The Same Birthday Riddle
How many people must be gathered together in a room, before you can be certain that there is a greater than 50/50 chance that at least two of them have the same birthday?
Hint:
Only twenty-three people need be in the room, a surprisingly small number. The probability that there will not be two matching birthdays is then, ignoring leap years, 365x364x363x...x343/365 over 23 which is approximately 0.493. this is less than half, and therefore the probability that a pair occurs is greater than 50-50. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days. Did you answer this riddle correctly?
YES NO
YES NO
Four Balls In A Bowl
This is a famous paradox probability riddle which has caused a great deal of argument and disbelief from many who cannot accept the correct answer.
Four balls are placed in a bowl. One is Green, one is Black and the other two are Yellow. The bowl is shaken and someone draws two balls from the bowl. He looks at the two balls and announces that at least one of them is Yellow. What are the chances that the other ball he has drawn out is also Yellow?
Four balls are placed in a bowl. One is Green, one is Black and the other two are Yellow. The bowl is shaken and someone draws two balls from the bowl. He looks at the two balls and announces that at least one of them is Yellow. What are the chances that the other ball he has drawn out is also Yellow?
Hint:
1/5
There are six possible pairings of the two balls withdrawn,
Yellow+Yellow
Yellow+Green
Green+Yellow
Yellow+Black
Black+Yellow
Green+Black.
We know the Green + Black combination has not been drawn.
This leaves five possible combinations remaining. Therefore the chances tbowl the Yellow + Yellow pairing has been drawn are 1 in 5.
Many people cannot accept tbowl the solution is not 1 in 3, and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as Yellow before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one. Did you answer this riddle correctly?
YES NO
There are six possible pairings of the two balls withdrawn,
Yellow+Yellow
Yellow+Green
Green+Yellow
Yellow+Black
Black+Yellow
Green+Black.
We know the Green + Black combination has not been drawn.
This leaves five possible combinations remaining. Therefore the chances tbowl the Yellow + Yellow pairing has been drawn are 1 in 5.
Many people cannot accept tbowl the solution is not 1 in 3, and of course it would be, if the balls had been drawn out separately and the color of the first ball announced as Yellow before the second had been drawn out. However, as both balls had been drawn together, and then the color of one of the balls announced, then the above solution, 1 in 5, must be the correct one. 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
The Last Cookie Riddle
Mike and James are arguing over who gets the last cookie in the jar, so their dad decides to create a game to settle their dispute. First, Mike flips a coin twice, and each time James calls heads or tails in the air. If James gets both calls right, he gets the last cookie. If not, Mike picks a number between one and six and then rolls a die. If he gets the number right, he gets the last cookie. If not, James picks two numbers between one and five, then spins a spinner with numbers one through five on it. If the spinner lands on one of James' two numbers, he gets the last cookie. If not, Mike does.
Who is more likely to win the last cookie, Mike or James? And what is the probability that person wins it?
Who is more likely to win the last cookie, Mike or James? And what is the probability that person wins it?
Hint: Their dad is a very smart person.
Believe it or not, both Mike and James have a 1/2 chance of winning.
James wins if:
-he calls both coin flips right = 1/2 x 1/2 = 1/4
OR
-he does not call both coin flips right, Mike does not call the die roll correctly, and he guesses the number on the spinner right = 3/4 x 5/6 x 2/5 = 30/120 = 1/4
1/4 + 1/4 = 1/2
Mike wins if:
-James does not call both coin flips right and he calls the die roll correctly = 3/4 x 1/6 = 3/24 = 1/8
OR
-James does not call both coin flips right, he does not call the die roll correctly, and Mike does not guess the number on the spinner right = 3/4 x 5/6 x 3/5 = 45/120 = 3/8
1/8 + 3/8 = 1/2
Of course, dad could have just flipped a coin Did you answer this riddle correctly?
YES NO
James wins if:
-he calls both coin flips right = 1/2 x 1/2 = 1/4
OR
-he does not call both coin flips right, Mike does not call the die roll correctly, and he guesses the number on the spinner right = 3/4 x 5/6 x 2/5 = 30/120 = 1/4
1/4 + 1/4 = 1/2
Mike wins if:
-James does not call both coin flips right and he calls the die roll correctly = 3/4 x 1/6 = 3/24 = 1/8
OR
-James does not call both coin flips right, he does not call the die roll correctly, and Mike does not guess the number on the spinner right = 3/4 x 5/6 x 3/5 = 45/120 = 3/8
1/8 + 3/8 = 1/2
Of course, dad could have just flipped a coin Did you answer this riddle correctly?
YES NO
100 Blank Cards Riddle
Someone offers you the following deal:
There is a deck of 100 initially blank cards. The dealer is allowed to write ANY positive integer, one per card, leaving none blank. You are then asked to turn over as many cards as you wish. If the last card you turn over is the highest in the deck, you win; otherwise, you lose.
Winning grants you $50, and losing costs you only the $10 you paid to play.
Would you accept this challenge?
There is a deck of 100 initially blank cards. The dealer is allowed to write ANY positive integer, one per card, leaving none blank. You are then asked to turn over as many cards as you wish. If the last card you turn over is the highest in the deck, you win; otherwise, you lose.
Winning grants you $50, and losing costs you only the $10 you paid to play.
Would you accept this challenge?
Hint: Perhaps thinking in terms of one deck is the wrong approach.
Yes!
A sample strategy:
Divide the deck in half and turn over all lower 50 cards, setting aside the highest number you find. Then turn over the other 50 cards, one by one, until you reach a number that is higher than the card you set aside: this is your chosen "high card."
Now, there is a 50% chance that the highest card is contained in the top 50 cards (it is or it isn't), and a 50% chance that the second-highest card is contained in the lower 50. Combining the probabilities, you have a 25% chance of constructing the above situation (in which you win every time).
This means that you'll lose three out of four games, but for every four games played, you pay $40 while you win one game and $50. Your net profit every four games is $10.
Obviously, you have to have at least $40 to start in order to apply this strategy effectively. Did you answer this riddle correctly?
YES NO
A sample strategy:
Divide the deck in half and turn over all lower 50 cards, setting aside the highest number you find. Then turn over the other 50 cards, one by one, until you reach a number that is higher than the card you set aside: this is your chosen "high card."
Now, there is a 50% chance that the highest card is contained in the top 50 cards (it is or it isn't), and a 50% chance that the second-highest card is contained in the lower 50. Combining the probabilities, you have a 25% chance of constructing the above situation (in which you win every time).
This means that you'll lose three out of four games, but for every four games played, you pay $40 while you win one game and $50. Your net profit every four games is $10.
Obviously, you have to have at least $40 to start in order to apply this strategy effectively. 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
The 3 Inch Cube Riddle
A 3 inch cube is painted on all sides with RED. The cube is then cut into small cubes of dimension 1 inch. All the so cut cubes are collected and thrown on a flat surface. What is the probability that all the top facing surfaces have RED paint on them?
Hint: Visualize the core of the cube.
ZERO.
The core of the 3 inch cube when cut, has all faces that are not painted. Hence at least one cube with no painted face always occurs. Did you answer this riddle correctly?
YES NO
The core of the 3 inch cube when cut, has all faces that are not painted. Hence at least one cube with no painted face always occurs. Did you answer this riddle correctly?
YES NO
The Traffic Light Riddle
There is a traffic light at the top of a hill. Cars can't see the light until they are 200 feet from the light.
The cycle of the traffic light is 30 seconds green, 5 seconds yellow and 20 seconds red.
A car is traveling 45 miles per hour up the hill.
What is the probability that the light will be yellow when the driver first crests the hill and that if the driver continues through the intersection at her present speed that she will run a red light?
The cycle of the traffic light is 30 seconds green, 5 seconds yellow and 20 seconds red.
A car is traveling 45 miles per hour up the hill.
What is the probability that the light will be yellow when the driver first crests the hill and that if the driver continues through the intersection at her present speed that she will run a red light?
Hint:
The probability of the driver encountering a yellow light and the light turning red before the car enters the intersection is about 5.5%.
At 45 mph the car is traveling at 66 feet/second and will take just over 3 seconds (3.03) to travel the 200 feet to the intersection. Any yellow light that is in the last 3.03 seconds of the light will cause the driver to run a red light.
The entire cycle of the light is 55 seconds. 3.03/55 = 5.5%. Did you answer this riddle correctly?
YES NO
At 45 mph the car is traveling at 66 feet/second and will take just over 3 seconds (3.03) to travel the 200 feet to the intersection. Any yellow light that is in the last 3.03 seconds of the light will cause the driver to run a red light.
The entire cycle of the light is 55 seconds. 3.03/55 = 5.5%. Did you answer this riddle correctly?
YES NO
I Have A Tail Riddle
Hint:
Soccer Kick Riddle
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
A Man Was Outside Taking A Walk When It Began To Rain
A man was outside taking a walk when it began to rain. He did not have an umbrella and he wasn't wearing a hat. His clothes were soaked, yet not a single hair on his head got wet. How could this happen?
Hint:
A Suit In A Deck Of Cards
I'm red but Im not a strawberry
I'm a shape but Im not a square
I'm part of your body but Im not your mouth
I'm a suit in a deck of cards but Im not a spade
I'm used to say I love you but Im not a diamond
I'm a?
I'm a shape but Im not a square
I'm part of your body but Im not your mouth
I'm a suit in a deck of cards but Im not a spade
I'm used to say I love you but Im not a diamond
I'm a?
Hint:
A London Scholar Riddle
There was a London scholar walking along.
He pulled of his gloves an Drew of his
head. What was that London scholars name?
He pulled of his gloves an Drew of his
head. What was that London scholars name?
Hint: The name was inside the sentence.
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:
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