Random Slamming Doors
This place has hardly any lights
But a lot of creaking floors
There are all kinds of strange noises
And some random slamming doors
Where is this place?
But a lot of creaking floors
There are all kinds of strange noises
And some random slamming doors
Where is this place?
Hint:
Found In A Graveyard Riddle
I have writing on me but Im not a newspaper
I mark a spot but Im not a treasure map
I have names and dates on me but Im not a birth certificate
Im not needed if youre alive but Im not a coffin
I can be found in a graveyard but Im not a bunch of flowers
I'm a...
I mark a spot but Im not a treasure map
I have names and dates on me but Im not a birth certificate
Im not needed if youre alive but Im not a coffin
I can be found in a graveyard but Im not a bunch of flowers
I'm a...
Hint:
The Blind Mammals Riddle
The fact this mammal has webbed wings
Makes it a one of a kind
And contrary to the saying
None of these creatures are blind
What are these mammals?
Makes it a one of a kind
And contrary to the saying
None of these creatures are blind
What are these mammals?
Hint:
A Bath Without Water
Hint:
Wet Coat Riddle
Hint:
Ears On An Engine Riddle
Hint:
The Same Birthday Riddle
How many people must be gathered together in a room, before you can be certain that there is a greater than 50/50 chance that at least two of them have the same birthday?
Hint:
Only twenty-three people need be in the room, a surprisingly small number. The probability that there will not be two matching birthdays is then, ignoring leap years, 365x364x363x...x343/365 over 23 which is approximately 0.493. this is less than half, and therefore the probability that a pair occurs is greater than 50-50. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days. Did you answer this riddle correctly?
YES NO
YES NO
Four Balls In A Bowl
This is a famous paradox probability riddle which has caused a great deal of argument and disbelief from many who cannot accept the correct answer.
Four balls are placed in a bowl. One is Green, one is Black and the other two are Yellow. The bowl is shaken and someone draws two balls from the bowl. He looks at the two balls and announces that at least one of them is Yellow. What are the chances that the other ball he has drawn out is also Yellow?
Four balls are placed in a bowl. One is Green, one is Black and the other two are Yellow. The bowl is shaken and someone draws two balls from the bowl. He looks at the two balls and announces that at least one of them is Yellow. What are the chances that the other ball he has drawn out is also Yellow?
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. Did you answer this riddle correctly?
YES NO
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. Did you answer this riddle correctly?
YES NO
The Secret Santa Exchange
A group of ten friends decide to exchange gifts as secret Santas. Each person writes his or her name on a piece of paper and puts it in a hat. Then each person randomly draws a name from the hat to determine who has him as his or her secret Santa. The secret Santa then makes a gift for the person whose name he drew.
When it's time to exchange presents, each person walks over to the person he made the gift for and holds his or her left hand in his right hand.
What is the probability that the 10 friends holding hands form a single continuous circle?
When it's time to exchange presents, each person walks over to the person he made the gift for and holds his or her left hand in his right hand.
What is the probability that the 10 friends holding hands form a single continuous circle?
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. Did you answer this riddle correctly?
YES NO
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. Did you answer this riddle correctly?
YES NO
Non Bouncing Ball Riddle
Hint:
Reindeer Virus Riddle
Hint:
A Dark Room Riddle
If you had only one match, and entered a dark room containing an oil lamp, some newspaper, and some kindling wood, which would you light first?
Hint:
Ticking In Its Weary Pace
Time is writ upon my face
My heart ticks in its weary pace.
Though wind and rain do leave their trace,
I watch them in their endless race.
What kind of clock is it, though?
My heart ticks in its weary pace.
Though wind and rain do leave their trace,
I watch them in their endless race.
What kind of clock is it, though?
Hint:
Two Brothers Riddle
Two brothers we are, great burdens we bear,
All day we are bitterly pressed;
Yet this we will say, we are full all the day
And empty when we go to rest.
What are we?
All day we are bitterly pressed;
Yet this we will say, we are full all the day
And empty when we go to rest.
What are we?
Hint: Your burdens are also our burdens, but greater by the measure of you.
Richest Air Riddle
Hint:
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