The Monty Hall Show

You are in Monty Hall’s TV show where in the final round the host gives you the option to open one of three boxes and to receive the reward inside. Two of the boxes contain just a penny, while the third box contains $1.000.000. In order to make the game more exciting, after you pick your choice, the rules require the host to open one of the two remaining boxes, such that it contains a penny inside. After that he asks you whether you want to keep your chosen box or to switch it with the third remaining one. What should you do?

This is the so called “Monty Hall” problem. The answer is that in order to maximize your chances of winning $1.000.000, you should switch your box. The reason is that if initially you picked a box with a penny, then after switching you will get a box with $1.000.000. If initially you picked a box with $1.000.000, then after switching you will get a box with a penny. Since in the beginning the chance to get a penny is 2/3, then after switching your chance to get $1.000.000 is also 2/3. If you stay with your current box, then your chance to get $1.000.000 will be just 1/3.

12 Balls, 1 Defective

You have 12 balls, 11 of which have the same weight. The remaining one is defective and either heavier or lighter than the rest. You can use a balance scale to compare weights in order to find which is the defective ball and whether it is heavier or lighter. How many measurement do you need so that will be surely able to do it?

It is easy to see that if we have more than 9 balls, we need at least 3 measurements. We will prove that 3 measurements are enough for 12 balls.

We place 4 balls on each side of the scale. Let balls 1, 2, 3, 4 be on the right side, and balls 5, 6, 7, 8 on the left side.

CASE 1. The scale does not tip to any side. For the second measurement we place on the left side balls 1, 2, 3, 9 and on the right side balls 4, 5, 10, 11.

If the scale again does not tip to any side, then the defective ball is number 12 and we can check whether it is heavier or lighter with our last measurement.

If the scale tips to the left side, then either the defective ball is number 9 and is heavier, or it is number 10/11 and is lighter. We measure up balls 10 and 11 against each other and if one of them is lighter than the other, then it is the defective one. If they have the same weight, then ball 9 is the defective one.

If the scale tips to the right side, the procedure is similar.

CASE 2. Let the scale tip to the left side during the first measurement. This means that either one of the balls 1, 2, 3, 4 is defective and it is heavier, or one of the balls 5, 6, 7, 8 is defective and it is lighter. Clearly, balls 9, 10, 11, 12 are all genuine. Next we place balls 1, 2, 5, 6 on one side and balls 3, 7, 9, 10 on the other side.

If the scale tips to the left, then either one of the balls 1, 2 is defective and it is heavier, or ball 8 is defective and lighter. We just measure up balls 1 and 2 against each other and find out which among the three is the defective one.

If the scale tips to the right, the procedure is similar.

If the scale does not tip to any side, then either the defective ball is 4 and it is heavier, or the defective ball is 8 and it is lighter. We just measure up balls 1 and 4 against each other and easily find the defective ball.

Pinned Men

The following game is played under very specific rules – no pinned piece checks the opposite king. How can White mate Black in 2 moves?

First, White plays f3 and threatens mate with Qxe2. Indeed, blocking with the black rook on d4 will not help, because it will become pinned, which means that the rook on d6 will become unpinned, which will make the bishop on b6 pinned, and that will unpin the knight on c7, resulting in a mate. Below are listed all variations of the game.

1. … Rd5 2. Qxe2#
2. … Bxa5 2. Kc8#
3. … Bxc7 2. Nxc7#
4. … Bxe8 2. Kxe8#
5. … Qxe7+ 2. Kxe7#
6. … Rd2 2. Bxd2#
7. … Rxd6+ 2. Qxd6#

Shark Attack

A man is in the center of a circular field which is encompassed by a narrow ring of water. In the water there is a shark which is swimming four times as fast as the man is running. Can the man escape the field and get past the water to safety?

Yes, he can. Let the radius of the field is R and its center I. First the man should start running along a circle with center I and radius R/4. His angular speed will be bigger than the angular speed of the shark, so he can keep running until gets opposite to it with respect to I. Then he should dash away (in a straight line) towards the water. Since he will need to cover approximately 3R/4 distance and the shark will have to cover approximately 3.14R distance, the man will have enough time to escape.

Ants on a Stick

On the ground there is a stick and 10 ants standing on top of it. All ants have the same constant speed and each of them can travel along the entire stick in exactly 1 minute (if it is left alone). The ants start moving simultaneously straightforward, either towards the left or the right end of the stick. When two ants collide with each other, they both turn around and continue moving in the opposite directions. How much time at most would it take until all ants fall off the stick?

Imagine the ants are just dots moving along the stick. Now it looks looks like all dots keep moving in their initially chosen directions and just occasionally pass by each other. Therefore it will take no more than a minute until they fall off the stick. If any of them starts at one end of the stick and moves towards the other end, then the time it will take for it to fall off will be exactly 1 minute.