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.

## Riddler Express

From Christopher Kovach, some Election Day math:

The midterm elections are less than 40 days away. FiveThirtyEight’s House forecast gives Democrats an 80 percent chance of winning the House. Our Senate forecast gives Republicans a 70 percent chance of holding the Senate.

Assume that those individual probabilities are correct, and assume (merely for the purposes of this problem) that we know *nothing* else about voters or how the outcomes of those legislative races are likely to interact with one another. (In reality, of course, the outcomes are correlated, but ignore that — this is just a math exercise.)

What is the lowest possible probability of a split Congress — where one party controls one chamber and the other party controls the other? And what is the highest possible probability of a split Congress?

## Riddler Classic

From Ricky Jacobson, rack ‘em robots!

You own a start-up, RoboRackers™, that makes robots that can rack pool balls. To operate the robot, you give it a template, such as the one shown below. (The template only recognizes the differences among stripes, solids and the eight ball. None of the other numbers matters.)

First, the robot randomly corrals all of the balls into the wooden triangle. From there, the robot can either swap the location of two balls *or* rotate the entire rack 120 degrees in either direction. The robot continues performing these operations until the balls’ formation matches the template, and it always uses the fewest number of operations possible to do so.

Using the template given above — a correct rack for a standard game of eight-ball — what is the maximum number of operations the robot would perform? What starting position would yield this? How about the average number of operations?

*Extra credit*: What is the maximum number of operations the robot would perform using *any* template? Which template and starting position would yield this?

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ Thomas Burris ÑÑâÐ of Charlotte, North Carolina, winner of last week’s Riddler Express!

It’s officially fall, and that means the Major League Baseball playoffs are nearly upon us. Huzzah! Last week brought two math questions to that effect: First, given the current playoff format, what is the best possible winning percentage a team can have in the playoffs *without* winning the World Series? And second, what is the worst possible winning percentage a team can have in the playoffs and still *win* the World Series? (For bonus points, how close to either of these extremes has a team come?)

The best possible winning percentage *without* winning a ring is **73.3 percent**. To achieve this dubious honor, enter the playoffs as a wild card and win the wild-card game. Then, sweep the division series 3-0. Then, sweep the league championship series 4-0. Then, lose the World Series in a nailbiter, winning as many games as you can, 4-3. That’s 11 wins and four loses (and 11/15 ≈ 0.733).

The worst possible winning percentage *while* winning a ring is **57.9 percent**. To do this, be good enough in the regular season to skip the wild-card game. Then, eke out the division series 3-2. Then, eke out the league championship series 4-3. Then, eke out the World Series 4-3. After all that eking, you’ve got 11 wins, eight losses and the Commissioner’s Trophy (and 11/19 ≈ 0.579). In baseball, you don’t have to be great to be the greatest.

The 2014 Kansas City Royals accomplished that first, dubious honor, winning a wild-card game against the Athletics, then sweeping the Angels, then sweeping the Orioles, then falling to the Giants in the World Series in seven games. The 2017 Houston Astros nearly completed the second, eke-out feat: They beat the Red Sox 3-1, the Yankees 4-3 and the Dodgers 4-3 in the World Series for a playoff winning percentage of 61.1 percent and a shiny new trophy.

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ Marc Harber ÑÑâÐ of Baltimore, winner of last week’s Riddler Classic!

Last week, you were hired by Riddler HQ and relocated to Riddler City, where you moved into a lovely apartment five blocks west and 10 blocks south of the office. The streets of Riddler City are laid out on a perfect grid, of course, and every morning and evening you walk your commute. Inquisitive such as you were, you preferred to walk a different path through the city every time. Assuming you didn’t take paths that were longer than necessary, how many days could you live in that apartment before you had to take a path twice?

There were 3,003 different routes from apartment to office, giving you **1,501 days** — or a little over four years — of unique walks. Start drafting your 214-week notice now.

Here’s one way to get there: To plot a path from home to work, intersection by intersection, you know you’ll need to go east five times and north 10 times. In other words, we’re looking for the number of different ways you can assemble a string of five “E”s and 10 “N”s. For example, you could walk straight east and then straight north: EEEEENNNNNNNNNN. Or you could walk straight north and then straight east: NNNNNNNNNNEEEEE. Or you could walk some more meandering path in between: NENNNENNENNNNEE, say. There are 15! (1,307,674,368,000) ways to order 15 letters, which we must then divide by 10! (3,628,800) and 5! (120). We need to do this division because it doesn’t matter how the individual “N”s and “E”s are arranged — they are all identical for our purposes, and describe equivalent paths. All together now:\(\)

\begin{equation*}\frac{15!}{10!\cdot 5!} = 3,003\end{equation*}

Solver Tyler Barron created a heatmap of your commute, showing the blocks you’ll visit most frequently over the course of those 3,003 walks.

Put another way, the total number of paths for our commute is the binomial coefficient “10+5 choose 5,” or 3,003. So if we lived M blocks west and N blocks south of the office, our number of paths would be “M+N choose N.” If we moved to a nicer apartment a bit farther away — 15 blocks west and 10 blocks south, say — we’d have 3,268,760 walking paths to choose from.

Gotta get those steps in!

## Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.