The Loaded Revolver Riddle
Henry has been caught stealing cattle, and is brought into town for justice. The judge is his ex-wife Gretchen, who wants to show him some sympathy, but the law clearly calls for two shots to be taken at Henry from close range. To make things a little better for Henry, Gretchen tells him she will place two bullets into a six-chambered revolver in successive order. She will spin the chamber, close it, and take one shot. If Henry is still alive, she will then either take another shot, or spin the chamber again before shooting.
Henry is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower. He steels himself as Gretchen loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Gretchen asks, "Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?"
What should Henry choose?
Henry is a bit incredulous that his own ex-wife would carry out the punishment, and a bit sad that she was always such a rule follower. He steels himself as Gretchen loads the chambers, spins the revolver, and pulls the trigger. Whew! It was blank. Then Gretchen asks, "Do you want me to pull the trigger again, or should I spin the chamber a second time before pulling the trigger?"
What should Henry choose?
Hint:
Henry should have Gretchen pull the trigger again without spinning.
We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.
If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position. Did you answer this riddle correctly?
YES NO
We know that the first chamber Gretchen fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber. So if Henry has Gretchen pull the trigger again, the probability that a bullet will be fired is 1/4.
If Gretchen spins the chamber again, the probability that she shoots Henry would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position. Did you answer this riddle correctly?
YES NO
Roll The Dice
A gambler goes to bet. The dealer has 3 dice, which are fair, meaning that the chance that each face shows up is exactly 1/6.
The dealer says: "You can choose your bet on a number, any number from 1 to 6. Then I'll roll the 3 dice. If none show the number you bet, you'll lose $1. If one shows the number you bet, you'll win $1. If two or three dice show the number you bet, you'll win $3 or $5, respectively."
Is it a fair game?
The dealer says: "You can choose your bet on a number, any number from 1 to 6. Then I'll roll the 3 dice. If none show the number you bet, you'll lose $1. If one shows the number you bet, you'll win $1. If two or three dice show the number you bet, you'll win $3 or $5, respectively."
Is it a fair game?
Hint: What will happen if there are 6 gamblers, each of whom bet on a different number?
It's a fair game. If there are 6 gamblers, each of whom bet on a different number, the dealer will neither win nor lose on each deal.
If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1.
If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1.
If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1.
In each case, the dealer neither wins nor loses. Hence it's a fair game. Did you answer this riddle correctly?
YES NO
If he rolls 3 different numbers, e.g. 1, 2, 3, the three gamblers who bet 1, 2, 3 each wins $1 while the three gamblers who bet 4, 5, 6 each loses $1.
If two of the dice he rolls show the same number, e.g. 1, 1, 2, the gambler who bet 1 wins $3, the gambler who bet 2 wins $1, and the other 4 gamblers each loses $1.
If all 3 dice show the same number, e.g. 1, 1, 1, the gambler who bet 1 wins $5, and the other 5 gamblers each loses $1.
In each case, the dealer neither wins nor loses. Hence it's a fair game. Did you answer this riddle correctly?
YES NO
Fighting In A Truel
Mr. Black, Mr. Gray, and Mr. White are fighting in a truel. They each get a gun and take turns shooting at each other until only one person is left. Mr. Black, who hits his shot 1/3 of the time, gets to shoot first. Mr. Gray, who hits his shot 2/3 of the time, gets to shoot next, assuming he is still alive. Mr. White, who hits his shot all the time, shoots next, assuming he is also alive. The cycle repeats. All three competitors know one another's shooting odds. If you are Mr. Black, where should you shoot first for the highest chance of survival?
Hint: Think from the points of view of Mr. Gray and Mr. White, not just Mr. Black.
He should shoot at the ground.
If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before. Did you answer this riddle correctly?
YES NO
If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr. Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr. Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White, Mr. Black has a better chance of winning than before. Did you answer this riddle correctly?
YES NO
Two In A Row Riddle
A certain mathematician, his wife, and their teenage son all play a fair game of chess. One day when the son asked his father for 10 dollars for a Saturday night date, his father puffed his pipe for a moment and replied, "Let's do it this way. Today is Wednesday. You will play a game of chess tonight, tomorrow, and a third on Friday. If you win two games in a row, you get the money."
"Whom do I play first, you or mom?"
"You may have your choice," said the mathematician, his eyes twinkling.
The son knew that his father played a stronger game than his mother. To maximize his chance of winning two games in succession, should he play father-mother-father or mother-father-mother?
"Whom do I play first, you or mom?"
"You may have your choice," said the mathematician, his eyes twinkling.
The son knew that his father played a stronger game than his mother. To maximize his chance of winning two games in succession, should he play father-mother-father or mother-father-mother?
Hint: Who does he need to beat to win?
Father-mother-father
To beat two games in a row, it is necessary to win the second game. This means that it would be to his advantage to play the second game against the weaker player. Though he plays his father twice, he has a higher chance of winning by playing his mother second. Did you answer this riddle correctly?
YES NO
To beat two games in a row, it is necessary to win the second game. This means that it would be to his advantage to play the second game against the weaker player. Though he plays his father twice, he has a higher chance of winning by playing his mother second. Did you answer this riddle correctly?
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
Three Rats Riddle
Three rats are sitting at the three corners of an equilateral triangle. Each rat starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the rats collide?
Hint:
So lets think this through. The rats can only avoid a collision if they all decide to move in the same direction (either clockwise or rati-clockwise). If the rats do not pick the same direction, there will definitely be a collision. Each rat has the option to either move clockwise or rati-clockwise. There is a one in two chance that an rat decides to pick a particular direction. Using simple probability calculations, we can determine the probability of no collision. Did you answer this riddle correctly?
YES NO
YES NO
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
The Gardners Riddle
Gretchen and Henry were discussing their new neighbors, the Gardners. Gretchen mentioned that she met two of the daughters, and they each had blond hair.
"I have met all of the sisters," replied Henry, "and the probability that both of the girls you met would have had blond hair, assuming you were equally likely to meet any of the sisters, is exactly 50%. Do you know how many children there are?"
After thinking for a minute, Gretchen asks if the family is abnormally large. When Henry replies that it is not, Gretchen tells him how many girls are in the family. What number did she say?
"I have met all of the sisters," replied Henry, "and the probability that both of the girls you met would have had blond hair, assuming you were equally likely to meet any of the sisters, is exactly 50%. Do you know how many children there are?"
After thinking for a minute, Gretchen asks if the family is abnormally large. When Henry replies that it is not, Gretchen tells him how many girls are in the family. What number did she say?
Hint:
Gretchen said that there were 4 girls in the family, three of whom were blond.
This would make the probability that she saw two blonds (3/4) * (2/3), which equals 1/2.
Other numbers would work, but the next pair would lead to a rather large family. Did you answer this riddle correctly?
YES NO
This would make the probability that she saw two blonds (3/4) * (2/3), which equals 1/2.
Other numbers would work, but the next pair would lead to a rather large family. Did you answer this riddle correctly?
YES NO
The Blue And Red Dice Riddle
Timothy and Urban play a game with two dice. But they do not use the numbers. Some of the faces are painted red and the others blue. Each player throws the dice in turn. Timothy wins when the two top faces are the same color. Urban wins when the colors are different. Their chances are even.
The first die has 5 red faces and 1 blue face. How many red and how many blue are there on the second die?
The first die has 5 red faces and 1 blue face. How many red and how many blue are there on the second die?
Hint:
Each die has 6 faces. When two dice are thrown, there are 36 equally possible results. For chances to be even, there must be 18 ways of getting the same color on top. Let X be the number of red faces on the second die. We have: 18 = 5X + 1(6 - X)
X = 3
The second die must have 3 red faces and 3 blue faces. Did you answer this riddle correctly?
YES NO
X = 3
The second die must have 3 red faces and 3 blue faces. Did you answer this riddle correctly?
YES NO
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
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
Blue Eyes Riddle
Both of my parents have brown eyes, as do I. My brother and my wife have blue eyes. Using the simple brown-blue model (two genes; a brown gene dominates blue gene), what are the chances of my first child having blue eyes?
Hint: Given my brother's blue eyes, what are the odds on my pair of eye-color genes?
1 in 3.
Since my brother has blue eyes (bb), both of my parents carry one brown and one blue gene (Bb). The three possibilities for my genotype, equally likely, are BB, Bb, and bB. Thus, there is a 2/3 chance that I carry a blue gene.
If I carry a blue gene, there is a 50% chance I will pass it on to my first child (and, obviously, 0% if I carry two brown genes).
Since my child will certainly get a blue gene from my wife, my gene will determine the eye color.
Multiplying the probabilities of those two independent events, there is a chance of 1/2 x 2/3 = 1/3 of my passing on a blue gene. Did you answer this riddle correctly?
YES NO
Since my brother has blue eyes (bb), both of my parents carry one brown and one blue gene (Bb). The three possibilities for my genotype, equally likely, are BB, Bb, and bB. Thus, there is a 2/3 chance that I carry a blue gene.
If I carry a blue gene, there is a 50% chance I will pass it on to my first child (and, obviously, 0% if I carry two brown genes).
Since my child will certainly get a blue gene from my wife, my gene will determine the eye color.
Multiplying the probabilities of those two independent events, there is a chance of 1/2 x 2/3 = 1/3 of my passing on a blue gene. 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
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:
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