Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the 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. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

From Peter Mowrey comes an elegant electoral enigma:

Riddler Township is having its quadrennial presidential election. Each of the town’s 10 “shires” is allotted a certain number of electoral votes: two, plus one additional vote for every 10 citizens (rounded to the nearest 10).

The names and populations of the 10 shires are summarized in the table below.

##### Which way will Riddler Township swing?

Shire | Population | Electoral votes |
---|---|---|

Oneshire | 11 | 3 |

Twoshire | 21 | 4 |

Threeshire | 31 | 5 |

Fourshire | 41 | 6 |

Fiveshire | 51 | 7 |

Sixshire | 61 | 8 |

Sevenshire | 71 | 9 |

Eightshire | 81 | 10 |

Nineshire | 91 | 11 |

Tenshire | 101 | 12 |

As you may know, under this sort of electoral system, it is quite possible for a presidential candidate to lose the popular vote and still win the election.

If there are two candidates running for president of Riddler Township, and every single citizen votes for one or the other, then what is the *lowest* percentage of the popular vote that a candidate can get while still winning the election?

## Riddler Classic

Riddler Pinball is a game with an infinitely long wall and a circle whose radius is 1 inch and whose center is 2 inches from the wall. The wall and the circle are both fixed and never move. A single pinball starts 2 inches from the wall and 2 inches from the center of the circle.

To play, you flick the pinball toward a spot of your choosing along the wall, specified by its distance *x* from the point on the wall that’s closest to the circle, as shown in the diagram below.

The goal of the game is simple: Get the ball to bounce as many times as possible.

(Note: This is a geometry problem, not a physics problem. In other words, assume the system is frictionless and that all collisions are perfectly elastic.)

Let’s take a look at some games to see how they play out.

If you aim too far to the right (i.e., your value of *x* is too small), the pinball will quickly bounce its way through the gap between the circle and the wall. That’s what happened in the game below, when *x* was 0.75 inches, resulting in a rather pedestrian four bounces.

But if you aim too far to the left (i.e., your value of *x* is too big), the pinball will quickly come back out the same side it went in. That’s what happened in the next game, when *x* was 0.9 inches, again yielding just four bounces.

As you can see, Riddler Pinball is an unforgiving game — the slightest error can tank your chances of victory. But if you strategize *just* right, it’s possible to do quite well.

What’s the greatest number of bounces you can achieve? And, more importantly, what value of *x* gets you the most bounces?

## Solution to last week’s Riddler Express

Congratulations to 👏 Rick Schubert 👏 of San Diego, California, winner of last week’s Riddler Express.

Last week, you set your sights on breaking baseball records, albeit in a shortened season. Your true batting average was .350, meaning you had a 35 percent chance of getting a hit with every at-bat. If you had four at-bats per game, what were your chances of batting at least .400 over the course of the 60-game season?

In a 60-game season, with four at-bats per game, there were 240 total at-bats. Forty percent of 240 was 96, so to bat at least .400 you needed at least 96 hits. Since each at-bat was independent, you could use the binomial distribution to determine the probability of each number of hits.

For example, the probability of getting exactly 96 hits in 240 at-bats was equal to 0.35^{96} (i.e., getting a hit in 96 at-bats) times 0.65^{144} (i.e., *not* getting a hit in the remaining 144 at-bats), times the number of ways you can order 96 hits among 240 at-bats (240 choose 96).

Again, that was just the probability of getting *exactly* 96 hits. To find the probability of *at least* 96 hits, you had to add up the probabilities of getting 96 hits, 97 hits, 98 hits, and so on, up to (the very unlikely) case of 240 hits. Alternatively, you could have added up the probabilities of getting between zero and 95 hits, and then subtracted this from 1. Whether you did this by hand, by spreadsheet or by computer code, the correct answer was **6.1 percent**.

It turned out that it was 16 times easier to bat .400 over 60 games than it was over a full slate of 162 games, where the probability was just **0.38 percent**.

The puzzle’s submitter, Taylor Firman, extended this further, looking at the probability of *anyone in the league* batting .400, based on each player’s own career batting average. For a 60-game season, Taylor found that this probability was about 3.4 percent, while it was just 0.002 percent for a 162-game season. So if no one catches Ted Williams this year, forget about it.

For extra credit, you were asked to find your chances of getting a hit in at least 56 consecutive games within the 60-game season, tying or breaking Joe DiMaggio’s record. It was helpful to first calculate the probability of keeping the streak alive — that is, getting at least one hit in a game. Again using the binomial distribution, this probability was 82.15 percent. But while it was likely you’d get a hit in any given game, pulling that off 56 times *in a row* was another story.

As Mark Girard explained, there were several distinct ways to achieve a streak of at least 56 games:

- Getting a hit in games 1 to 56
- Not getting a hit in game 1 and getting a hit in games 2 to 57
- Not getting a hit in game 2 and getting a hit in games 3 to 58
- Not getting a hit in game 3 and getting a hit in games 4 to 59
- Not getting a hit in game 4 and getting a hit in games 5 to 60

Each of these five cases was very unlikely. And together, they were still very unlikely. Overall, your chances of having a 56-game hitting streak in a 60-game season stood at just **0.0028 percent**. And remember, that was assuming you were a lifetime .350 hitter! Over a 162-game season, these chances improved by about tenfold, to **0.033 percent**. It would appear that Joe DiMaggio’s record is quite safe.

Finally, for *extra* extra credit, you had to find your chances of both batting at least .400 *and* getting a hit in at least 56 games. This was especially tricky, since these events were *not* independent. In other words, if you hit .400, you were more likely to have a very long hitting streak, and vice versa.

Paul Wright worked it out analytically, starting with streaks of 56, 57, 58, 59 or 60 games, each of which had their own corresponding probabilities of occurring. Then, for each streak, you knew that the games *within* the streak had one, two, three or four hits, while the remaining handful of games could also have zero hits. Putting the streak and non-streak games together, the probability of batting at least .400 *and* having a hitting streak of at least 56 games was about **0.0027 percent**. This was very close to the answer for the extra credit, meaning If you tied or broke Joe DiMaggio’s record in 60 games, then it was *very* likely that you’d also bat at least .400.

Finally, Angela Zhou calculated your chances of reaching both milestones over a 162-game season. It wasn’t likely.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Jim Boyce 👏 of Kensington, Connecticut, winner of last week’s Riddler Classic.

Last week, the tortoise and the hare were about to begin a 10-mile race along a “stretch” of road. The tortoise was driving a car that traveled 60 miles per hour, while the hare was driving a car that traveled 75 miles per hour. (For the purposes of this problem, you were asked to assume that both cars accelerated from 0 miles per hour to their cruising speed instantaneously.)

The hare did a quick mental calculation and realized if it waited until two minutes had passed, they would cross the finish line at the exact same moment. And so, when the race began, the tortoise drove off while the hare patiently waited.

But one minute into the race, after the tortoise had driven 1 mile, something extraordinary happened. The road turned out to be magical and instantaneously stretched by 10 miles! As a result of this stretching, the tortoise was now *2* miles ahead of the hare, who remained at the starting line.

At the end of every subsequent minute, the road stretched by 10 miles. With this in mind, the hare did some more mental math.

How long after the race began should the hare have waited so that both the tortoise and the hare crossed the finish line at the same exact moment?

At first, it might have seemed like neither the tortoise nor the hare would ever finish the race. How could they, when they drove a mile or so per minute, while the road stretched *10* miles per minute? The key was to realize that the road stretched *uniformly*, which meant it could carry each car along as it stretched.

To better understand this, let’s take a closer look at the tortoise over time. After one minute, it traveled 1 mile, or 10 percent of the total distance. Then, the road stretched to a length of 20 miles. But because the stretching was uniform, the tortoise was still 10 percent of the way across, meaning it was 2 miles down the road. After another minute, it was 3 miles down the road. Then, after the road stretched to a length of 30 miles, the tortoise was 4.5 miles down the road.

But rather than focusing on the distance the tortoise traveled, you should have zeroed in on the *fraction* of the total distance it covered each minute. In the first minute, the tortoise drove 1/10 of the total distance. In the second minute, it drove 1/20. In the third minute, it drove 1/30, In the fourth minute, it was 1/40. After *N* minutes, it drove 1/10 · (1/1 + 1/2 + 1/3 + 1/4 + … + 1/*N*). As many solvers noticed, that sum inside the parentheses is none other than the *N*^{th} harmonic number! Once this sum exceeded 10, the tortoise was guaranteed to have finished the race. This happened 12,367 minutes — more than eight days — into the competition.

Calculating this was no small feat. But alas, much like the idling hare, you still had your work cut out for you.

The tortoise didn’t finish *exactly *12,367 minutes into the race — the sum of the first 12,367 terms of the harmonic series is in fact *greater than* 10. The precise time turned out to be approximately 12,366.47 minutes.

But as solvers Hector Pefo and Josh Silverman cleverly observed, you didn’t *need* to calculate this exact time. As long as the hare started running the moment the tortoise completed 20 percent of the race, they’d finish together. That’s because the hare would then travel a distance that was 25 percent longer over the same amount of time, perfectly balancing the fact that the hare traveled 25 percent faster.

At this point, you just needed to determine when the tortoise had finished 20 percent of the race. After just four minutes, when the race was 40 miles long, the tortoise had traveled about 8.33 miles — just a shade over 20 percent of the way. Since the tortoise drove 1 mile per minute, it must have passed the 20 percent mark a third of a minute (i.e., 20 seconds) prior. And so the hare took off exactly **3 minutes and 40 seconds** into the race.

You can see this epic tie in all its glory, courtesy of Allen Gu:

Solver Hypergeometricx extended the puzzle further, looking at what would happen if the road stretched *continuously* over time, rather than discretely every minute, finding that the hare would then wait *e*^{2}−1 minutes (about 6 minutes and 23 seconds) into the race.

And the moral of all this? Slow and stretchy wins the race.

## 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.