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]

Triangle Trek

This week’s Riddler Classic is a problem involving traversing a triangle.

Amare the ant is traveling within Triangle ABC, as shown below. Angle A measures 15 degrees, and sides AB and AC both have length 1.

Amare must:

  • Start at point B.
  • Second, touch a point — any point — on side AC.
  • Third, touch a point — any point — back on side AB.
  • Finally, proceed to a point — any point — on side AC (not necessarily the same point he touched earlier).

What is the shortest distance Amare can travel to complete the desired path?

I solved the problem in two different ways. The elegant solution:
[Show Solution]

And the more complicated solution:
[Show Solution]

Squid game

This week’s Riddler Classic is Squid Game-themed!

There are 16 competitors who must cross a bridge made up of 18 pairs of separated glass squares. Here is what the bridge looks like from above:

To cross the bridge, each competitor jumps from one pair of squares to the next. However, they must choose one of the two squares in a pair to land on. Within each pair, one square is made of tempered glass, while the other is made of normal glass. If you jump onto tempered glass, all is well, and you can continue on to the next pair of squares. But if you jump onto normal glass, it will break, and you will be eliminated from the competition.

The competitors have no knowledge of which square within each pair is made of tempered glass. The only way to figure it out is to take a leap of faith and jump onto a square. Once a pair is revealed — either when someone lands on a tempered square or a normal square — all remaining competitors take notice and will choose the tempered glass when they arrive at that pair.

On average, how many of the 16 competitors will make it across the bridge?

Here is my solution.
[Show Solution]

And here is a much better solution!
[Show Solution]

Evil twin

This week’s Riddler Classic is a pursuit problem with a twist. Here is the problem, paraphrased.

You are walking in a straight line (moving forward at all times) near a lamppost. Your evil twin begins opposite you, hidden from view by the lamppost, as illustrated in the figure below.

Assume your evil twin moves precisely twice as fast as you at all times, and they remain obscured from your view by the lamppost at all times. What is the farthest your evil twin can be from the lamppost after you’ve walked the 200 feet as shown?

Here is my solution.
[Show Solution]

Perfect pursuit

This week’s Riddler Classic is about catching

Hames Jarrison has just intercepted a pass at one end zone of a football field, and begins running — at a constant speed of 15 miles per hour — to the other end zone, 100 yards away.

At the moment he catches the ball, you are on the very same goal line, but on the other end of the field, 50 yards away from Jarrison. Caught up in the moment, you decide you will always run directly toward Jarrison’s current position, rather than plan ahead to meet him downfield along a more strategic course.

Assuming you run at a constant speed (i.e., don’t worry about any transient acceleration), how fast must you be in order to catch Jarrison before he scores a touchdown?

Here is the solution.
[Show Solution]

For a detailed derivation (warning: calculus!) click below.
[Show Solution]