## Friends and Enemies

Show that in each group of 6 people, there are either 3 who know each other, or 3 who do not know each other.

Let’s call the people A, B, C, D, E, F. Person A either knows at least 3 among B, C, D, E, F, or does not know at least 3 among B, C, D, E, F.

Assume the first possibility – A knows B, C, D. If B and C know each other, C and D know each other, or B and D know each other, then we find a group of 3 people who know each other. Otherwise, B, C, and D form a group in which no-one knows the others.

If A doesn’t know at least 3 among B, C, D, E, F, the arguments are the same.

## Fathers, Sons, and Fish

Two fathers and two sons went out fishing. Each of them catches two fish. However, they brought home only six fish. How so?

They were a son, his father, and his grandfather – 3 people in total.

## Burn the Ropes

You have two ropes and a lighter. Each of the ropes burns out in exactly 60 minutes, but not at a uniform rate – it is possible for example that half of a rope burns out in 40 minutes and the other half in just 20. How can you measure exactly 45 minutes using the ropes and the lighter?

First, you light up both ends of the first rope and one of the ends of the second rope. Exactly 30 minutes later the first rope will burn out completely and then you have to light up the other end of the second rope. It will take 15 more minutes for the second rope also to burn out completely, for a total of 30 + 15 = 45 minutes.

## Gambling Problems

One very common type of questions in Quant interviews is Random Walks. In this post, we will start with a simple example of a random walk question and gradually will generalize it, also discussing possible variations.

Problem: Two tennis players, A and B, are at deuce (40-40) during a game of a tennis match. If A is serving and has a 60% probability to win a point, what is the probability that he wins the game?

Remark: The game is won when one of the two players gets a 2-point lead.

Let us denote by P the probability that A wins the game. There are three possibilities after playing the next 2 points in the game:

1. A wins both points and the game. The probability for this is 60% × 60% = 36%.
2. B wins both points and the game. The probability for this is 40% × 40% = 16%.
3. A and B win one point each and they are at the same place they started from. The probability for this is 100% – 36% – 16% = 48%.

Therefore, we can set the equation:

P = 0.36+0.48P \implies 0.52P=0.36 \implies P=9/13 \sim 69\%

Remark: Another approach for solving the problem above is computing the probabilities for A winning the game after 2 points (48%), 4 points (36% × 48%), 6 points (36% × 36% × 48%), etc, and summing them up using the geometric progression formula. However, the recursive approach above is usually the shorter and better choice in problems like this one.