Your friend pulls out a perfectly circular table and a sack of quarters, and proposes a game.
"We'll take turns putting a quarter on the table," he says. "Each quarter must lay flat on the table, and cannot sit on top of any other quarters. The last person to successfully put a quarter on the table wins."
He gives you the choice to go first or second. What should you do, and what should your strategy be to win?

You should go first, and put a quarter at the exact center of the table.
Then, each time your opponent places a quarter down, you should place your next quarter in the symmetric position on the opposite side of the table.
This will ensure that you always have a place to set down our quarter, and eventually your oppponent will run out of space.

There are n coins in a line. (Assume n is even). Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.
Would you rather go first or second? Does it matter?
Assume that you go first, describe an algorithm to compute the maximum amount of money you can win.
Note that the strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.
Example 18 20 15 30 10 14
First Player picks 18, now row of coins is
20 15 30 10 14
Second player picks 20, now row of coins is
15 30 10 14
First Player picks 15, now row of coins is
30 10 14
Second player picks 30, now row of coins is
10 14
First Player picks 14, now row of coins is
10
Second player picks 10, game over.
The total value collected by second player is more (20 + 30 + 10) compared to first player (18 + 15 + 14). So the second player wins.

Going first will guarantee that you will not lose. By following the strategy below, you will always win the game (or get a possible tie).
(1) Count the sum of all coins that are odd-numbered. (Call this X)
(2) Count the sum of all coins that are even-numbered. (Call this Y)
(3) If X > Y, take the left-most coin first. Choose all odd-numbered coins in subsequent moves.
(4) If X < Y, take the right-most coin first. Choose all even-numbered coins in subsequent moves.
(5) If X == Y, you will guarantee to get a tie if you stick with taking only even-numbered/odd-numbered coins.
You might be wondering how you can always choose odd-numbered/even-numbered coins. Let me illustrate this using an example where you have 6 coins:
Example
18 20 15 30 10 14
Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.
Since the sum of even coins is more, the first player decides to collect all even coins. He first picks 14, now the other player can only pick a coin (10 or 18). Whichever is picked the other player, the first player again gets an opportunity to pick an even coin and block all even coins.

There are 5 pirates in a ship. Pirates have hierarchy C1, C2, C3, C4 and C5. C1 designation is the highest and C5 is the lowest.
These pirates have three characteristics:
a. Every pirate is so greedy that he can even take lives to make more money.
b. Every pirate desperately wants to stay alive.
c. They are all very intelligent.
There are total 100 gold coins on the ship. The person with the highest designation on the deck is expected to make the distribution. If the majority on the deck does not agree to the distribution proposed, the highest designation pirate will be thrown out of the ship (or simply killed). 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?

To understand the answer,we need to reduce this problem to only 2 pirates. So what happens if there are only 2 pirates. Pirate 2 can easily propose that he gets all the 100 gold coins. Since he constitutes 50% of the pirates, the proposal has to be accepted leaving Pirate 1 with nothing.
Now let's look at 3 pirates situation, Pirate 3 knows that if his proposal does not get accepted, then pirate 2 will get all the gold and pirate 1 will get nothing. So he decides to bribe pirate 1 with one gold coin. Pirate 1 knows that one gold coin is better than nothing so he has to back pirate 3. Pirate 3 proposes {pirate 1, pirate 2, pirate 3} {1, 0, 99}. Since pirate 1 and 3 will vote for it, it will be accepted.
If there are 4 pirates, pirate 4 needs to get one more pirate to vote for his proposal. Pirate 4 realizes that if he dies, pirate 2 will get nothing (according to the proposal with 3 pirates) so he can easily bribe pirate 2 with one gold coin to get his vote. So the distribution will be {0, 1, 0, 99}.
Smart right?
Now can you figure out the distribution with 5 pirates? Let's see. Pirate 5 needs 2 votes and he knows that if he dies, pirate 1 and 3 will get nothing. He can easily bribe pirates 1 and 3 with one gold coin each to get their vote. In the end, he proposes {1, 0, 1, 0, 98}. This proposal will get accepted and provide the maximum amount of gold to pirate 5.

One day, Emperor Akbar posed a question to Birbal. He asked him what Birbal would choose if he offered either justice or a gold coin.
"The gold coin," said Birbal without hesitation.
On hearing this, Akbar was taken aback. "You would prefer a gold coin to justice?" he asked, not believing his own ears.
"Yes," said Birbal.
The other courtiers were amazed by Birbal's display of idiocy. They were full of glee that Birbal had finally managed himself to do what these courtiers had not been able to do for a long time - discredit Birbal in the emperor's eyes!
"I would have been disappointed if this was the choice made even by my lowliest of servants," continued the emperor. "But coming from you it's not only disappointing, but shocking and sad. I did not know you were so debased!"
How did Birbal justify his answer to the enraged and hurt Emperor?

"One asks for what one does not have, Your Majesty." said Birbal, smiling gently and in quiet tones.
"Under Your Majesty´s rule, justice is available to everybody. But I am a spendthrift and always short of money and therefore I said I would choose the gold coin."
The answer immensely pleased the emperor and respect for Birbal was once again restored in the emperor's eyes.

An archeologist claims he found a Roman coin dated 46 B.C. in Egypt. How much should Louvre Museum pay for the coin?
Note: Roman coins can really be found in Egypt

Nothing. That coin is as phony as a three dollar bill. In 46 B.C., they wouldn't have known how many years before Christ it was.

You are blindfolded and 10 coins are place in front of you on table. You are allowed to touch the coins, but can't tell which way up they are by feel. You are told that there are 5 coins head up, and 5 coins tails up but not which ones are which.
How do you make two piles of coins each with the same number of heads up?
You can flip the coins any number of times.

Make 2 piles with equal number of coins. Now, flip all the coins in one of the pile.
How this will work? lets take an example.
So initially there are 5 heads, so suppose you divide it in 2 piles.
Case:
P1 : H H T T T
P2 : H H H T T
Now when P1 will be flipped
P1 : T T H H H
P1(Heads) = P2(Heads)
Another case:
P1 : H T T T T
P2 : H H H H T
Now when P1 will be flipped
P1 : H H H H T
P1(Heads) = P2(Heads)

If you were to put a coin into an empty bottle and then insert a cork into the neck, how could you remove the coin without taking out the cork or breaking the bottle?

Push the cork into the bottle and shake the coin out.