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-sized 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 if you have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

From Steven Pratt, a real-life electoral problem:

In Steven’s hometown, 11 fine folks are running in a primary for three at-large seats on the City Commission. Each voter may vote for up to three candidates. This election will reduce the field of candidates from 11 to six.

- How many different (legal) ways may a voter cast his or her ballot?
- How many different outcomes (excluding ties) are there for who advances to November’s general election?

## Riddler Classic

From Àlex Sierra, a cloak-and-dagger puzzle:

Twice before, I’ve pitted Riddler Nation against itself in a battle royale for national domination. War must wage once more. Here are the rules. There are two warlords: you and your archenemy, with whom you’re competing to conquer castles and collect the most victory points. Each of the 10 castles 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 between 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.) Whoever ends up with the most points wins.

But now, you have a spy! You know how many soldiers your archenemy will send to each castle. The bad news, though, is that you no longer have 100 soldiers — your army suffered some losses in a previous battle.

What is the value of the spy?

That is, how many soldiers do you need to have in order to win, no matter the distribution of your opponent’s soldiers? Put another way: What k is the minimum number such that, for any distribution of 100 soldiers in the 10 castles by your opponent, you can distribute k soldiers and win the battle?

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ Elaine Hou ÑÑâÐ of Tampa, Florida, winner of the previous Express puzzle!

You and your two older siblings are sharing two extra-large pizzas and decide to cut them in an unusual way. You overlap the pizzas so that the crust of one touches the center of the other (and vice versa since they are the same size). You then slice both pizzas around the area of overlap. Two of you will each get one of the crescent-shaped pieces, and the third will get both of the football-shaped cutouts. Which should you choose to get more pizza: one crescent or two footballs?

You’ll get more pizza by eating the **two footballs**.

To show why, solver Sai Rijal began with the following shape in the middle of the pizzas:

Since sides AB, BC, CD, DA, and BD are all radii of one of the circular pizzas, Sai explained, they form two equilateral triangles: ABD and CDB. Because of this, the angles ABC and ADC are each 120 degrees. Therefore, the slice of the red pizza bound by ABC, and the slice of the blue pizza bound by ADC, have the area \((1/3)\pi r^2\) — they’re just one third of each pizza. Let’s remove these 1/3 slices from each of the two pieces of football slices you have. You have 2/3 of a pizza. This is your fair share, because there were two pizzas and three eaters. The remaining segments are extra pizza you have earned through mathematics!

Although it wasn’t necessary for answering the question, you could also go further and calculate the specific areas of the crescents and the footballs. Assume, for simplicity, that each pizza has a radius of 1. It turns out that the two footballs have an area of \(\frac{4\pi -3\sqrt{3}}{3}\), or about 2.46; one crescent has an area of \(\frac{2\pi+3\sqrt{3}}{6}\), or about 1.91. Solver Zack Segel shared his lovely pen and paper work to that end:

GitHub user mimno even created an interactive Monte Carlo simulation of the pizzas, again finding that you’re better off going for the footballs. And others, I’m told, solved the problem by ordering two actual extra-large pizzas. Bon appétit!

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ Neema Salimi ÑÑâÐ of Atlanta, winner of the previous Classic puzzle!

In the National Squishyball league, the owner of the top-seeded team (i.e., you) gets to select the length of the championship series in advance of the first game, so you could decide to play a single game, a best two out of three series, a three out of five series, etc., all the way up to a 50 out of 99 series. The owner of the winning team gets $1 million minus $10,000 for each of the victories required to win the series, regardless of how many games the series lasts in total. Thus, if the top-seeded team’s owner selects a single-game championship, the winning owner will collect $990,000. If he or she selects a four out of seven series, the winning team’s owner will collect $960,000. The owner of the losing team gets nothing. Your team has a 60 percent chance of winning any individual game. How long a series should you select in order to maximize your expected winnings?

You should select a **best-of-25 series**, where the first team to 13 wins takes the title. You stand to win about $736,222 on average.

This problem was most commonly approached computationally, with most solvers turning to Excel to guide them through the postseason strategizing. Solver Andrew Hoffman explained how he built such a spreadsheet:

At any given point in the series, each team has a certain number of games left that they need to win in order to win the series (call this number A for Acme and B for Boondocks). They start with the same number. After each game, there’s a 60 percent chance A decreases by 1 and a 40 percent chance B decreases by 1. If either team has 0 games to go, that team has won. You can then build a table recursively for the win probability at cell (A, B), which we’ll call P(A, B) = 0.6 * P(A-1, B) + 0.4 * P(A, B-1). For example, P(2, 1) = 0.6 * P(1, 1) + 0.4 * P(1, 0) = 0.6 * (0.6 * P(0, 1) + 0.4 * P(1, 0)) + 0.4 * 0 = 0.6 * (0.6 * 1 + 0.4 * 0) = 0.6 * 0.6 = 0.36. Build the table through P(50, 50). Then for each N, multiply P(N, N) by the winnings if N games are required: 1,000,000 – 10,000N. This gives the expected winnings for Acme for each N. The maximum value of $736222.04 occurs at N = 13, referring to a 13-out-of-25 series.

Andrew also submitted his graphical results of this project, showing that the expected value (EV) peaks at a series requiring 13 wins.

Others, such as Tyler Barron, Chris Ketelsen and Justin Brookman, turned to Python code, and you can find their alternate solutions above.

## Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.