The Almighty Sphere Of Death
We are a family. We live in an alley on oily ground. We all look exactly the same, and when we get hurt we get pushed out of the way and replaced. We fear the almighty sphere of death.
What are we?
What are we?
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:
Gun Fighting Riddle
Kangwa, Rafael and Ferdinand plans for gun fighting.
They each get a gun and take turns shooting at each other until only one person is left.
History suggests:
Kangwa hits his shot 1/3 of the time, gets to shoot first.
Rafael, hits his shot 2/3 of the time, gets to shoot next if still living.
Ferdinand having perfect record at shooting(100% accuracy) shoots last , if alive.
The cycle repeats. If you are Kangwa, where should you shoot first for the highest chance of survival?
They each get a gun and take turns shooting at each other until only one person is left.
History suggests:
Kangwa hits his shot 1/3 of the time, gets to shoot first.
Rafael, hits his shot 2/3 of the time, gets to shoot next if still living.
Ferdinand having perfect record at shooting(100% accuracy) shoots last , if alive.
The cycle repeats. If you are Kangwa, where should you shoot first for the highest chance of survival?
Hint:
He should shoot at the ground.
If Kangwa shoots the ground, it is Rafael's turn. Rafael would rather shoot at Ferdinand than Kangwa, because he is better.
If Rafael kills Ferdinand, it is just Kangwa and Rafael left, giving Kangwa a fair chance of winning.
If Rafael does not kill Ferdinand, it is Ferdinand's turn. He would rather shoot at Rafael and will definitely kill him. Even though it is now Kangwa against Ferdinand, Kangwa has a better chance of winning than before. Did you answer this riddle correctly?
YES NO
If Kangwa shoots the ground, it is Rafael's turn. Rafael would rather shoot at Ferdinand than Kangwa, because he is better.
If Rafael kills Ferdinand, it is just Kangwa and Rafael left, giving Kangwa a fair chance of winning.
If Rafael does not kill Ferdinand, it is Ferdinand's turn. He would rather shoot at Rafael and will definitely kill him. Even though it is now Kangwa against Ferdinand, Kangwa has a better chance of winning than before. Did you answer this riddle correctly?
YES NO
Knights Of The Round Table Riddle
King Arthur, Merlin, Sir Lancelot, Sir Gawain, and Guinevere decide to go to their favorite restaurant to share some mead and grilled meats. They sit down at a round table for five, and as soon as they do, Lancelot notes, "We sat down around the table in age order! What are the odds of that?"
Merlin smiles broadly. "This is easily solved without any magic." He then shared the answer. What did he say the odds were?
Merlin smiles broadly. "This is easily solved without any magic." He then shared the answer. What did he say the odds were?
Hint: Does it matter if they are sitting clockwise or counterclockwise? Or where the oldest sits?
The odds are 11:1. (The probability is 1/12.)
Imagine they sat down in age order, with each person randomly picking a seat. The first person is guaranteed to pick a seat that "works". The second oldest can sit to his right or left, since these five can sit either clockwise or counterclockwise. The probability of picking a seat that works is thus 2/4, or 1/2. The third oldest now has three chairs to choose from, one of which continues the progression in the order determined by the second person, for a probability of 1/3. This leaves two seats for the fourth oldest, or a 1/2 chance. The youngest would thus be guaranteed to sit in the right seat, since there is only one seat left. This gives 1 * 1/2 * 1/3 * 1/2 * 1 = 1/12, or 11:1 odds against. Did you answer this riddle correctly?
YES NO
Imagine they sat down in age order, with each person randomly picking a seat. The first person is guaranteed to pick a seat that "works". The second oldest can sit to his right or left, since these five can sit either clockwise or counterclockwise. The probability of picking a seat that works is thus 2/4, or 1/2. The third oldest now has three chairs to choose from, one of which continues the progression in the order determined by the second person, for a probability of 1/3. This leaves two seats for the fourth oldest, or a 1/2 chance. The youngest would thus be guaranteed to sit in the right seat, since there is only one seat left. This gives 1 * 1/2 * 1/3 * 1/2 * 1 = 1/12, or 11:1 odds against. Did you answer this riddle correctly?
YES NO
Three People In A Room
Three people enter a room and have a green or blue hat placed on their head. They cannot see their own hat, but can see the other hats.
The color of each hat is purely random. They could all be green, or blue, or any combination of green and blue.
They need to guess their own hat color by writing it on a piece of paper, or they can write 'pass'.
They cannot communicate with each other in any way once the game starts. But they can have a strategy meeting before the game.
If at least one of them guesses correctly they win $50,000 each, but if anyone guess incorrectly they all get nothing.
What is the best strategy?
The color of each hat is purely random. They could all be green, or blue, or any combination of green and blue.
They need to guess their own hat color by writing it on a piece of paper, or they can write 'pass'.
They cannot communicate with each other in any way once the game starts. But they can have a strategy meeting before the game.
If at least one of them guesses correctly they win $50,000 each, but if anyone guess incorrectly they all get nothing.
What is the best strategy?
Hint:
Simple strategy: Elect one person to be the guesser, the other two pass. The guesser chooses randomly 'green' or 'blue'. This gives them a 50% chance of winning.
Better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down 'pass'.
It works like this ('-' means 'pass'):
Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose
Result: 75% chance of winning! Did you answer this riddle correctly?
YES NO
Better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down 'pass'.
It works like this ('-' means 'pass'):
Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose
Result: 75% chance of winning! Did you answer this riddle correctly?
YES NO
Little Billy's Calculator
Little Billy has a calculator with 15 buttons. He has 10 keys for 0-9, a key for addition, multiplication, division, and subtraction. Finally, he has an = sign. However, Mark the Meanie messed up the programming on Billy's calculator. Now, whenever Billy presses any of the number keys, it comes up with a random single-digit number. The same goes for the four operations keys (+,-,x, /). So whenever Billy tries to press the + button, the calculator chooses randomly between addition, multiplication, subtraction, and division. The only key left untouched was the = sign.
Now, if Billy were to press one number key, one operation key, then another number key, then the = button, what are the chances the answer comes out to 6?
Now, if Billy were to press one number key, one operation key, then another number key, then the = button, what are the chances the answer comes out to 6?
Hint: Think about how many ways he could possibly get 6.
There is a 4% chance.
There are 16 possible ways to get 6.
0+6
1+5
2+4
3+3
6+0
5+1
4+2
9-3
8-2
7-1
6-0
1x6
2x3
6x1
3x2
6/1
There are 400 possible button combinations.
When Billy presses any number key, there are 10 possibilities; when he presses any operation key, there are 4 possibilities.
10(1st#)x4(Operation)x10(2nd#)=400
16 working combinations/400 possible combinations= .04 or 4% Did you answer this riddle correctly?
YES NO
There are 16 possible ways to get 6.
0+6
1+5
2+4
3+3
6+0
5+1
4+2
9-3
8-2
7-1
6-0
1x6
2x3
6x1
3x2
6/1
There are 400 possible button combinations.
When Billy presses any number key, there are 10 possibilities; when he presses any operation key, there are 4 possibilities.
10(1st#)x4(Operation)x10(2nd#)=400
16 working combinations/400 possible combinations= .04 or 4% Did you answer this riddle correctly?
YES NO
Billie's Birthday Riddle
Billie was born on December 28th, yet her birthday always falls in the summer. How is this possible?
Hint:
Egg Yolk 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
An Electric Train Riddle
An electric train is moving north at 100mph and a wind is blowing to the west at 10mph. Which way does the smoke blow?
Hint:
An Absentminded Philosopher Riddle
An absentminded philosopher forgot to wind up the only clock in his house. He had no radio, television, telephone, internet, or any other means of ascertaining the time. He therefore decided to travel by foot to his friend's house, a few miles down a straight desert road. He stayed there for the night and when he came back home the following morning, he was able to set his clock to the correct time. Assuming the philosopher always walks at the same speed, how did he know the exact time upon his return? Note: this is not a trick question. The Philosopher did not bring anything to his friend's house, nor did he bring anything back with him on his trip home.
Hint: We can assume that the journey to his friend's and back took exactly the same amount of time.
He Philosopher winds the grandfather clock to a random time right before leaving, 9:00 for example. Although this is not the right time, the clock can now be used to measure elapsed time. As soon as he arrives at his friend's house, the Philosopher looks at the time on his friend's clock. Let's say the time is 7:15. He stays overnight and then, before leaving in the morning, he looks at the clock one more time. Let's say the time is now 10:15 (15 hours later). When the Philosopher arrives home, he looks at his grandfather clock. Let's say his clock reads 12:40. By subtracting the time he set it to when he left (9:00) from the current time (12:40) he knows that he has been gone for 15 hours and 40 minutes. He knows that he spent 15 hours at his friends house, so that means he spent 40 minutes walking. Since he walked at the same speed both ways, it took him 20 minutes to walk from his friend's home back to his place. So the correct time to set the clock to in this example would therefore be 10:15 (the time he left his friend's house) + 20 minutes (the time it took him to walk home) = 10:35. Did you answer this riddle correctly?
YES NO
YES NO
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:
Never Asks Questions Riddle
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
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