Solve fun **Probability Riddles**! Tease your brain with these cool mind boggling puzzles and jokes that will stump you.

## Probability Riddles And Answers To Solve

### Die Toss Riddle

If you toss a die and it comes up with the number one 9 times in a row, what is the probability that it will come up with one on the next throw?

One in six. A die has no memory of what it last showed.

### 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?

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%.

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%.

### 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?

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

Expected number of odds is 2/5 * 90 = 36

### 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?

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.

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.

### 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?

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.

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.

### The Coin Toss Riddle

You are in a bar having a drink with an old friend when he proposes a wager.

“Want to play a game?” he asks.

“Sure, why not?” you reply.

“Ok, here’s how it works. You choose three possible outcomes of a coin toss, either HHH, TTT, HHT or whatever. I will do likewise. I will then start flipping the coin continuously until either one of our combinations comes up. The person whose combination comes up first is the winner. And to prove I’m not the cheating little weasel you’re always making me out to be, I’ll even let you go first so you have more combinations to choose from. So how about it? Is $10.00 a fair bet?”

You know that your friend is a skilled trickster and usually has a trick or two up his sleeve but maybe he’s being honest this time. Maybe this is a fair bet. While you try and think of which combination is most likely to come up first, you suddenly hit upon a strategy which will be immensely beneficial to you. What is it?

“Want to play a game?” he asks.

“Sure, why not?” you reply.

“Ok, here’s how it works. You choose three possible outcomes of a coin toss, either HHH, TTT, HHT or whatever. I will do likewise. I will then start flipping the coin continuously until either one of our combinations comes up. The person whose combination comes up first is the winner. And to prove I’m not the cheating little weasel you’re always making me out to be, I’ll even let you go first so you have more combinations to choose from. So how about it? Is $10.00 a fair bet?”

You know that your friend is a skilled trickster and usually has a trick or two up his sleeve but maybe he’s being honest this time. Maybe this is a fair bet. While you try and think of which combination is most likely to come up first, you suddenly hit upon a strategy which will be immensely beneficial to you. What is it?

The answer is to let your friend go first. This puzzle is based on an old game/scam called Penny Ante. No matter what you picked, your friend would be able to come up with a combination which would be more likely to beat yours. For example, if you were to choose HHH, then unless HHH was the first combination to come up you would eventually lose since as soon as a Tails came up, the combination THH would inevitably come up before HHH. The basic formula you can use for working out which combination you should choose is as follows. Simply take his combination (eg. HHT) take the last term in his combination, put it at the front (in this case making THH) and your combination will be more likely to come up first. Try it on your friends!

### The Cheap Mp3 Player

My MP3 player is cheap ‘n’ nasty and has now broken: it is stuck on ‘Shuffle’. In this mode it starts with whatever track you put it on, but then plays tracks in a random order. The only restriction is it never plays a song that’s already been played that day.

I purchased my favourite murder mystery book in audio format, and put the first 6 chapters on my MP3 player. (Each chapter is exactly 1 track.) There’s nothing else on my player at the moment. What is the probability that I will hear the 6 chapters in order as I listen today, without having to change tracks at all? (Obviously, I will ensure it plays chapter 1 first.)

The next day I empty the player before putting on the next 6 chapters. This time I also transfer a CD of mine with 11 songs on. I don’t mind songs coming in between the chapters of my book, as long as the chapters are in order. What’s the probability of that happening now?

I purchased my favourite murder mystery book in audio format, and put the first 6 chapters on my MP3 player. (Each chapter is exactly 1 track.) There’s nothing else on my player at the moment. What is the probability that I will hear the 6 chapters in order as I listen today, without having to change tracks at all? (Obviously, I will ensure it plays chapter 1 first.)

The next day I empty the player before putting on the next 6 chapters. This time I also transfer a CD of mine with 11 songs on. I don’t mind songs coming in between the chapters of my book, as long as the chapters are in order. What’s the probability of that happening now?

With only 6 tracks on the player:

The first chapter has been set to play first. The probability of the next 5 chapters playing in order is 1/5! = 1/120.

With the music on the player as well:

Seeing as I don’t care about when the music plays, it doesn’t change anything. The answer is still 1/120.

The first chapter has been set to play first. The probability of the next 5 chapters playing in order is 1/5! = 1/120.

With the music on the player as well:

Seeing as I don’t care about when the music plays, it doesn’t change anything. The answer is still 1/120.

### 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?

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.

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.

### 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?)

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.

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.

### 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?

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.

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.

### 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?

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.

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.

### 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?

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

1×6

2×3

6×1

3×2

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%

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

1×6

2×3

6×1

3×2

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%

### 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?

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.

X = 3

The second die must have 3 red faces and 3 blue faces.

### 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?”

“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?

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.

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.

### 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?

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

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

### Chances Of A 2nd Girl Riddle

Tipli and Pikli are a married couple (dont ask me who he is and who she is)

They have two children, one of the child is a boy. Assume safely that the probability of each gender is 1/2.

What is the probability that the other child is also a boy?

They have two children, one of the child is a boy. Assume safely that the probability of each gender is 1/2.

What is the probability that the other child is also a boy?

1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Boy – Girl

Girl – Boy

Boy – Boy

Girl – Girl

Since we know one of the children is a boy, we will drop the girl-girl possibility from the sample space.

This leaves only three possibilities, one of which is two boys. Hence the probability is 1/3

This is a famous question in understanding conditional probability, which simply means that given some information you might be able to get a better estimate.

The following are possible combinations of two children that form a sample space in any earthly family:

Boy – Girl

Girl – Boy

Boy – Boy

Girl – Girl

Since we know one of the children is a boy, we will drop the girl-girl possibility from the sample space.

This leaves only three possibilities, one of which is two boys. Hence the probability is 1/3

### 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?

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!

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!

### 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?

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.

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.

### Matching Socks Riddle

Mismatched Joe is in a pitch dark room selecting socks from his drawer. He has only six socks in his drawer, a mixture of black and white. If he chooses two socks, the chances that he draws out a white pair is 2/3. What are the chances that he draws out a black pair?

He has a ZERO chance of drawing out a black pair.

Since there is a 2/3 chance of drawing a white pair, then there MUST be 5 white socks and only 1 black sock. The chances of drawing two whites would thus be: 5/6 x 4/5 = 2/3 . With only 1 black sock, there is no chance of drawing a black pair.

Since there is a 2/3 chance of drawing a white pair, then there MUST be 5 white socks and only 1 black sock. The chances of drawing two whites would thus be: 5/6 x 4/5 = 2/3 . With only 1 black sock, there is no chance of drawing a black pair.

### 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?

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.

### 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?

Take out 5000 pearls. If the remaining pearl is white, then you’ve won 5000 gold bars!

### 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?

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.

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.

### 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?

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.

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.

### The Emperor’s Proposition Riddle

You are a prisoner sentenced to death. The Emperor offers you a chance to live by playing a simple game. He gives you 50 black marbles, 50 white marbles and 2 empty bowls. He then says, “Divide these 100 marbles into these 2 bowls. You can divide them any way you like as long as you use all the marbles. Then I will blindfold you and mix the bowls around. You then can choose one bowl and remove ONE marble. If the marble is WHITE you will live, but if the marble is BLACK… you will die.”

How do you divide the marbles up so that you have the greatest probability of choosing a WHITE marble?

How do you divide the marbles up so that you have the greatest probability of choosing a WHITE marble?

Place 1 white marble in one bowl, and place the rest of the marbles in the other bowl (49 whites, and 50 blacks).

This way you begin with a 50/50 chance of choosing the bowl with just one white marble, therefore life! BUT even if you choose the other bowl, you still have ALMOST a 50/50 chance at picking one of the 49 white marbles.

This way you begin with a 50/50 chance of choosing the bowl with just one white marble, therefore life! BUT even if you choose the other bowl, you still have ALMOST a 50/50 chance at picking one of the 49 white marbles.

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