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.
From Zack Beamer comes a baffling brain teaser of basketball, just in time for the NBA playoffs:
Once a week, folks from Blacksburg, Greensboro, and Silver Spring get together for a game of pickup basketball. Every week, anywhere from one to five individuals will show up from each town, with each outcome equally likely.
Using all the players that show up, they want to create exactly two teams of equal size. Being a prideful bunch, everyone wears a jersey that matches the color mentioned in the name of their city. However, since it might create confusion to have one jersey playing for both sides, they agree that the residents of two towns will combine forces to play against the third town’s residents.
What is the probability that, on any given week, it’s possible to form two equal teams with everyone playing, where two towns are pitted against the third?
Extra credit: Suppose that, instead of anywhere from one to five individuals per town, anywhere from one to N individuals show up per town. Now what’s the probability that there will be two equal teams?
The solution to this Riddler Express can be found in the following week’s column.
This month, the Tour de France is back, and so is the Tour de FiveThirtyEight!
For every mountain in the Tour de FiveThirtyEight, the first few riders to reach the summit are awarded points. The rider with the most such points at the end of the Tour is named “King of the Mountains” and gets to wear a special polka dot jersey.
At the moment, you are racing against three other riders up one of the mountains. The first rider over the top gets 5 points, the second rider gets 3, the third rider gets 2, and the fourth rider gets 1.
All four of you are of equal ability — that is, under normal circumstances, you all have an equal chance of reaching the summit first. But there’s a catch — two of your competitors are on the same team. Teammates are able to work together, drafting and setting a tempo up the mountain. Whichever teammate happens to be slower on the climb will get a boost from their faster teammate, and the two of them will both reach the summit at the faster teammate’s time.
As a lone rider, the odds may be stacked against you. In your quest for the polka dot jersey, how many points can you expect to win on this mountain, on average?
The solution to this Riddler Classic can be found in the following week’s column.
Last week’s Riddler
Congratulations to 👑 Brendan Hill 👑 of Edmond, Oklahoma, winner of last week’s Riddler and the new ruler of Riddler Nation!
Last week was the fifth Battle for Riddler Nation, and things were a little different this time around.
In a distant, war-torn land, there were 13 castles — three more than the usual 10 from prior battles. There were two warlords: you and your archenemy. Each castle had its own strategic value for a would-be conqueror. Specifically, the castles were worth 1, 2, 3, …, 12, and 13 victory points. You and your enemy each had 100 soldiers to distribute, any way you liked, to fight at any of the 13 castles. Whoever sent more soldiers to a given castle conquered that castle and won its victory points. If sent the same number of troops as your opponent, you split the points. You didn’t know what distribution of forces your enemy had chosen until the battles began. Whoever won the most points won the war.
I received a total of 970 battle plans. Of those, I excluded ones that were not valid, including any that had in excess of 100 troops, or, like the strategy submitted by Lowell Vaughn, tried to sneak in 101 troops to Castles 2 through 13 by having -1,112 (yes, a negative number) troops at Castle 1. Also, to keep things fair, whenever anyone submitted multiple strategies, I only counted the last strategy they submitted. In the end, there were 821 valid strategies.
Next, I ran all 336,610 one-on-one matchups, awarding one victory to each victor. In the event of a tie, both warlords were granted half a victory. Brendan Hill was the overall winner, tallying 630 wins against just 186 losses and 4 ties. Here’s a rundown of the 10 strongest warlords, along with how many soldiers they deployed to each castle:
Who were Riddler Nation’s strongest warlords?
The top 10 finishers in FiveThirtyEight’s Battle for Riddler Nation, with their distribution of soldiers for each castle and overall record
|Soldiers per castle||Record|
|4||Fivey The Swing Voter||0||0||1||1||3||12||13||3||18||2||27||1||19||592||22||206|
In previous battles, when there were just 10 castles, there were 55 points in play. As long as you won more than half them — that is, at least 28 points — you were guaranteed a victory. The top strategies clustered soldiers into a small number of castles worth exactly 28 points. It took at least four castles to achieve 28 points, and there were several ways to do it: 4+5+9+10, 3+6+9+10, etc.
This time around, with 13 castles, there were 91 points in play, which meant you needed at least 46 points to secure a victory. Two-time Battle of Riddler Nation victor Vince Vatter was the one who suggested increasing the number of castles to 13, since there was only a single way to reach 46 points by winning exactly four castles: 10+11+12+13. Vince was curious whether that strategy would prevail or instead a strategy that targeted more castles would win the day.
Our winner, Brendan, placed at least five soldiers at a whopping seven castles. Adding the values of these castles gave 3+4+5+6+7+9+12, which was indeed exactly 46 points.
Vince did fine, by the way, coming in 69th place with 532 wins against 266 losses and 22 ties. Our previous champion, David Love, was a little lower down, coming in 335th with 438 wins, 352 losses and 30 ties.
The complete data set of strategies will be posted in the coming weeks. In the meantime, the following graph summarizes all the strategies. Each column represents a different castle, while each row is a strategy, with the strongest performers on top and the weakest on the bottom. The shading of a cell indicates the number of soldiers placed. It’s a lot to take in, but at the very least you might see a few “bands” — for example, the 10+11+12+13 strategies are clustered together in a few places, since they were similarly (somewhat) successful.
Finally, I was delighted to see that there was a fierce competition in Iowa’s Sheldon Community School District. Three classrooms — Sheldon Middle School Advanced Math, Sheldon Middle School TAG, and Sheldon High School STEM — submitted strategies. Among these, Sheldon Middle School TAG was the strongest, coming in 282nd place, with 458 wins, 347 losses and 15 ties. Definitely a talented group, there.
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 firstname.lastname@example.org
CORRECTION (Sept. 11, 2020, 3:36 p.m.): In an earlier version of this article, the table about Riddler Nation’s strongest warlords transposed the columns for ties and losses.