Five Haystacks Riddle
A farmer has five haystacks in one field and four haystacks in another. How many haystacks would he have if he combined them all in one field?
Hint:
One. If he combines all his haystacks, they all become one big stack. Did you answer this riddle correctly?
YES NO
YES NO
The Butcher Shop Clerk Riddle
A clerk at a butcher shop stands five feet ten inches tall and wears size 13 sneakers. What does he weigh?
Hint:
Exposed To A Disease Riddle
A boy and his father have been exposed to a disease. Sadly, the father rapidly develops a tumor and dies. The boy survives, but desperately needs an operation and is rushed to hospital. A surgeon is called. Upon entering the room and seeing the patient, the surgeon exclaims, Oh no! I cant do the operation. Thats my son!
Hint:
Something I Seek
There is something I seek.
While it is bound, it chooses kings and peasants.
When it is freed, it foretells war or woe.
While it bound, it propels men's lusts and furies.
When it is freed, it tumbles, falls, and fades.
While it is bound, life will often thrive.
When it is freed, death will often follow.
What do I seek?
While it is bound, it chooses kings and peasants.
When it is freed, it foretells war or woe.
While it bound, it propels men's lusts and furies.
When it is freed, it tumbles, falls, and fades.
While it is bound, life will often thrive.
When it is freed, death will often follow.
What do I seek?
Hint:
A Town With No Houses
Hint:
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
The Prime Number Riddle
Two hundred people in an auditorium are asked to think of a single digit number from 1 to 9 inclusive and write it down. All those who wrote down a prime number are now asked to leave. Ninety people remain behind in the hall. How many of these are expected to have written down an odd number?
Hint: Remember that 1 is not a prime number.
Those that remain behind must have written {1,4,6,8,9} and from this only {1,9} are odd. The probability of an odd number is thus 2/5.
Expected number of odds is 2/5 * 90 = 36 Did you answer this riddle correctly?
YES NO
Expected number of odds is 2/5 * 90 = 36 Did you answer this riddle correctly?
YES NO
Russian Roulette Riddle
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?
Hint:
Russian Roulette
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Puzzle ID: #17681
Fun: *** (2.59)
Difficulty: ** (2.07)
Category: Probability
Submitted By: JMCLEOD****
Corrected By: cnmne
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?
Answer
Label the chambers 1 through 6. Chambers 1 through 3 have bullets and chambers 4 through 6 are empty. After you spin the cylinder there are six possible outcomes:
1. Chamber 1 is fired first: Player 1 loses
2. Chamber 2 is fired first: Player 1 loses
3. Chamber 3 is fired first: Player 1 loses
4. Chamber 4 is fired first: Player 2 loses (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)
5. Chamber 5 is fired first: Player 1 loses (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)
6. Chamber 6 is fired first: Player 2 loses (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)
Therefore player 2 has an 4/6 or 2/3 chance of winning. Did you answer this riddle correctly?
YES NO
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Puzzle ID: #17681
Fun: *** (2.59)
Difficulty: ** (2.07)
Category: Probability
Submitted By: JMCLEOD****
Corrected By: cnmne
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?
Answer
Label the chambers 1 through 6. Chambers 1 through 3 have bullets and chambers 4 through 6 are empty. After you spin the cylinder there are six possible outcomes:
1. Chamber 1 is fired first: Player 1 loses
2. Chamber 2 is fired first: Player 1 loses
3. Chamber 3 is fired first: Player 1 loses
4. Chamber 4 is fired first: Player 2 loses (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)
5. Chamber 5 is fired first: Player 1 loses (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)
6. Chamber 6 is fired first: Player 2 loses (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)
Therefore player 2 has an 4/6 or 2/3 chance of winning. 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
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
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
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
The Miracle Mountain Riddle
A hiker climbs all day up a steep mountain path and arrives at the mountain top where he camps overnight. The next day he begins the descent down the same trail to the bottom of the mountain when suddenly he looks at his watch and exclaims, "That is amazing! I was at this very same spot at exactly the same time of day yesterday on my way up."
What is the probability that a hiker will be at exactly the same spot on the mountain at the same time of day on his return trip, as he was on the previous day's hike up the mountain?
Is the probability closest to (A) 99% or (B) 50% or (C) 0.1% ?
What is the probability that a hiker will be at exactly the same spot on the mountain at the same time of day on his return trip, as he was on the previous day's hike up the mountain?
Is the probability closest to (A) 99% or (B) 50% or (C) 0.1% ?
Hint: This is not a trick. His watch works perfectly well. He does not sit in the same spot all day or any other such device, although it would not change the answer if he did!
The answer is (A). Since it must happen, the probability is actually 1 (100%).
Explanation: Firstly, consider 2 men, one starting from the top of the mountain and hiking down while the other starts at the bottom and hikes up. At some time in the day, they will cross over. In other words they will be at the same place at the same time of day.
Now consider our man who has walked up on one day and begins the descent the next day. Imagine there is someone (a second person) shadowing his exact movements from the day before. When he meets his shadower (it must happen) it will be the exact place that he was the day before, and of course they are both at this spot at the same time.
Contrary to our common sense, which seems to say that this is an extremely unlikely event, it is a certainty.
NOTE: There is one unlikely event here, and that is that he will notice the time when he is at the correct location on both days, but that was not what the question asked. Did you answer this riddle correctly?
YES NO
Explanation: Firstly, consider 2 men, one starting from the top of the mountain and hiking down while the other starts at the bottom and hikes up. At some time in the day, they will cross over. In other words they will be at the same place at the same time of day.
Now consider our man who has walked up on one day and begins the descent the next day. Imagine there is someone (a second person) shadowing his exact movements from the day before. When he meets his shadower (it must happen) it will be the exact place that he was the day before, and of course they are both at this spot at the same time.
Contrary to our common sense, which seems to say that this is an extremely unlikely event, it is a certainty.
NOTE: There is one unlikely event here, and that is that he will notice the time when he is at the correct location on both days, but that was not what the question asked. Did you answer this riddle correctly?
YES NO
Pearl Problems Riddle
"I'm a very rich man, so I've decided to give you some of my fortune. Do you see this bag? I have 5001 pearls inside it. 2501 of them are white, and 2500 of them are black. No, I am not racist. I'll let you take out any number of pearls from the bag without looking. If you take out the same number of black and white pearls, I will reward you with a number of gold bars equivalent to the number of pearls you took."
How many pearls should you take out to give yourself a good number of gold bars while still retaining a good chance of actually getting them?
How many pearls should you take out to give yourself a good number of gold bars while still retaining a good chance of actually getting them?
Hint: If you took out 2 pearls, you would have about a 50% chance of getting 2 gold bars. However, you can take even more pearls and still retain the 50% chance.
Take out 5000 pearls. If the remaining pearl is white, then you've won 5000 gold bars! Did you answer this riddle correctly?
YES NO
YES NO
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
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