Tower of goats

This week’s Riddler classic is a counting problem. Can the goats fit in the tower?

A tower has 10 floors, each of which can accommodate a single goat. Ten goats approach the tower, and each goat has its own (random) preference of floor. Multiple goats can prefer the same floor. One by one, each goat walks up the tower to its preferred room. If the floor is empty, the goat will make itself at home. But if the floor is already occupied by another goat, then it will keep going up until it finds the next empty floor, which it will occupy. But if it does not find any empty floors, the goat will be stuck on the roof of the tower. What is the probability that all 10 goats will have their own floor, meaning no goat is left stranded on the roof of the tower?

My solution:
[Show Solution]

Catch the grasshopper

This week’s Riddler classic is a probability problem about a grasshopper!

You are trying to catch a grasshopper on a balance beam that is 1 meter long. Every time you try to catch it, it jumps to a random point along the interval between 20 centimeters left of its current position and 20 centimeters right of its current position. If the grasshopper is within 20 centimeters of one of the edges, it will not jump off the edge. For example, if it is 10 centimeters from the left edge of the beam, then it will randomly jump to anywhere within 30 centimeters of that edge with equal probability (meaning it will be twice as likely to jump right as it is to jump left). After many, many failed attempts to catch the grasshopper, where is it most likely to be on the beam? Where is it least likely? And what is the ratio between these respective probabilities?

My solution:
[Show Solution]

Desert escape

This week’s Riddler classic is about geometry and probability, and desert escape! Here is the (paraphrased) problem:

There are $n$ travelers who are trapped on a thin and narrow oasis. They each independently pick a random location in the oasis from which to start and a random direction in which to travel. What is the probability that none of their paths will intersect, in terms of $n$?

My solution:
[Show Solution]

Tetrahedral dice game

This week’s Riddler Classic is a game of four-sided dice:

You have four fair tetrahedral dice whose four sides are numbered 1 through 4.

You play a game in which you roll them all and divide them into two groups: those whose values are unique, and those which are duplicates. For example, if you roll a 1, 2, 2 and 4, then the 1 and 4 will go into the “unique” group, while the 2s will go into the “duplicate” group.

Next, you reroll all the dice in the duplicate pool and sort all the dice again. Continuing the previous example, that would mean you reroll the 2s. If the result happens to be 1 and 3, then the “unique” group will now consist of 3 and 4, while the “duplicate” group will have two 1s.

You continue rerolling the duplicate pool and sorting all the dice until all the dice are members of the same group. If all four dice are in the “unique” group, you win. If all four are in the “duplicate” group, you lose.

What is your probability of winning the game?

My solution:
[Show Solution]

Frustrating elevator

This weeks Riddler Express is a problem about a frustrating elevator! Here it goes:

You are on the 10th floor of a tower and want to exit on the first floor. You get into the elevator and hit 1. However, this elevator is malfunctioning in a specific way. When you hit 1, it correctly registers the request to descend, but it randomly selects some floor below your current floor (including the first floor). The car then stops at that floor. If it’s not the first floor, you again hit 1 and the process repeats.

Assuming you are the only passenger on the elevator, how many floors on average will it stop at (including your final stop, the first floor) until you exit?

My solution:
[Show Solution]

Cone crawling

This week’s Riddler Classic is a geometry problem about traversing the surface of a cone

The circular base of the cone has a radius of 2 meters and a slant height of 4 meters. We start on the base, a distance of 1 meter away from the center. The goal is to reach the point half-way up the cone, 90 degrees around the cone’s central axis from the start, as shown. What is the shortest path?

Here is my solution:
[Show Solution]

Visualize the vertex

This week’s Riddler Classic is a neat geometry problem about

Suppose you have two distinct points anywhere on the coordinate plane. If I tell you that a parabola with a vertical line of symmetry passes through those two points, where on the plane could that parabola’s vertex be?

Here is my solution:
[Show Solution]

The luckiest coin

This week’s Riddler Classic is about finding the “luckiest” coin!

I have in my possession 1 million fair coins. I first flip all 1 million coins simultaneously, discarding any coins that come up tails. I flip all the coins that come up heads a second time, and I again discard any of these coins that come up tails. I repeat this process, over and over again. If at any point I am left with one coin, I declare that to be the “luckiest” coin.

But getting to one coin is no sure thing. For example, I might find myself with two coins, flip both of them and have both come up tails. Then I would have zero coins, never having had exactly one coin.

What is the probability that I will at some point have exactly one “luckiest” coin?

Here is my solution:
[Show Solution]

A cube of primes

This week’s Riddler Classic is a question about prime numbers.

Consider a cube, which has eight vertices, or corners. Suppose I assign a prime number to each vertex. A “face sum” is the value I get when I add up all four prime numbers on one of the six faces.

Can you find eight distinct primes and arrange them on a cube so that the six face sums are all equal?

Extra credit: Can you find another set of eight distinct primes that can similarly be arranged on the vertices of a cube? How many more can you find?

Extra Extra credit: Same puzzle for the other four platonic solids.

Here is my solution:
[Show Solution]

Polarization Puzzle

This week’s Riddler Classic is about light polarization.


When light passes through a polarizer, only the light whose polarization aligns with the polarizer passes through. When they aren’t perfectly aligned, only the component of the light that’s in the direction of the polarizer passes through. For example, here is what happens if you use two polarizers, the first at 45 degrees, and the second at 90 degrees. The length of the original vector is decreased by a factor of 1/2.

I have tons of polarizers, and each one also reflects 1 percent of any light that hits it — no matter its polarization or orientation — while polarizing the remaining 99 percent of the light. I’m interested in horizontally polarizing as much of the incoming light as possible. How many polarizers should I use?

Here is my solution:
[Show Solution]