Ill Mannered Reindeer
Hint:
A Leprechauns Gold Riddle
Hint:
Chocolatey Men Riddle
Hint:
Because they are sweet, smooth, and they usually head right for your hips. Did you answer this riddle correctly?
YES NO
YES NO
Reindeer Loving Scrooge
Hint:
Foil Wrapped Chocolate
I am a Jewish chocolate candy that is wrapped in foil. I sometimes have a picture of a menorah on me or the Star of David. What am I?
Hint:
Pot O Gold Riddle
Hint:
Golf And Pizza Riddle
Robert and David played several golf matches against each other in a week. They played for a pizza at each match, but no pizzas were purchased until the end of the week. If at any time Robert and David had the same number of wins, those pizzas were canceled. Robert won four matches (but no pizzas), and David won three pizzas. How many rounds of golf were played?
Hint:
Eleven, David won 7 matches, 4 to cancel out Robert's 4 wins, and 3 more to win the pizzas. Did you answer this riddle correctly?
YES NO
YES NO
Golfer Socks Riddle
Hint:
A Bad Golfer
Hint:
A bad golfer goes 'Whack, Dang.' and a bad sky diver goes 'Dang, Whack'. Did you answer this riddle correctly?
YES NO
YES NO
9 Golf Balls
You know that out of your 9 golf balls there is one that is lighter than the others. But, the light ball is to small a difference for the "hand balance!" You're in luck, the course you're at has a scale, but you can only afford to use it 2 times. How do you figure it out?
Hint:
STEP 1: Divide it into sets of three. If you put one set on either side, it'll either equal the same, or you'll see which set is lighter. Either way, you narrow it down to three balls.
STEP 2: put two of the balls on the scale. If they're equal, the last ball is defective, but if not, the scale will tell you which one is lighter. Did you answer this riddle correctly?
YES NO
STEP 2: put two of the balls on the scale. If they're equal, the last ball is defective, but if not, the scale will tell you which one is lighter. Did you answer this riddle correctly?
YES NO
Goal With Pole Riddle
With depression I am pitted
My goal with pole has been fitted
From wood to grass I must pass
Frequency will make you last
What am I?
My goal with pole has been fitted
From wood to grass I must pass
Frequency will make you last
What am I?
Hint:
Golf Ball Cake Riddle
Hint:
Goats On The Piano
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
Reindeer Virus Riddle
Hint:
Add Your Riddle Here
Have some tricky riddles of your own? Leave them below for our users to try and solve.