A MAN IN A CAR HE SEES A DOOR A GOLD DOOR A DIAMOND DOOR A SILVER DOOR WHICH DOOR D RIDDLES WITH ANSWERS TO SOLVE - PUZZLES & BRAIN TEASERS

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Riddles and Answers © 2024

Two Tablets Of Stone

Hint:
Moses
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Around The Yard Riddle

Hint:
A fence
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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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The Loaded Revolver Riddle

Hint:
Henry should have Gretchen pull the trigger again without spinning.

We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.

If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position.
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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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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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Matching Socks Riddle

Hint: Three pairs of matching socks... maybe not!!!
He has a ZERO chance of drawing out a black pair.

Since there is a 2/3 chance of drawing a white pair, then there MUST be 5 white socks and only 1 black sock. The chances of drawing two whites would thus be: 5/6 x 4/5 = 2/3 . With only 1 black sock, there is no chance of drawing a black pair.
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Four Balls In A Bowl

Hint:
1/5

There are six possible pairings of the two balls withdrawn,
Yellow+Yellow
Yellow+Green
Green+Yellow
Yellow+Black
Black+Yellow
Green+Black.

We know the Green + Black combination has not been drawn.

This leaves five possible combinations remaining. Therefore the chances tbowl the Yellow + Yellow pairing has been drawn are 1 in 5.

Many people cannot accept tbowl the solution is not 1 in 3, and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as Yellow before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one.
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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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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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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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