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]

Vehicular trouble

This week’s Riddler Classic is about steady-state mixing of fluids. Here is the paraphrased problem.

Your old van holds 12 quarts of transmission fluid. At the moment, all 12 quarts are “old.” But changing all 12 quarts at once carries a risk of transmission failure. Instead, you decide to replace the fluid a little bit at a time. Each month, you remove one quart of old fluid, add one quart of fresh fluid and then drive the van to thoroughly mix up the fluid. Unfortunately, after precisely one year of use, what was once fresh transmission fluid officially turns “old.” You keep up this process for many, many years. One day, immediately after replacing a quart of fluid, you decide to check your transmission. What percent of the fluid is old?

Here is my solution:
[Show Solution]

Inscribed hexagons

This week’s Riddler Classic is a geometry problem involving inscribed hexagons.

The larger regular hexagon in the diagram below has a side length of 1. What is the side length of the smaller regular hexagon?
If you look very closely, there are two more, even smaller hexagons on top. What are their side lengths?

Here is my solution:
[Show Solution]

Optimal Wordle

This week’s Riddler Classic is about the viral word game Wordle.

Find a strategy that maximizes your probability of winning Wordle in at most three guesses.

Here is my solution:
[Show Solution]