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
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 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
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
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
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
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
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
Kindness And Cruelty
Capable of Kindness and cruelty, I take victims when I sour. I can be on your side or wrong you. I bring gifts though you already have me. What am I?
Hint:
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:
Skunk Christmas Riddle
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:
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:
Over London Bridge
As I went over London Bridge
I met my sister Jenny
I broke her neck and drank her blood
And left her standing empty
Who is Jenny?
I met my sister Jenny
I broke her neck and drank her blood
And left her standing empty
Who is Jenny?
Hint:
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