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
Snowman Baby Crib Riddle
Hint:
Crossing Safety Riddle
Two boys and a man need to cross a river. They can only use the canoe. It will hold only the man OR the two boys' weight. How can they all get across safely?
Answer:
Answer:
Hint:
The two boys go across. One of them get out. The other one goes back. He gets out and the man gets in. He goes across. Then the man gets out and the other boy gets in and goes across. Then the boy that was left gets in and now they both go across together. Did you answer this riddle correctly?
YES NO
YES NO
Prints In The Sand
Im something with five digits
But I am not a hand
When you walk along the beach
I leave prints in the sand
What could I be?
But I am not a hand
When you walk along the beach
I leave prints in the sand
What could I be?
Hint:
I Come As A Pair
I come as a pair but Im not jeans
I have several digits but Im not a cellphone
I have an arch but Im not a bridge
I have nails but I dont have a hammer
Im part of the body but Im not a hand
What part of the body am I?
I have several digits but Im not a cellphone
I have an arch but Im not a bridge
I have nails but I dont have a hammer
Im part of the body but Im not a hand
What part of the body am I?
Hint:
A 10 Foot Rope Ladder
A 10 foot rope ladder hangs over the side of a boat with the bottom rung on the surface of the water. The rungs are one foot apart, and the tide goes up at the rate of 6 inches per hour. How long will it be until three rungs are covered?
Hint:
An Island That Has 3 Gods
There is an Island that has 3 gods. One god always tells a lie, and the other always tells the truth. The third god has a random behavior. To top it off, these three gods, being jerks, answer in their own languages such that you are unable to tell which word, between "ja" or "da", means "no" or "yes". You have 3 questions to work out the True god, the false god, and the Random god.
Hint:
Question 1: (To any of the three gods) If I were to ask you "Is that the random god," would your answer be "ja?" (This questions, no matter the answer, will enable you to tell which god is not random i.e. the god who is either False or True)
Question 2: (To either the True or False god) If I asked you "are you false," would your answer be "ja?"
Question 3: (To the same god you asked the second question) If I asked you "whether the first god I spoke to is random," would your answer be "ja?" Did you answer this riddle correctly?
YES NO
Question 2: (To either the True or False god) If I asked you "are you false," would your answer be "ja?"
Question 3: (To the same god you asked the second question) If I asked you "whether the first god I spoke to is random," would your answer be "ja?" Did you answer this riddle correctly?
YES NO
Unwilling To Kiss
First think of the person who lives in disguise,
Who deals in secrets and tells naught but lies.
Next, tell me whats always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard
During the search for a hard-to-find word.
Now string them together, and answer me this,
Which creature would you be unwilling to kiss?
Who deals in secrets and tells naught but lies.
Next, tell me whats always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard
During the search for a hard-to-find word.
Now string them together, and answer me this,
Which creature would you be unwilling to kiss?
Hint:
The Quietest Whimper
I talk, but I do not speak my mind
I hear words, but I do not listen to thoughts
When I wake, all see me
When I sleep, all hear me
Many heads are on my shoulders
Many hands are at my feet
The strongest steel cannot break my visage
But the softest whisper can destroy me
The quietest whimper can be heard.
What am I?
I hear words, but I do not listen to thoughts
When I wake, all see me
When I sleep, all hear me
Many heads are on my shoulders
Many hands are at my feet
The strongest steel cannot break my visage
But the softest whisper can destroy me
The quietest whimper can be heard.
What am I?
Hint:
An Old Relative Riddle
Hint:
Never Asks Questions Riddle
Hint:
Catching A Bullet Riddle
Alan fires a bullet from his hand gun and his friend Wade catches the bullet with his bare hands. The gun shoots actual, deadly bullets. The bullet does not touch anything but air after it leaves the gun and until it reaches Wades hand. Wade is uninjured. How does he do it?
Hint:
Alan fires his bullet from a .25 ACP (Automatic Colt Pistol), which will reach a maximum height of 2,287 feet. He shoots directly upward while standing at the base of Burj Khalifa, a 2,722 foot tall building.
Wade is a window cleaner at that building, waiting at 2,287 feet. When the bullet reaches that height and is about to go back down again, he reaches out with his bare hands and catches it. Did you answer this riddle correctly?
YES NO
Wade is a window cleaner at that building, waiting at 2,287 feet. When the bullet reaches that height and is about to go back down again, he reaches out with his bare hands and catches it. Did you answer this riddle correctly?
YES NO
A Sharpshooter Riddle
A sharpshooter hangs up his hat, turns around and walks 5000 meters, then turns around and shoots his gun, putting a hole right through his hat. How did he do it?
Hint:
Age Of Three Daughters Riddles
I was visiting a friend one evening and remembered that he had three daughters. I asked him how old they were. The product of their ages is 72, he answered. Quizzically, I asked, Is there anything else you can tell me? Yes, he replied, the sum of their ages is equal to the number of my house. I stepped outside to see what the house number was. Upon returning inside, I said to my host, Im sorry, but I still cant figure out their ages. He responded apologetically, Im sorry, I forgot to mention that my oldest daughter likes strawberry shortcake. With this information, I was able to determine all three of their ages. How old is each daughter?
Hint:
3, 3, and 8. The only groups of 3 factors of 72 to have non-unique sums are 2 6 6 and 3 3 8 (with a sum of 14). The rest have unique sums:
2 + 2 + 18 = 22
2 + 3 + 12 = 18
2 + 4 + 9 = 15
3 + 4 + 6 = 13
The house number alone would have identified any of these groups. Since more information was required, we know the sum left the answer unknown. The presence of a single oldest child eliminates 2 6 6, leaving 3 3 8 as the only possible answer. Did you answer this riddle correctly?
YES NO
2 + 2 + 18 = 22
2 + 3 + 12 = 18
2 + 4 + 9 = 15
3 + 4 + 6 = 13
The house number alone would have identified any of these groups. Since more information was required, we know the sum left the answer unknown. The presence of a single oldest child eliminates 2 6 6, leaving 3 3 8 as the only possible answer. Did you answer this riddle correctly?
YES NO
Banana Clock Riddle
Hint: 1. Look closely at the clock.
2. Number of Bananas.
3. Some thing regarding the sides of the Shapes figure.
38
Logic :
From 1st the hexagon shape has value 15.
The shape has 15 edges(6 of hexagon,5 of Pentagon and 4 of Square)
From 2nd we get value of one bunch of bananas is 4.
So each banana has value of 1.
From 3rd we get that each clock has value of 3.
Which resembles the time on the clock which is 3.
Hence by using these insights, we get the last required values as
Clock = 2 (2 in the clock)
Bananas = 3 (3 bananas in the bunch)
Hexagon = 11 (Hexagon[6 sides] and pentagon[5 sides], so 6+5=11)
So required value is
2+3+3x11=?
2+3+33=? (Multiply first - Bodmas rule)
5+33=38 and hence the answer is 38. Did you answer this riddle correctly?
YES NO
Logic :
From 1st the hexagon shape has value 15.
The shape has 15 edges(6 of hexagon,5 of Pentagon and 4 of Square)
From 2nd we get value of one bunch of bananas is 4.
So each banana has value of 1.
From 3rd we get that each clock has value of 3.
Which resembles the time on the clock which is 3.
Hence by using these insights, we get the last required values as
Clock = 2 (2 in the clock)
Bananas = 3 (3 bananas in the bunch)
Hexagon = 11 (Hexagon[6 sides] and pentagon[5 sides], so 6+5=11)
So required value is
2+3+3x11=?
2+3+33=? (Multiply first - Bodmas rule)
5+33=38 and hence the answer is 38. Did you answer this riddle correctly?
YES NO
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