## Ceo Riddles To Solve

## Solving Ceo Riddles

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## 12 Islanders Teeter Totter Riddle

There is an island with 12 islanders. All of the islanders individually weigh exactly the same amount, except for one, who either weighs more or less than the other 11.

You must use a see-saw to figure out whose weight is different, and you may only use the see-saw 3 times. There are no scales or other weighing device on the island.

How can you find out which islander is the one that has a different weight?

You must use a see-saw to figure out whose weight is different, and you may only use the see-saw 3 times. There are no scales or other weighing device on the island.

How can you find out which islander is the one that has a different weight?

Hint:

Six on one side - six on the other = one side is heavier.

Take the heavier six men, divide them into three and three (random).

Three on one side - three on the other = one side will one heavier.

Divide that three men from the heavier side side, have one on one side - one on the other.

Two results can determine which of the last three men weight is a different weight than each other.

With the last group of three men, have two men go head-to-head. The see-saw will either weight different: one weights more than the other man meaning the heavier man is the "12th man" or the see-saw will balance between the two men because they are the same weight. That means the third man standing on the sidelines by default weights more than the last two men weighted. Thus making that man on the sidelines the "12th man" that weights more than other 11.

Heavier wins 6v6; winner gets divided. Heavier wins 3v3; winner gets divided. Heavier wins 1v1 (12th man) or Equal 1v1 = third man weight more, he's the 12th man.

You could find the same results changing the process and picking from the lighter group three times. You’re only trying to find the difference in weight. Not the exact weight (more or less) of that "12th man."

Lightest 6v6; Lightest 3v3; Lightest 1v1 or Equal 1v1 = third man weight less.

YES NO

Take the heavier six men, divide them into three and three (random).

Three on one side - three on the other = one side will one heavier.

Divide that three men from the heavier side side, have one on one side - one on the other.

Two results can determine which of the last three men weight is a different weight than each other.

With the last group of three men, have two men go head-to-head. The see-saw will either weight different: one weights more than the other man meaning the heavier man is the "12th man" or the see-saw will balance between the two men because they are the same weight. That means the third man standing on the sidelines by default weights more than the last two men weighted. Thus making that man on the sidelines the "12th man" that weights more than other 11.

Heavier wins 6v6; winner gets divided. Heavier wins 3v3; winner gets divided. Heavier wins 1v1 (12th man) or Equal 1v1 = third man weight more, he's the 12th man.

You could find the same results changing the process and picking from the lighter group three times. You’re only trying to find the difference in weight. Not the exact weight (more or less) of that "12th man."

Lightest 6v6; Lightest 3v3; Lightest 1v1 or Equal 1v1 = third man weight less.

*Did you answer this riddle correctly?*YES NO

## Pierce Ones Ears Riddle

Hint:

## The Fourth Column Riddle

The fourth column has question marks in the place of numbers.

Do you know which one of the given options will take place of the fourth column?

Do you know which one of the given options will take place of the fourth column?

Hint:

The correct choice will be the option 'd'.

In each succeeding row, the previous column is reversed and the lowest digits are omitted.

YES NO

In each succeeding row, the previous column is reversed and the lowest digits are omitted.

*Did you answer this riddle correctly?*YES NO

## Piece Of Burned Wood Riddle

Hint:

## Aliens Favorite Place On A Computer Riddle

Hint:

## Sweet And Bakes Riddle

I have eyes but I cant see

I have skin but I cant feel anything

I can be sweet but Im not a piece of candy

I can be baked but Im not a cake

I can be peeled but Im not a carrot

What could I be?

I have skin but I cant feel anything

I can be sweet but Im not a piece of candy

I can be baked but Im not a cake

I can be peeled but Im not a carrot

What could I be?

Hint:

## Associated With Cob

Im yellow but Im not the sun

I grow in a field but Im not a sunflower

Im found on an ear but Im not a piece of jewelry

I go well with butter but Im not a slice of toast

Im associated with cob but Im not a web

What am I?

I grow in a field but Im not a sunflower

Im found on an ear but Im not a piece of jewelry

I go well with butter but Im not a slice of toast

Im associated with cob but Im not a web

What am I?

Hint:

## A Piece On A Chessboard

This is in a deck of cards

And a piece on a chessboard

Shes the monarch of Britain

Who knights people with a sword

Who is she?

And a piece on a chessboard

Shes the monarch of Britain

Who knights people with a sword

Who is she?

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:

## Safe And Secure Riddle

As a whole, I am both safe and secure.

Behead me, and I become a place of meeting.

Behead me again, and I am the partner of ready.

Restore me, and I become the domain of beasts.

What am I?

Behead me, and I become a place of meeting.

Behead me again, and I am the partner of ready.

Restore me, and I become the domain of beasts.

What am I?

Hint:

## 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

## 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?

Hint:

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!

YES NO

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!

*Did you answer this riddle correctly?*YES NO

## Pearl Problems Riddle

"I'm a very rich man, so I've decided to give you some of my fortune. Do you see this bag? I have 5001 pearls inside it. 2501 of them are white, and 2500 of them are black. No, I am not racist. I'll let you take out any number of pearls from the bag without looking. If you take out the same number of black and white pearls, I will reward you with a number of gold bars equivalent to the number of pearls you took."

How many pearls should you take out to give yourself a good number of gold bars while still retaining a good chance of actually getting them?

How many pearls should you take out to give yourself a good number of gold bars while still retaining a good chance of actually getting them?

Hint: If you took out 2 pearls, you would have about a 50% chance of getting 2 gold bars. However, you can take even more pearls and still retain the 50% chance.

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

YES NO

*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.

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

## 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.

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

## Add Your Riddle Here

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