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Riddles and Answers © 2020

## 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?

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%.

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

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.

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

## 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.

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 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?

Hint: Think what would be most likely to happen if you chose HHH, would this be a good decision?

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!

YES NO

*Did you answer this riddle correctly?*YES NO

## 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?

Hint:

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.

YES NO

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.

*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.

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

## 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.

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 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.

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

## Post Your Probability Riddles Below

Can you come up with a cool, funny or clever Probability Riddles of your own? Post it below (without the answer) to see if you can stump our users.