I AM A KIND OF COAT THAT CAN BE ONLY BE PUT ON WHEN WET RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS

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Riddles and Answers © 2024

Five Prom Couples Riddle

Hint:
Mark and Susan wore red.
Quintin and Jessica wore blue.
Jim and Amanda wore pink.
Bob and Betty wore green.
James and Jasmin wore yellow.
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Dracula's Dog Riddle

Hint:
A bloodhound
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Vampire Money Riddle

Hint:
In a bloodbank
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Random Slamming Doors

Hint:
It is in a haunted house
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Stockings Hung Over Me

Hint:
A fireplace
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The Blind Mammals Riddle

Hint:
They are bats!
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A Dry Shower Riddle

Hint:
A baby shower!
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Never Eaten Riddle

Hint:
A pack of cards
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A Bath Without Water

Hint:
A sun bath
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Ears On An Engine Riddle

Hint:
Engineers.
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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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Roll The Dice

Hint: What will happen if there are 6 gamblers, each of whom bet on a different number?
It's a fair game. If there are 6 gamblers, each of whom bet on a different number, the dealer will neither win nor lose on each deal.

If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1.

If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1.

If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1.

In each case, the dealer neither wins nor loses. Hence it's a fair game.
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The Miracle Mountain Riddle

Hint: This is not a trick. His watch works perfectly well. He does not sit in the same spot all day or any other such device, although it would not change the answer if he did!
The answer is (A). Since it must happen, the probability is actually 1 (100%).

Explanation: Firstly, consider 2 men, one starting from the top of the mountain and hiking down while the other starts at the bottom and hikes up. At some time in the day, they will cross over. In other words they will be at the same place at the same time of day.

Now consider our man who has walked up on one day and begins the descent the next day. Imagine there is someone (a second person) shadowing his exact movements from the day before. When he meets his shadower (it must happen) it will be the exact place that he was the day before, and of course they are both at this spot at the same time.

Contrary to our common sense, which seems to say that this is an extremely unlikely event, it is a certainty.

NOTE: There is one unlikely event here, and that is that he will notice the time when he is at the correct location on both days, but that was not what the question asked.
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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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The Cheap Mp3 Player

Hint:
With only 6 tracks on the player:
The first chapter has been set to play first. The probability of the next 5 chapters playing in order is 1/5! = 1/120.

With the music on the player as well:
Seeing as I don't care about when the music plays, it doesn't change anything. The answer is still 1/120.
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