Dead In Central Park
Anne was found dead in the central park of London.
There are six suspects "hazard", "Costa", "Pedro", "Willian", "Terry" and "Courtois".
Anne has written the murdered name in cipher on the floor as "dqvxf".
Police were unable to solve the mystery so they called Sherlock.
After a minute, Sherlock was able to decipher the cipher and ask the police to capture the murderer.
Who is the murderer?
There are six suspects "hazard", "Costa", "Pedro", "Willian", "Terry" and "Courtois".
Anne has written the murdered name in cipher on the floor as "dqvxf".
Police were unable to solve the mystery so they called Sherlock.
After a minute, Sherlock was able to decipher the cipher and ask the police to capture the murderer.
Who is the murderer?
Hint:
Costa
Explanation:
c + 1 characters-> d
o + 2 characters-> q
s + 3 characters-> v
t + 4 characters-> x
a + 5 characters-> f
=> dqvxf = costa Did you answer this riddle correctly?
YES NO
Explanation:
c + 1 characters-> d
o + 2 characters-> q
s + 3 characters-> v
t + 4 characters-> x
a + 5 characters-> f
=> dqvxf = costa Did you answer this riddle correctly?
YES NO
Dead In A Forest
Hint:
There was a fire in the forest,the man was swimming, while he swimming a helicopter scooped down to pick up some water, it picked the man up with the water and then dropped him and the water in the burning forest. Did you answer this riddle correctly?
YES NO
YES NO
Blonde And The Bowling Ball
Hint:
3 Men Hunting
A girl and a boy were out one night. They were in the woods, and they saw 3 men hunting. Next day the girl and boy were found dead. Why is this?
Hint:
It was Halloween night and they were dressed up as deer. Did you answer this riddle correctly?
YES NO
YES NO
Find Me In Your Body
This is found in your body
From your feet up to your head
It is a type of liquid
And it is the color red
What am I?
From your feet up to your head
It is a type of liquid
And it is the color red
What am I?
Hint:
I Can Fly Riddle
Im colorful but Im not a rainbow
Im sometimes found on a shoulder but Im not a bag strap
I have two legs but Im not an ostrich
I can talk but Im not a person
I can fly but Im not an airplane
I'm a...
Im sometimes found on a shoulder but Im not a bag strap
I have two legs but Im not an ostrich
I can talk but Im not a person
I can fly but Im not an airplane
I'm a...
Hint:
Logs That Are Aflame
This can be found inside some homes
But can you work out its name?
Surrounded by a mantelpiece
It has logs that are aflame
Its...
But can you work out its name?
Surrounded by a mantelpiece
It has logs that are aflame
Its...
Hint:
Stockings Hung Over Me
Whats the answer to this rhyme?
I am found beneath a chimney
And when it is Christmas time
I have stockings hung over me
What am I?
I am found beneath a chimney
And when it is Christmas time
I have stockings hung over me
What am I?
Hint:
Ultraviolet Sun Rays Riddle
Hint:
Electrically Charged Particles Riddle
Hint:
Winged Creatures Riddle
These winged creatures often scare people
If you touch one you are brave
Theyre often found just hanging out
Upside down inside a cave
What are these creatures?
If you touch one you are brave
Theyre often found just hanging out
Upside down inside a cave
What are these creatures?
Hint:
Adams Rib Riddle
This woman was created
So Adam wouldnt be alone
And the way that God made her
Was by using Adams rib bone
Who could this lady be?
So Adam wouldnt be alone
And the way that God made her
Was by using Adams rib bone
Who could this lady be?
Hint:
Hung Around Peoples Necks
It can look like a T
It can look like an X
They can be found on chains
Hung around peoples necks
What is it?
It can look like an X
They can be found on chains
Hung around peoples necks
What is it?
Hint:
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
Yahtzee Riddle
The game of Yahtzee is played with five dice. On the first turn, a player rolls all five dice, and then may decide to keep any, all, or none of the dice aside before rolling again. Each player has a maximum of three rolls to try to get a favorable combination of dice "kept" on the side.
If a player rolls two 2s and two 4s on his/her first roll, and keeps all four of these dice aside, what is the probability of getting a full house (three of one value and two of another) in one of his/her next two rolls? (ie what is the probability of getting either a 2 or a 4 in one of the next two rolls?)
If a player rolls two 2s and two 4s on his/her first roll, and keeps all four of these dice aside, what is the probability of getting a full house (three of one value and two of another) in one of his/her next two rolls? (ie what is the probability of getting either a 2 or a 4 in one of the next two rolls?)
Hint: Think of the probability of NOT getting a full house.
5/9
The answer is NOT 2/3 because you cannot add probabilities. On each roll, the probability of getting a 2 or a 4 is 1/3, so therefore, the probability of not getting a 2 or a 4 is 2/3. Since the die is being rolled twice, square 2/3 to get a 4/9 probability of NOT getting a full house in two rolls. The probability of getting a full house is therefore 1 - 4/9, or 5/9. Did you answer this riddle correctly?
YES NO
The answer is NOT 2/3 because you cannot add probabilities. On each roll, the probability of getting a 2 or a 4 is 1/3, so therefore, the probability of not getting a 2 or a 4 is 2/3. Since the die is being rolled twice, square 2/3 to get a 4/9 probability of NOT getting a full house in two rolls. The probability of getting a full house is therefore 1 - 4/9, or 5/9. Did you answer this riddle correctly?
YES NO
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