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The Sixth Battle For Riddler Nation

Welcome to The Riddler. Most weeks, I offer up two problems related to the things we hold dear around here: math, logic and probability.

But this week is special. The time has flown, and somehow it’s been two years since I (peacefully) took over this column from my predecessor, Oliver Roeder.

It is only fitting that we continue one of Ollie’s (and my) favorite traditions here at The Riddler: the Battle for Riddler Nation. In order to have a chance at рџ‘‘ winning рџ‘‘ and becoming the next ruler, I need to receive your battle plans before 11:59 p.m. Eastern time on Monday. Have a great weekend!

This week’s Riddler

Some readers may be familiar with the first, second, third, fourth and fifth Battles for Riddler Nation. If you missed out, you may want to consult the thousands of attack distributions from these previous contests.

I am pleased to say that this week marks the sixth such competition — but like last year, I’m tweaking 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 phalanxes of soldiers to distribute, any way you like, to fight at any of the 10 castles. Whoever sends more phalanxes to a given castle conquers that castle and wins its victory points. If you each send the same number of phalanxes, 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.

Unlike previous iterations of this challenge, you can split up phalanxes into tenths. In other words, the number of phalanxes you assign to any given castle can go to one decimal place. For example, you could assign 5 phalanxes to a castle, but 4.9 and 5.1 are also valid assignments. But remember — your total across all 10 castles must still add up to 100.

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

Who can steal the crown from рџ‘‘ Brendan Hill рџ‘‘ of Edmond, Oklahoma, who currently sits atop the throne?

Perhaps you have the cunning and logic to be the next ruler of Riddler Nation!

The results of this Riddler can be found in the following column.

Solution to last week’s Riddler Express

Congratulations to рџ‘Џ Vasco Villas-Boas рџ‘Џ of Berkeley, California, winner of last week’s Riddler Express.

Last week, you and your two best friends were in a three-person fantasy football league. You were drafting just three positions each for your teams: quarterback, running back and wide receiver. (Yes, this was a simplified version of fantasy football.)

The following table showed the top three athletes in each position, as well as the number of fantasy points they were expected to earn over the course of the season. You and your friends each had to select exactly one player from each position.

Can you build a Riddler fantasy football dream team?

Top players in Riddler fantasy football for each position, by expected number of fantasy points over the course of the season

Quarterback Running Back Wide Receiver
Matrick Pahomes 400 Caffrey McChristian 300 Avante Dadams 250
Osh Jallen 350 Calvin Dook 225 Hyreek Till 225
Myler Kurray 300 Herrick Denry 200 Defon Stiggs 175

The draft was a “snake draft.” If person A drafted first, B drafted second and C drafted third, then the order of the picks was as follows: A-B-C-C-B-A-A-B-C.

Your kind friends agreed that you could choose your pick number. Which draft position should you have chosen to maximize your expected fantasy score?

Here’s what Riddler Nation had to say:

  • 19.9 percent of readers said they would draft first.
  • 62.3 percent of readers said they would draft second.
  • 17.7 percent of readers said they would draft third.

The majority opted to draft second. But was that the best strategy?

At first glance, some readers thought Osh Jallen (350 points) was a better pick than Caffrey McChristian (300 points). However, missing out on Jallen meant you could grab Myler Kurray, who was only 50 points lower. Meanwhile, missing out on McChristian meant that your running back, at best, was 75 points lower. So while Jallen had more expected points, McChristian was the more valuable pick in the draft.

To better make sense of this, it was helpful to rescale all the expected point values, as solver Johnny of Toronto, Canada, did. All that mattered were the relative differences between the scores. One way to rescale them was to say that the lowest-scoring player at each position was worth 0 points, and that every 25 “original” points were worth 1 “rescaled” point. That meant the quarterbacks (QBs) were respectively worth 4, 2 and 0 points; the running backs (RBs) were worth 4, 1 and 0 points; and the wide receivers (WRs) were worth 3, 2 and 0 points.

Person A’s best move was to grab one of the 4-pointers — either the top QB or the top RB. Suppose they picked the top QB, which meant B picked the top RB, giving them 4 points apiece. From there, C would pick the top WR for 3 points as well as the second QB for 2 points, netting them 5 points in total (since the third picks were all worth 0 points). Because of the snake draft ordering, B got the next pick. They would have snagged the top remaining player, which was the second WR for another 2 points. Finally, A would have drafted the second RB for 1 point. During the final round, each person would have drafted the position they needed to complete their roster. For this scenario, the final result was: 5 points for A, 6 points for B and 5 points for C.

Drafting second was looking good so far. But what if person A had opted for the other 4-pointer, the top RB?

In this case, B would have taken the top QB. The picks for C would have remained the same: the top WR and the second QB, for a total of 5 points. Once again, B then picked the second WR. At this point, the top player left on the board was the second RB. However, A had already drafted an RB, meaning they had to select either the 0-pointer QB or WR. In the end, the result was: 4 points for A, 7 points for B and 5 points for C. As noted by solver Jane Steele of Santa Clara, California, this outcome was worse for A, so we expected A to take the top QB rather than the top RB with their first pick.

In the end, person B came out on top, which meant picking second in the draft was indeed your best option. Returning to the original point scale, that meant A and C both had 800 points, while B had 825 points. With a team of McChristian, Kurray and Hyreek Till, drafting second would indeed have made you unstoppable.

Solution to last week’s Riddler Classic

Congratulations to рџ‘Џ Christian Wolters рџ‘Џ of San Jose, California, winner of last week’s Riddler Classic

Last week, Hames Jarrison had just intercepted a pass at one end zone of a football field and began running — at a constant speed of 15 miles per hour — to the other end zone, 100 yards away.

At the moment he caught the ball, you were on the very same goal line but on the other side of the field, 50 yards away from Jarrison. Caught up in the moment, you decided you would always run directly toward Jarrison’s current position, rather than plan ahead to meet him downfield along a more strategic course.

Assuming you ran at a constant speed (i.e., you didn’t have to worry about any transient acceleration), how fast did you have to be in order to catch Jarrison before he scored a touchdown?

First off, as noted by solvers Christopher Green and Jenny Mitchell, this pursuit problem was very similar to a prior riddle about an angry ram that chased you out of a pen. Yours truly almost took home that week’s coolest extension award, losing out to Lex K.

Solver Chris Sears used approximation methods to numerically determine the speed that would just catch Jarrison. As you can see from the animation below, Chris found your speed had to be somewhere around 19.2 miles per hour.

Several solvers, like Dan Swenson, Rajeev Pakalapati and Laurent Lessard, were able to find more precise solutions by setting up differential equations to describe the motion of you and Jarrison and then solving these equations using calculus.

As with the riddle of the ram from five years ago, it was also possible to solve without using calculus, as demonstrated by Allen Gu. Allen noted the similar triangles formed by your velocity and Jarrison’s velocity at any given moment, shown below.

Graph showing Jarrison moving vertically upward from the point (1, 0), while you pursue him, starting at the point (0, 0). You are currently at (x(t), y(t)). Your speed toward Jarrison is v, and the velocity has horizontal and vertical components x_dot(t) and y_dot(y).

Meanwhile, Jarrison's current position is (1, t). His velocity is upward with speed 1, which is broken down into components of y_dot(t)/v away from you and x_dot(t)/v in the perpendicular direction.

This meant the ratio of your vertical speed (i.e., in the direction from one end zone to the other) to his total speed was equal to the ratio of Jarrison’s speed away from you to his actual speed running down the sideline.

Before moving further, let’s rescale the field and the speeds. Suppose your starting position is (0, 0), while Jarrison starts at (1, 0) and is moving toward (0, 2). Then suppose Jarrison’s speed is 1 and your speed is v. (The final solution will then be v multiplied by 15 miles per hour.)

The initial distance between you and Jarrison was 1, which fell to zero by the time you caught him at the goal line at time 2. During this time, you pursued him at speed v, bringing you closer by a distance 2v. However, Jarrison was also running away from you by a net distance of 2/v. (You can show this by integrating your vertical speed over time, which had to come out to a distance of 2, and then scaling down by a factor of v.) Writing these changes in distance as an equation, Allen found that it was quadratic: 1 = 2v−2/v, or 2v2v−2 = 0.

Applying the quadratic formula, v came out to be (1+√17)/4. Scaling back to the original values of the problem, your speed had to be this many times greater than 15 miles per hour, or about 19.2116 miles per hour — very nearly the result from Chris’s approximation.

Jarrison had better hope that your name isn’t Usain Bolt or Tyreek Hill. But maybe they’ll be too busy racing each other to worry about catching Jarrison.

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

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.