Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,^{1} and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Quick announcement: Have you enjoyed the puzzles in this column? If so, I’m pleased to tell you that we’ve collected many of the best, along with some that have never been seen before, in a real live book! It’s called “The Riddler,” and it will be released in October — just in time for loads of great holidays. It’s a physical testament to the mathematical collaboration that you, Riddler Nation, have helped build here, which in my estimation is the best of its kind. So I hope you’ll check out the book, devour the puzzles anew, and keep adding to our nation by sharing the book with loved ones.

And now, to this week’s puzzles!

## Riddler Express

From Brett Andersen, some numismatic numerology:

I was recently traveling in Europe and struck by the number of coins the euro uses. They have 2 euro, 1 euro, 50 cent, 20 cent, 10 cent, 5 cent, 2 cent and 1 cent coins. This got me thinking: If Riddler Nation needed to make change (anywhere from 0.01 to 0.99) and was establishing its own mint, what values of coins would be ideal to yield the smallest number of coins in any transaction? When picking values, let’s say we’re ditching the Europeans and limiting our mint to four different coin denominations — replacing the current common American ones of penny, nickel, dime and quarter.

## Riddler Classic

From Mike Donner, a puzzling piece of wartime naval navigation:

An enemy submarine determined to sink your ship is sitting, torpedoes armed, exactly halfway between you and your home port. The submarine submerges to some fixed depth, and your ship now has no further information about its position. The enemy sub has to be directly underneath your ship to sink it — but the sub can track your moves with precision and respond efficiently. If your ship is fast enough, though, you will be able to set a wide course around the sub and reach port safely.

How much faster than the sub does your ship have to be to guarantee you can avoid the sub and get home?

## Solution to last week’s Riddler Express

Congratulations to 👏 David Diamondstone 👏 of San Francisco, winner of last week’s Riddler Express!

Alas, the World Cup is over, but luckily for us we can still puzzle over its mathematics. Last week, you were asked what the chances were that a penalty shootout went past each team’s fifth kick, assuming a 75 percent success rate for each shot.

Those chances are about **29 percent**.

One good way to get to the answer is to break the problem down into a few smaller parts. Specifically, break it into the chances that the teams tie the shootout 0-0, 1-1, … , 5-5, and add up those chances.

The chances that they tie 0-0, for example, are the chances that each team misses every shot, or \(\left(0.25^5\right)^2\), or about 1 in a million. The chances that they tie 5-5, similarly, are \(\left(0.75^5\right)^2\), or nearly 6 in 100.

Calculating the other possibilities — 1-1, 2-2, 3-3 and 4-4 — is only slightly trickier, and we must insert one extra term into the calculations. The chances of a 1-1 draw, for example, are \(\left((0.25^4 \cdot 0.75^1){5 \choose 1}\right)^2\) — each team must miss four shots and make one, and there are “5 choose 1” ways to select the shot that is made. Similarly, the chances of a 2-2 draw are \(\left((0.25^3 \cdot 0.75^2){5 \choose 2}\right)^2\) — each team must miss three shots and make two, and there are “5 choose 2” ways to select the shots that are made.

Finally, we can add up all the chances of a draw after five kicks: 0.290203, or about 29 percent.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Sam Reicks 👏 of Pella, Iowa, winner of last week’s Riddler Classic!

Last week brought a simple but tricky challenge: Given two rectangles of any size stuck together to form an “L” shape like the ones shown below, could you draw a straight line that cut that L perfectly in half, using only a straightedge (no rulers, protractors or compasses allowed)?

Indeed you could, and here’s how. As this puzzle’s submitter, Adam Wagner, explained and illustrated in the images below, we do know how to slice an individual rectangle in half: We simply draw any line that passes through its center. So let’s reframe the puzzle in terms of individual rectangles.

First, complete the “L” into one big rectangle by drawing a couple of imaginary line segments along the edges, like so^{2}:

Any line that passes through the center of this new large rectangle cuts it in half. Also, any line that passes through the center of the new little white negative-space rectangle cuts that in half. And — crucially for us — the line that passes through both of these center points cuts in half both the small rectangle *and* the big rectangle that contains it. Therefore, it must also cut in half all the “rest of the stuff” — i.e., our original L.

We can find the center of the big rectangle like so^{3}:

And the center of the little rectangle like so^{4}:

Then we connect these center points on a line, and we’re done! The bold line is the bisector of the L, and our final answer^{5}:

Solver Laurent Lessard also noted that this bisecting strategy “works for *any* shape that is the sum or difference of two rectangles! So it’s not restricted to L-shapes.” Moreover, it also works with parallelograms, and even in three dimensions!

Happy bisecting.

## Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.