Welcome to The Riddler. Every week, we offer up problems related to the things we hold dear around here: math, logic and probability.

Before we puzzle this week, some Riddler news: This is my final Riddler column as editor. I’m stepping away from FiveThirtyEight to spend a year researching how journalists should investigate and write about advanced artificial intelligence. Algorithms and robots, beware.

Editing this column for the past four years has been a pure joy, thanks entirely to you, its readers. You submit the innovative puzzles, you labor over the elegant solutions and you power this whole inventive mathematical enterprise — the exciting, challenging and educational numerical community that has become known as Riddler Nation. I have been merely a curator of your quantitative creations. Thank you.

And fear not. Riddler Nation shall persist.

I’m thrilled to introduce FiveThirtyEight’s new puzzle editor, Zach Wissner-Gross.

Zach studied physics and biology at MIT before completing a doctorate in physics at Harvard. These days he leads a team that’s developing a K–12 math curriculum, and by night he’s a proud member of Riddler Nation. I’m excited for his leadership of this column and community.

Regular readers of this column will already recognize the contributions Zach has made as a citizen of Riddler Nation. He’s authored puzzles about robot pizza cutters, misanthropic settlers and swindling car salesmen, and his many beautiful solutions have tackled alien invasions, railroad construction and martini glass spillage, to name but a few. Who knows where he’ll take us next?

You can find Zach on Twitter, and do be sure to email him all your favorite puzzles. As for me, I may no longer be editing but I will surely be reading — and solving. (Or, at least, always trying to do the latter.)

## This week’s Riddler

For a long time, my favorite Riddler has been the Battle for Riddler Nation, first suggested a couple of years ago by Joel Baker. It brings together all of what makes this column what it is: a fantastical story, mathematics, large-scale interaction, the power of computer simulation, game theory and fierce (yet good-natured) competition. So let us battle once more. Here are the rules:

In a distant, war-torn land, there are 10 castles. There are two warlords: you and your archenemy. Each castle has its own strategic value for a would-be conqueror. Specifically, the castles are worth 1, 2, 3, …, 9, and 10 victory points. You and your enemy each have 100 soldiers to distribute, any way you like, to fight at any of the 10 castles. Whoever sends more soldiers to a given castle conquers that castle and wins its victory points. If you each send the same number of troops, you split the points. You don’t know what distribution of forces your enemy has chosen until the battles begin. Whoever wins the most points wins the war.

Submit a plan distributing your 100 soldiers among the 10 castles. Once we receive all your battle plans, we’ll adjudicate all the possible one-on-one matchups. Whoever wins the most wars wins the battle royale and is crowned king or queen of Riddler Nation!

For your strategizing benefit, you may consult Riddler Nation’s chosen attack distributions from the first, second and third sanctionings of this battle royale, which had the same rules. A summary of them is in the table below. Adapt and conquer!

##### Troop deployment strategies

How Riddler Nation distributed its forces in each battle royale

Average number of soldiers sent to castle … | ||||||||||
---|---|---|---|---|---|---|---|---|---|---|

Battle | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

May 2019 | 2 | 3 | 4 | 7 | 9 | 12 | 13 | 17 | 17 | 16 |

May 2017 | 3 | 4 | 6 | 8 | 10 | 12 | 15 | 17 | 14 | 12 |

Feb. 2017 | 3 | 3 | 4 | 7 | 9 | 13 | 16 | 19 | 16 | 11 |

Finally, it should be noted that your work is royally cut out for you. There is currently a *two-time* defending champion, Vince Vatter. Can anyone dethrone the king?

## Solution to last week’s Riddler Express

Congratulations to 👏 Stephen Wilson 👏 of Nashville, Tennessee, winner of last week’s Riddler Express!

Last week found you employed as an expert counterfeiter, specializing in forging $100 bills. You knew from experience that the bank could only spot your fakes 25 percent of the time, and that trying to deposit only counterfeit bills would be a ticket to jail. However, if you combined fake and real notes, there was a chance the bank will accept your money. You had $2,500 in real hundreds and an unlimited supply of counterfeits. The bank scrutinized cash deposits carefully: They randomly selected 5 percent of the notes they received, rounded up to the nearest whole number, for close examination. If they identified any note in a deposit as fake, they would confiscate the entire sum, leaving you only enough time to flee. How many fake notes should you have added to the $2,500 in order to maximize the expected value of your bank account? How much free money were you likely to make?

You should add **55 fake notes** to your 25 real notes for a total deposit of $8,000. Your expected gain is **$1,256**.

The full counterfeit solution (counterfeit *bills*, that is … the solution itself is very real) comes to us from this puzzle’s submitter, Jason Ash:

Your expected gain is the weighted average of expected profit from successful deposits and expected losses from bank seizures. With 55 fake notes, there is a 47 percent chance — a combination of the chances of the bank selecting our fake notes for examination and of the examination revealing any fake notes — we avoid detection and collect a profit of $5,500, the value of the fake notes we were able to sneak into circulation. (Remember we started with $2,500, so only the fake notes count as profit.) There is a 53 percent chance we are caught by the bank and lose the $2,500 in real dollars we used as decoys. Therefore, our expected profit is (0.47)(5,500) – (0.53)(2,500) = 1,256.

How can we be sure no other strategy produces a higher profit? For example, suppose we combine 30 fake notes with 25 real notes instead. The bank will select three notes for its audit — that is, 5 percent of 55, rounded up to the nearest whole number. Depending on our luck, the bank could choose all three fake notes, all three real notes, or some combination in between.

Let’s assume the bank randomly chooses two fake notes and one real note for its audit. This occurs with roughly 41.5 percent probability, given by the formula (3 choose 2)(30/55)(29/54)(25/53). Each fake note is detected 25 percent of the time, which means at least one fake note from the pool of two is detected \(1-0.75^2\) = 43.75 percent of the time. We use similar logic to solve for the likelihood and detection rate of the audits with three fake, one fake and zero fake notes. The overall detection rate is equal to the weighted average across each potential audit.

For the 30 fake, 25 real strategy, the probability of success is 64.3 percent — again, we need to either avoid their selecting our fake notes for examination or, if they do select them, avoiding identification of them as such — and the probability of detection is 35.7 percent. Therefore, the expected profit is (0.643)(3,000) – (0.357)(2,500) = 1,038. That’s a decent payday, but we can do better. The chart below shows the profit for strategies with up to 200 fake notes, and it illustrates that the maximum is achieved when we use 55 of them.

We can see two patterns above. First, as we move to the right, we enter the “greedy danger zone,” in which the bank becomes more likely to discover our fraud and seize the starting capital, resulting in larger and larger expected losses. Second, we see “sawtooth” behavior caused by the bank’s practice of auditing 5 percent of deposited notes. The effect is significant: 55 is the ideal answer because if we deposit 80 total notes, the bank will audit four. If we use 56 fake notes for a total of 81, then the bank audits five notes instead, which cuts the expected gain in half!

A life of crime only pays if you’re good with numbers. But with great power comes great responsibility, Riddler Nation.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Jake Wiley 👏 of Wilmington, Delaware, winner of last week’s Riddler Classic!

Last week also found you employed in your second job as the resident mathematician of the Puzzling Pizza pizzeria. Your boss had purchased the equipment necessary to design the machine below.

Your job was to calculate the exact path and flow rate the sauce-dispensing arm should use to fill a 12-inch circular pizza with sauce as fully and evenly as possible. The dough sat on a platform rotating at a constant speed, and the arm traveled in a straight line across the pizza, distributing sauce in a circular shape with a 0.5-inch diameter. Once the arm started pouring sauce, it couldn’t stop until the pizza is covered, and the sauce can only be poured onto the dough. At the end, the pizza was meant to have an even layer of sauce on as much of its surface as possible. What path and flow rate should the sauce-dispensing arm take to give you the best pizza?

The full solution comes to us from this puzzle’s submitter, Tyler Barron:

Let’s start with the path the arm would take. The best method is to start at the middle and move outwards at a constant velocity until you reached the end. (It’s also possible to travel this route from the outside in.) This shape it would create is known as an Archimedean spiral. Saucing a pizza like so would give you a path that looks like this — the spiral stops where it does simply because we’ve run out of room for more sauce:

This curve should move outward at a rate of 0.5 inches per rotation for 23.5 rotations, until the arm is 11.75 inches out. To introduce some formal notation to our pizzeria, this can be written as \(r(\theta)=\frac{0.5}{2\pi}(\theta)\).

Finding the flow rate is more difficult. To start, let’s get a baseline of what our sauce would look like if we distributed it at a constant rate through the entire path. If we did this, we would get a pizza that looks like this:

As you can see, the sauce is thinner on the edges and thicker in the center. Why? Imagine the final rotation where the arm travels around almost the entire circumference, compared to one earlier where it travels through the middle of the pizza. The distance traveled per rotation decreased but the amount of sauce per rotation stayed the same, so we’re left with a higher density of pizza sauce on the inner ring.

Therefore, we need to adjust the flow of sauce to the speed of the pizza traveling under the arm. This can be found by taking the derivative of the arc length equation. That beast — specifically, the Archimedean spiral arc length equation — is written mathematically as \(s(\theta)=\frac{b}{2}(\theta \sqrt{1+\theta^2}+\sinh^{-1}(\theta))\). Its derivative gives us the rate that the length changes as we rotate, and that is \(\frac{ds}{d\theta}=b\sqrt{1+\theta^2}\).

Finally (we promise) this can be converted to the relative flow rate by dividing the derivative by the rate of change at the edges, giving you the saucer’s arm speed compared to its maximum speed. That formula is: \(F(\theta)=\frac{b\sqrt{1+\theta^2}}{b\sqrt{1+(47\pi)^2}}=\frac{\sqrt{1+\theta^2}}{147.66}\).

That gives us a (nearly linear) increasing sauce-flow rate that looks like so. As the dough moves through its rotations, we increase the flow from almost nothing at the center to full strength at the edge:

Using this flow rate will give you an even spread of sauce on the pizza, which means we should program the pizza maker to use a path of \(r(\theta)=\frac{0.5}{2\pi}(\theta)\) and a relative flow rate of \(F(\theta)=\frac{\sqrt{1+\theta^2}}{147.66}\). The final result is the rather delicious-looking image below.

Now *that’s* a pizza Archimedes would be proud of.

## Want more riddles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

## Want to submit a riddle?

Email Zach Wissner-Gross (!) at riddlercolumn@gmail.com.