### #1 - Monty Hall Problem Puzzle

The host of a game show, offers the guest a choice of three doors. Behind one is a expensive car, but behind the other two are goats.
After you have chosen one door, he reveals one of the other two doors behind which is a goat (he wouldn't reveal a car).

Now he gives you the chance to switch to the other unrevealed door or stay at your initial choice. You will then get what is behind that door.

You cannot hear the goats from behind the doors, or in any way know which door has the prize.

Should you stay, or switch, or doesn't it matter?

You better switch!

Your first choice has a 1/3 chance of having the car, and that does not change.
The other two doors HAD a combined chance of 2/3, but now a Goat has ben revealed behind one, all the 2/3 chance is with the other door.

### #2 - Logic Interview Question

The first box has two white balls. The second box has two black balls. The third box has a white and a black ball.

Boxes are labeled but all labels are wrong!

You are allowed to open one box, pick one ball at random, see its color and put it back into the box, without seeing the color of the other ball.

How many such operations are necessary to correctly label the boxes?

Just One!

Because we know all labels are wrong.
So the BW box must be either BB or WW. Selecting one ball from BW will let you know which.
And the other two boxes can then be worked out logically.

### #3 - Measuring Time Interview Problem

I possess two wires. Both of them have an inconstant thickness but both of them burns completely in sixty minutes. The problem is that I want to measure forty-five minutes while using these two wires.

How can I measure if cutting the wire in half is not possible due to non-homogeneous construction ?

45minutes

It is not as hard as it seems. I will burn one wire on both ends and the other wire at one end only. The first wire will burn completely in thirty minutes and at that very moment, I will burn the other end of the second wire and it will burn in fifteen minutes. Thus both of the wires will be burnt in 30 + 15 = 45 minutes.

### #4 - Tricky Probability Interview Puzzle

I have two coins.
* One of the coin is a faulty coin having tail on both side of it.
* The other coin is a perfect coin (heads on side and tail on other).

I blind fold myself and pick a coin and put the coin on table. The face of coin towards the sky is tail.

What is the probability that other side is also tail ?

2/3

2 possible scenario are :
case number side shown other side
1 A1 (H) A2 (H)
2 A2 (H) A1 (H)
3 B1 (H) B2 (T)
4 B2 (T) B1 (H)

case 4 is not possible so ans is 2/3

### #5 - Weighing Balance Puzzle

You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg.

For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

### #6 - Hard River Crossing Logic Puzzle

You must have heard of so many river crossing riddles however this one is a bit tricky one. We have a dysfunctional family on one side of the river which includes mom and 2 daughters, dad and 2 sons, a maid and a dog. Like usual, there is a boat that can hold only two persons at a time (dog counts as one person as well). Obviously, the kids canâ€™t operate the boat and we need an adult for that task.

Here comes the difficulties. The maid must remain with the dog so she can control it or it will head up for a violent biting. The dad cannot be left with the daughters without mom and nor can the mother be left alone with the sons without dad.

Can you help them cross the river?

Suppose that everyone is standing at A corner and they have to reach the B corner

Let the housemaid and Dog go to corner B first and housemaid returns back to corner A.

Next, the housemaid and the first son go to B and she comes back along with the dog.

Now the father and the second son go to B and father comes back.

Mother and Father go to B and the mother comes back.

Housemaid and Dog go to B and the father comes back.

Father and mother goes to B and Mother returns back.

Mother and first daughter go towards B, Housemaid and Daughter returns back.

Housemaid and second Daughter go to B and Housemaid comes back.

Housemaid and Dog goes to B.

So, everyone has reached the other side of the river successfully.

### #7 - Infosys Logic Interview Question

This was asked in an interview of Infosys.

Interviewer has given me 100 marbles(50 white and 50 black) and two empty boxes.
He then told me that he will leave the room and i need to place all the marbles in two boxes.

And When he come back, he will draw a marble from any of the two box and if the marble is black i will be hired.

Also
* No box can be empty.
* All 100 marbles must be placed in one of the two boxes.

So whats should i have done ?

One black marble in one box, the other 99 marbles in the other box

### #8 - Card Riddle

I have mixed 13 decks of cards.
What is the minimum number of cards must be taken out from the above mixed cards to guarantee at least one 'four of a kind'.

13 x 3 + 1 = 40

Explanation:

The number of decks is irrelevant i.e any number of decks can be mixed and still the answer would be same.
Any card drawn will be a Ace,2,3,4,5,6,7,8,9,10,Jack,Queen or King, so there are 13 possibilities each time a card is drawn.
The fastest and luckiest way would need just 4 cards of the same kind.

The slowest way is our solution as it will guarantee a four of a kind.
i.e draw 3 of each kind =>39 cards will be fetched. now next card will guarantee 4 of a kind.

### #9 - Maths Handshake Puzzle

At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?

12

In general, with n+1 people, the number of handshakes is the sum of the first n consecutive numbers: 1+2+3+ ... + n.
Since this sum is n(n+1)/2, we need to solve the equation n(n+1)/2 = 66.
This is the quadratic equation n2+ n -132 = 0. Solving for n, we obtain 11 as the answer and deduce that there were 12 people at the party.

Since 66 is a relatively small number, you can also solve this problem with a hand calculator. Add 1 + 2 = + 3 = +... etc. until the total is 66. The last number that you entered (11) is n.

### #10 - Friday The 13th Riddle

Many people would think Friday the 13th will be an unlucky day. Is it possible that there is no Friday on 13th through the whole year? How many Fridays at 13th can we have in a year at most? Can you calculate it out?

We can calculate out how many days there will be for the 13th on each month if we count from the beginning of the year (January 1). Then we divide total days by 7 to get the remainders. We also need to consider the leap year. Through the whole year we had all kinds of remainders, from 0 to 6. The minimum of occurence for all the unique remainders was 1. It means that we have at least one Friday on 13th. In a regular year, the best chance you can get 3 Fridays on 13th, which are in February, March and December because the remainders of these 3 months are 2. In a leap year, the best chance you also can get 3 Fridays on 13th, which are in January, April and July because the remainders of these 3 months are 6.