Easy | General |

100 pirates stand in a circle. They start shooting alternately in a cycle such that 1st pirate shoots 2nd, 3rd shoots 4th and so on. They continue in circle till only one pirate is left. Which position should Captain Jack Sparrow stand to survive?

Solution

This is not a puzzle at all, this is apti/maths question, but has become quite common. You can calculate for a N=100, 73rd position is safe. For a general N, find largest m such that 2^m < N. Take t = N-2^m. Then Safe position is 2*t + 1. Please learn this formula. No need to prove it using strong induction. Proof: http://en.wikipedia.org/wiki/Josephus_problem#k.3D2

Source: Written Tests

Enable Like and Comment Easy | Probability |

There is a regular die and a special invisible die. You know that regular die has integers 1 to 6, but don't know what's on the invisible dice. After tossing, I speak the sum of outcome of both die. It so happens that the outcome is an interger between 1 to 12, with equal probability (1/12 each). Can you guess what are the numbers printed on special invisible dice?

Hint

For the sum to be 1, we need atleast 0, and for sum to be 12, what do we need? Can you guess the digits of invisible dice now?

Answer

0 0 0 6 6 6

Solution

Faces of cube have numbers: 0 0 0 6 6 6 . If we compute conditional probability, we get 0 with probabilty (1/2) and 6 with probability (1/2). Hence the sum is 1,2....12 with probability of (1/12) each

Source: Technical Interview

Enable Like and Comment Easy | Probability |

Sheldon says "Suppose I have two children. Younger one is a girl". What is the probability that both children are girls?

"Forget all that, and suppose I have two children, and atleast one of them is a boy". Find probability of two boys.

"Suppose I have two kids, lets call them Bouba and Kiki", says Dr. Cooper, "and suppose Bouba is a girl !" What is the probability that I have two daughters?"

"Forget all that, and suppose I have two children, and atleast one of them is a boy". Find probability of two boys.

"Suppose I have two kids, lets call them Bouba and Kiki", says Dr. Cooper, "and suppose Bouba is a girl !" What is the probability that I have two daughters?"

Answer

1/2, 1/3, 1/3

Solution

First and second are un-ambiguous. Third is debatable as per the interpretation. I believed 1/3, as the names Cooper used were anonymous, and should have given no extra information. But it turns out that Cooper did provide some extra information. He had actually fixed a person (A). Also, calculating the probability in part 3 using conditionals will give 1/2.

Source: Written Test

Enable Like and Comment Easy | General |

In a dark room, there is a deck of 52 cards, with exactly 10 cards facing up, rest facing down. You need to split this into two decks with equal number of cards facing up!

Hint

The question can be solved even when up facing cards are 11. The split decks need not have equal number of up-facing cards.

Solution

Create a deck of k cards randomly from the original 52-card deck, (k=10 here) and then turn over the k-card deck. Goal is achieved!

Source: Quant Interview

Enable Like and Comment Easy | Strategy |

A certain town comprises of 100 married couples. Some husbands secretly cheat on their wives. All wives know about the nature of every husband except their own. When a wife concludes that her husband cheated, she kicks her husband into the street at midnight. All husbands remain silent about their secret. One day, the mayor of the town announces to the whole town that there is at least 1 cheating husband in the town. After announcement, no one talks, waiting for someone to get kicked. Till 9th night from announcement, no husband was kicked, but on the 10th night, some husbands got kicked out simultaneously. How many are they?

Hint

What happens if only one husband cheated?

Solution

It must be 10 husbands kicked out.

If there was only 1 cheating husband in the town, there will be 99 women who know exactly who the cheater is. The 1 remaining woman, who is being cheated on, would have assumed there are no cheaters. But now that the mayor has confirmed that there is at least one cheater, she realizes that her own husband must be cheating on her. So her husband gets kicked on the day of the announcement.

Now let’s assume there are 2 cheaters in the town. There will be 98 women in the town who know who the 2 cheaters are. The 2 wives, who are being cheated on, would think that there is only 1 cheater in the town. Since neither of these 2 women know that their husbands are cheaters, they both do not report their husbands in on the day of the announcement. The next day, when the 2 women see that no husband was kicked, they realize that there could only be one explanation – both their husbands are cheaters. Thus, on the second day, 2 husbands are kicked.

Through induction, it can be proved that when this logic is applied to n cheating husbands, they are all kicked on the n th day after the mayor’s announcement. Hence it must be 10 husbands kicked in our case.

If there was only 1 cheating husband in the town, there will be 99 women who know exactly who the cheater is. The 1 remaining woman, who is being cheated on, would have assumed there are no cheaters. But now that the mayor has confirmed that there is at least one cheater, she realizes that her own husband must be cheating on her. So her husband gets kicked on the day of the announcement.

Now let’s assume there are 2 cheaters in the town. There will be 98 women in the town who know who the 2 cheaters are. The 2 wives, who are being cheated on, would think that there is only 1 cheater in the town. Since neither of these 2 women know that their husbands are cheaters, they both do not report their husbands in on the day of the announcement. The next day, when the 2 women see that no husband was kicked, they realize that there could only be one explanation – both their husbands are cheaters. Thus, on the second day, 2 husbands are kicked.

Through induction, it can be proved that when this logic is applied to n cheating husbands, they are all kicked on the n th day after the mayor’s announcement. Hence it must be 10 husbands kicked in our case.

Source: Common

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