Puzzles from real interviews

Timepieces

noreply@blogger.com (Administrator) — Wed, 30 Jun 2010 13:15:00 +0000

A sundial is a timepiece that has the fewest number of moving parts. Which timepiece has the most moving parts?

Answer
An Hourglass

Band Camp

noreply@blogger.com (Administrator) — Tue, 18 May 2010 18:19:00 +0000

Mogmatt's marching band played their show and then listened to it. They played the show almost perfectly and yet when they listened to it, it was all messed up. How could that have happened?

Ans

The band was hearing their echo. Because they were marching at the time, they were all different distances from the wall and their echoes played at different times.

Bird on the Moon

noreply@blogger.com (Administrator) — Tue, 18 May 2010 18:18:00 +0000

If a light oxygen tank were tied to a bird so that it can breathe on the moon, would the bird fly faster, slower or the same speed as it does on earth? (Remember that there is less gravity on the moon)

Ans

A bird cannot fly on the moon because there is no air to suspend it.

Water in the Cup

noreply@blogger.com (Administrator) — Tue, 18 May 2010 18:17:00 +0000

A man in a restaurant asked a waiter for a juice glass, a dinner plate, water, a match, and a lemon wedge. The man poured enough water onto the plate to cover it.
"If you can get the water on the plate into this glass without touching or moving this plate, I will give you $100," the man said. "You can use the match and lemon to do this."
A few minutes later, the waiter walked away with $100 in his pocket. How did the waiter get the water into the glass?

Ans
First, the waiter stuck the match into the lemon wedge, so that it would stand straight. Then he lit the match, and put it in the middle of the plate with the lemon. Then, he placed the glass upside-down over the match. As the flame used up the oxygen in the glass, it created a small vacuum, which sucked in the water through the space between the glass and the plate. Thus, the waiter got the water into the glass without touching or moving the plate.

Trapdoors

noreply@blogger.com (Administrator) — Thu, 29 Apr 2010 20:00:00 +0000

There are five doors, one leads to the exit, the others lead to traps. They are in a line. The clues tell you which position the doors are in the line and where the door to freedom is. All the clues are true. Each door has a clue written on it. The clues read:

The blue door: This door is two spots away from the door to freedom.
The red door: This door is at the far right, and is two spots away from the blue door.
The purple door: This door is not next to the door to freedom.
The green door: This door is left of the blue door.
The orange door: This door is not next to the red or blue doors.

Which door leads to freedom?

Ans

The orange door leads to freedom.

Tinman's Pick

noreply@blogger.com (Administrator) — Tue, 27 Apr 2010 20:46:00 +0000

Puzzle

Tinman was in quite a pickle. He was choosing a vehicle that he had to drive for the next 7 years, and he had to make the best decision possible. He had only five choices to choose from, and the choices each had different qualities that made them better or worse.

If the third choice was worse than the first choice and the second choice as good as the fifth, but the fifth choice was only as good as the worst choice leaving the fourth choice a little better than the third but not as good as the first, and the second was the worst choice to go with, which choice should Tinman go with if he wanted the best vehicle?

Answer

Tinman should go with the first choice. In order from worst to best, the choices are:

5th/2nd, 3rd, 4th, 1st.

Five Cards

noreply@blogger.com (Administrator) — Tue, 27 Apr 2010 20:45:00 +0000

Puzzle:

I was playing a game of five card draw poker with a bunch of logicians. By the time we had finished bidding and were just about to reveal our cards, I was pretty confident I would win of the four of us remaining. I had three nines, some face card (I can't remember what suit or even whether it was a jack, queen, or king) and a four. (Or was it a five? I can't remember.)

I was even more sure when two of my opponents laid down their cards. One had a pair of fours and a pair of sevens, the other had a pair of twos and a pair of eights. My third opponent, however, laid down his five cards face down in a row. He said, "I have a straight, and the cards are, from lowest to highest: a ten, a jack, a queen, a king, and an ace. I have at least one card of each of the four suits: clubs, spades, hearts, and diamonds. I am fairly certain that this is the winning hand, but I'm feeling generous today, and I will give a third of the pot to whoever can determine which suit I have two cards of.

Now I know you can't figure it out without some clues. Here they are:

1. The king is next to at least one diamond.
2. The queen is next to exactly one heart.
3. The jack is next to at least one spade, but is not next to any hearts.
4. The ten is next to at least one club.
5. The ace does not border any black cards, nor does it border any diamonds.
6. My two cards of the same suit are not next to each other.
7. Of the ten possible pairings of cards, only one pair, when removed, leaves three cards in ascending order from left to right.
8. My ace is not the card on the far left."

There was a minute's silence. One of the other logicians said, "I give up! There's no way to figure that out!"
The other agreed. But I didn't. I had just figured out which suit he had two of.

Which suit is it?

Solution:

The two cards on the ends and the card in the middle border a combined number of two suits. So at least one pair of them should be listed as bordering a card of the same suit in clues 1-5. Of the five, only the ace and the queen meet this criterion, so they must be two cards apart. One of them must be in the center, but it can't be the ace (the two cards on the ends would only border hearts - and one of them has to be bordering something else from clues 1-4), so it must be the queen. The ace is not on the left end from clue 8, so it must be on the right end, and the card next to it must be a heart. So far we have this(X represents unknown):
X X Q X A
X X X H X
At least one of the two cards to the left of the queen is lower than the queen (two of the remaining cards are lower, and only one can be right of the queen), so the triplet that is in ascending order is the queen, the ace, and the single card to the left of the queen that is lower than the queen. That means the king is left of the queen, and is also left of both the jack and the ten (otherwise, jack king ace or ten king ace would be a second triplet), so it must be on the far left. The jack must be left of the ten (or else ten jack ace would be a second triplet), so it must be second from the left, and the ten must be second from the right. The only card next to the king is the jack of diamonds (it's a diamond from clue 1). So far we have(T is ten):
K J Q T A
X D X H X
Now we seem to be stuck. We don't have any more clues that can be used. So how did I figure it out?
I held a face card (I said so myself in the intro). I must have known it cannot be that card and deduced the correct answer from there.
But which card did I have? We already know the suit of the jack, so my holding a jack would not help. Could I have held a queen? No, because I know the queen to either be a spade or club (clue 6), so one of the cards next to the queen would have its clue satisfied (see clues 3 and 4), and I could not determine the suit of the card on the other side of it.
Therefore, I must have held a king. But which king? I know the king is not a diamond (clue 6) or a heart (clue 3). If I held the king of clubs, then the king next to the jack would have satisfied the jack's clue, and I could not have determine the suit of either card next to the ten.
Therefore, I held the king of spades. The king of clubs must have been on the far left (only suit available), so the only card next to the jack that could be a spade is the queen, so the only card next to the ten that could be a club is the ace. In summary:
K J Q T A
C D S H C
Therefore there are two CLUBS.

Spirit Search

noreply@blogger.com (Administrator) — Tue, 27 Apr 2010 20:43:00 +0000

Puzzle:
You are an expert on paranormal activity and have been hired to locate a spirit haunting an old resort hotel. Strong signs indicate that the spirit lies behind one of four doors. The inscriptions on each door read as follows:

Door A: It's behind B or C
Door B: It's behind A or D
Door C: It's in here
Door D: It's not in here

Your psychic powers have told you three of the inscriptions are false, and one is true. Behind which door will you find the spirit?

Answer:

The spirit lies behind Door D.

If the spirit is behind Door A, then both B and D are true.

If the spirit is behind Door B, then both A and D are true.

If the spirit is behind Door C, then A, C, and D are all true.

If the spirit is behind Door D, then the statements on all the doors are false, except for that on Door B. This matches the rules, and therefore, the resort hotel spirit lurks behind Door D.

Marbles, Coin and Die

noreply@blogger.com (Administrator) — Wed, 21 Apr 2010 19:31:00 +0000

Puzzle

If I have:
a normal coin with a heads and tails;
a 6-sided die;
and a bag containing 4 blue and 2 red marbles,
what is the probability of me flipping a heads, rolling a 4, and picking out a red marble?

Solution

Probability of flipping a heads: 1/2
Probability of rolling a 4: 1/6
Probability of selecting a red marble: 2/6

Then multiply the results, so:
1/2 x 1/6 x 2/6= 1/36 or 0.027 to 3 decimal places

The Gardner Sisters

noreply@blogger.com (Administrator) — Mon, 19 Apr 2010 20:08:00 +0000

Puzzle:

Gretchen and Henry invited the four Gardner sisters over to their house for an afternoon tea. Henry went to the cabinet to take out some plates (they have both blue and green plates in the cabinet), and the first two plates he took out were blue. "What are the odds?" asked Martina Gardner, the youngest.

Henry thought for a moment, and then replied, "Knowing how many plates of each color I have, the probability that I would pull out two blue ones is exactly 1/2!" Martina then asked if Henry could feed two dozen people if he used all of his plates. "Not quite," he replied.

She then told him how many plates he had. What number did she say?

Solution:

Henry has 21 plates, and 15 of them are blue.

This makes the probability of drawing two plates (15/21) * (14/20), which equals 1/2.

Martina had to ask if he could feed two dozen people because there are other (larger) numbers that work. For example, if he had 85 blue plates out of 120 total, the probability that he would pull out two blue ones would have been (85/120) * (84/119), or 1/2. Other numbers work, as well, but all are greater than 120.

Another Game of Dice

noreply@blogger.com (Administrator) — Sun, 18 Apr 2010 19:30:00 +0000

Puzzle
Your friend offers to play a game of dice with you. He explains the game to you.

"We each get one die, the highest die wins. If we tie, I win, but since you always lose when you roll a one, if you roll a one you can roll again. If you get a one the second time you have to keep it."

What is each person's probability of winning?

What are the probabilities of winning if you can keep rolling until you get something besides a one?

Solution
In the first game the probabilities of winning are 37/72 for your friend and 35/72 for you. In the second game each player has a 50% probability of winning.

There are 36 combinations (6 x 6) for your die and your friend's die on the first roll. In the first game if you don't roll a one on the first roll then that roll is equally likely to be a 2, 3, 4, 5 or 6, which can beat 1, 2, 3, 4, 5 different numbers, respectively. That gives a

(1+2+3+4+5)/36 = 15/36

probability of winning on the first roll. There is a 1/6 chance of rolling a one, in which case you have a 15/36 probability of winning on the second roll, which gives a

1/6 * 15/36 = 15/216 = 5/72

probability of winning on the second roll for a total probability of winning of

15/36 + 5/72 = 30/72 + 5/72 = 35/72 = .486111...

In the second game we never keep a one, so there are only 30 combinations (5 x 6) for the final roll. As shown above, there are 15 ways to win, which gives a probability of

15/30 = 1/2

of winning the second game.

Duplicate Lottery Picks

noreply@blogger.com (Administrator) — Sun, 18 Apr 2010 19:28:00 +0000

Puzzle
In the Massachusetts Megabucks lottery, six different numbers from 1 to 42 (inclusive) are selected. When you buy a ticket, you can ask for a "quick pick" in which the computer chooses the numbers for you, and you can purchase up to five games on a single ticket. We'll assume that the computer's random number generator is fair, giving each possible combination an equal probability of being chosen.

1. If I "quick pick" for two games, what are the chances that the two games have the same combination of numbers?

2. If I "quick pick" for five games (one five-game ticket), what are the chances that there are two games on that ticket with the same combination?

3 (The toughie). How many five-game quick-pick tickets would I have to buy in order to have a greater than 50% chance of having at least one ticket with two games on it that match exactly?

Solution
1. 1/5245786. The first game on the ticket will be some combination. Then you just calculate the chances that the second game on the ticket will match it. This number of combinations is (42 choose 6) or

42! / 6! (42-6)! = 5245786

So the chances that they match is 1 over this number.

2. For this sort of problem, where you are asking what are the chances of something happening at least once out of several opportunities to happen, you first calculate the opposite -- the chances of it NOT happening in all the tries -- and subtract from 1. So, what are the chances that the 5 games on the ticket are all different?

We know that there are 5245786 possibilities for any one game. In the first game, we choose one of them. The second game now has a 5245785 / 5245786 chance of being different from that first one. Now the third game has a 5245784 / 5245786 chance of being different from either of the first two; and the fourth has a 5245783 / 5245786 chance of being a new selection. When you have the chances of individual events occurring, and you want to know the chance of them ALL occurring, you just multiply. We multiply the chances of all these events occurring (that is, each choice being different) to get:

(5245785 * 5245784 * 5245783 * 5245782) / 5245786 ^ 4

This equals

0.99999809370925005714289758986902

Remember, this is the chance of all of the games on a five-game ticket being different. So the chance of at least two of them being the same is 1 minus this number, or

0.0000019062907499428571024101309848736

which is a really tiny number.

3. We take a similar approach with this calculation that we took before: figure out the chance of it not happening for N tickets, and subtract that value from 1. Then we set that chance to 0.5 and solve for N.

We already know the chance of it happening in one try: the tiny number above, which, for now, we'll call p. So the chance of it NOT happening in one try is (1-p). The chance of it NOT happening in n tries is (1-p)^n, so the chance of it happening at least once in n tries is [1 - (1-p)^n]. We set this formula to 0.5 and solve for n.

Of course, solving for n is tricky, unless you are comfortable with logarithms. I start with the equation

0.5 = [1 - (1-p)^n]

Simplify
0.5 = (1-p)^n

Take natural logarithm of both sides
ln(0.5) = ln( (1-p)^n)

Use logarithm magic
ln(0.5) = n * ln(1-p)

Divide both sides by ln(1-p)
n = ln(0.5) / ln(1-p)

Plug in the number (which we already know) for p and let the calculator do what it's good at
n = 363610.07359999192796640483226154

which is how many tickets we would have to buy to have a 50% chance of seeing one ticket with a match. Since we can't buy fractional tickets, we round up, to make sure we have a greater than 50% chance.

So we need to buy 363611 five-game tickets to have a better than 50% chance of having at least one ticket on which two games match exactly.

St. Petersburg Paradox

noreply@blogger.com (Administrator) — Sun, 18 Apr 2010 19:26:00 +0000

Puzzle:
You are offered a game to play with a single fair coin. It costs 20 dollars to play this game, but you can win much more than that. The way it works is that you continue to flip the coin until you get tails. For every heads you get before that, your payoff doubles. For example, if you get:
Heads
Heads
Tails, then you would earn 4 dollars.
In other words, you get: 2^heads dollars after you play. The question is: would you come out with more or less money after you played this game an INFINITE number of times? Remember, each game costs 20 dollars!

Solution
Neither!
You would come out with an INFINITE amount of money! Here's why:

The way to calculate an expected value of a game=(the probability of event1)*(the payoff from event1)+(the probability of event2)*(the payoff from event2)...

Let's say:
event1=Tails
event2=Heads,Tails
event3=Heads,Heads,Tails, and so on.

The probability of these events are:
event1=1/2
event2=1/2*1/2=1/4
event3=1/2*1/2*1/2=1/8, and so on.

The payoff of these events are:
event1=1
event2=2
event3=4
event4=8, and so on.

Plugging this into the expected value formula, we get:
EV=(1/2*1)+(1/4*2)+(1/8*4)+(1/16*8)...

This simplifies to:
EV=1/2+1/2+1/2+1/2...
Any number added an infinite number of times will sum to infinity, so your expected value of this game is infinity.

100 Prisoners in Solitary Cells

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:38:00 +0000

100 prisoners are stuck in the prison in solitary cells. The warden of the prison got bored one day and offered them a challenge. He will put one prisoner per day, selected at random (a prisoner can be selected more than once), into a special room with a light bulb and a switch which controls the bulb. No other prisoners can see or control the light bulb. The prisoner in the special room can either turn on the bulb, turn off the bulb or do nothing. On any day, the prisoners can stop this process and say “Every prisoner has been in the special room at least once”. If that happens to be true, all the prisoners will be set free. If it is false, then all the prisoners will be executed. The prisoners are given some time to discuss and figure out a solution. How do they ensure they all go free?

Trailing Zeros in 100 Factorial

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:38:00 +0000

How many zeros are there in 100! (100 factorial)?

Apples and Oranges?

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:37:00 +0000

You have 3 baskets, one with apples, one with oranges and one with both apples and oranges mixed. Each basket is closed and is labeled with ‘Apples’, ‘Oranges’ and ‘Apples and Oranges’. However, each of these labels is always placed incorrectly. How would you pick only one fruit from a basket to place the labels correctly on all the baskets?

5 Pirates Fight for 100 Gold Coins

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:37:00 +0000

Five pirates discover a chest containing 100 gold coins. They decide to sit down and devise a distribution strategy. The pirates are ranked based on their experience (Pirate 1 to Pirate 5, where Pirate 5 is the most experienced). The most experienced pirate gets to propose a plan and then all the pirates vote on it. If at least half of the pirates agree on the plan, the gold is split according to the proposal. If not, the most experienced pirate is thrown off the ship and this process continues with the remaining pirates until a proposal is accepted. The first priority of the pirates is to stay alive and second to maximize the gold they get. Pirate 5 devises a plan which he knows will be accepted for sure and will maximize his gold. What is his plan?

How Strong is an Egg?

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:36:00 +0000

You have two identical eggs. Standing in front of a 100 floor building, you wonder what is the maximum number of floors from which the egg can be dropped without breaking it. What is the minimum number of tries needed to find out the solution?

What’s Your Eye Color?

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:36:00 +0000

On a certain island there are people with assorted eye colors. There are 100 people with blue eyes and 100 people with brown eyes. Since there are no mirrors on this island, no person knows the color of their own eyes. The people on the island are not allowed to talk or communicate with each other in any way. They are also NOT aware of the number of blue or brown eyed people on the island. For all they know, they could have red eyes too. But they are allowed to observe other people and keep count of the number of people with a certain eye color. There is a rule that the people on the island have to follow – any person who is sure of their eye color has to leave the island immediately.

One day, an outsider comes to the island and announces to the people that he sees someone with blue eyes. What do you think happens?

Bulb Or No Bulb?

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:35:00 +0000

There are 100 bulbs arranged in a row. Each bulb has its own switch and is currently turned off. In the first round, you turn every switch on. In the second round, you flip the switch of every second bulb (i.e. bulb 2, 4, 6, 8 and so on). In the third round, you flip the switch of every third bulb and so on. What is the state of bulb 9 after 100 rounds? Also, how many bulbs are on after 100 rounds?

6 Pirates Fight for 1 Gold Coin

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:35:00 +0000

Six pirates discover a chest containing 1 gold coin. They decide to sit down and devise a distribution strategy. The pirates are ranked based on their experience (Pirate 1 to Pirate 6, where Pirate 6 is the most experienced). The most experienced pirate gets to propose a plan and then all the pirates vote on it. If at least half of the pirates agree on the plan, the gold is split according to the proposal. If not, the most experienced pirate is thrown off the ship and this process continues with the remaining pirates until a proposal is accepted. The first priority of the pirates is to stay alive and second to maximize the gold they get. Pirate 6 devises a plan which he knows will keep him alive. What is his plan?

Chess Squares

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:34:00 +0000

ow many squares are on a chess board?

9 Minutes

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:34:00 +0000

You are given two hourglasses. One measures 4 minutes and one measures 7 minutes. How would you measure exactly 9 minutes?

Three Switches

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:33:00 +0000

You are standing outside a room next to three switches, all of which are off. Each switch operates a different light bulb in the room. The room door is closed, so you cannot see which switch operates which bulb. You are only allowed to go into the room once. Determine which switch operates which bulb.

Farmer’s Dilemma

noreply@blogger.com (Administrator) — Thu, 15 Apr 2010 22:33:00 +0000

A farmer bought a goat, a wolf and a cabbage from the market. On his way home, he has to cross a river. He has a small boat which only allows him to take one thing with him at a time. The farmer cannot leave the cabbage and the goat together (the goat would eat the cabbage) nor can he leave the goat and the wolf together (the wolf would eat the goat). How does he cross the river without losing any of the things he bought?