Hint: Write down the possibilities. Remember that there are only three cups, so if the rightmost cup wasn't touched...
The rightmost cup.

The rightmost cup has a half chance of holding the coin, and the other cups have a quarter chance.

Pretend that Os represent cups, and Q represents the cup with the coin.

The game starts like this:

OOQ

Then your friend switches the rightmost cup with another, giving two possibilities, with equal chance:

OQO
QOO

Your friend then moves the cups again, but doesn't touch the rightmost cup. The only switch possible is with the leftmost cup and the middle cup. This gives two possibilities with equal chance:

QOO
OQO

Lastly, your friend switches the rightmost cup with another cup. If the first possibility shown above was true, there would be two possibilities, with equal chance:

OOQ
QOO

If the second possibility shown above (In the second switch) was true, there would be two possibilities with equal chance:

OOQ
OQO

This means there are four possibilities altogether, with equal chance:

OOQ
QOO
OOQ
OQO

This means each possibility equals to a quarter chance, and because there are two possibilities with the rightmost cup having the coin, there is a half chance that the coin is there.
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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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Three Rats Riddle

Hint:
So lets think this through. The rats can only avoid a collision if they all decide to move in the same direction (either clockwise or rati-clockwise). If the rats do not pick the same direction, there will definitely be a collision. Each rat has the option to either move clockwise or rati-clockwise. There is a one in two chance that an rat decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision.
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Chances Of A 2nd Girl Riddle

Hint: It is not 1/2 as you would first think.
1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Boy - Girl
Girl - Boy
Boy - Boy
Girl - Girl

Since we know one of the children is a boy, we will drop the girl-girl possibility from the sample space.
This leaves only three possibilities, one of which is two boys. Hence the probability is 1/3
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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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