Always On The Lookout Riddle
I have a ginger beard
And green clothing I am told
I am always on the lookout
For coins for my pot of gold
Who am I?
And green clothing I am told
I am always on the lookout
For coins for my pot of gold
Who am I?
Hint:
Santa's Eyeless Reindeer Riddle
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Sleeping Skywalker Riddle
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Santa Standing Still Riddle
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Karate Claus Riddle
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Santa's Arrived Riddle
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Santa's Helpers Riddle
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Calling Santa Riddle
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Santas Suit Riddle
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Santa Money Riddle
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Laundry Detergent Santa Riddle
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I Go Up And I Go Down Towards The Sky And The Ground Riddle
towards the sky and the ground.
I'm present and past tense too,
Let's go for a ride, me and you.
What am I?
Hint:
Mr And Mrs Mustard Have 6 Daughters Riddle
Hint:
There are nine Mustards in the family. Since each daughter shares the same brother, there are six girls, one boy and Mr. and Mrs. Mustard. Did you answer this riddle correctly?
YES NO
YES NO
I Touch The Earth I Touch The Sky Riddle
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
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