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.

The Riddler has returned after a week off for vacation, which I hope you all enjoyed as well. This week, we’re back strong with two cool summer puzzles about sports.

## Riddler Express

From Josh Streeter, a participatory tie-breaker:

Suppose every Riddler reader, you included, is a coach of a soccer team in the Riddler Official Football League, or ROFL. They are instructing their players on strategies for taking penalty kicks to decide a soccer game that has been tied through regulation and extra time.

The goal is divided into two rows (upper and lower) and three columns (left, center and right). The player taking the penalty selects one of these six areas at which to aim their shot, and the goalie chooses one of the six to defend. If the goalie guesses the same area as the shot, she automatically saves the ball. But even if the goalie does *not* guess right, aiming at the outer regions of the goal carries some risk of the shot missing the goal altogether. For example, an undefended shot to the upper left corner will score 70 percent of the time. (See the illustration below.)

Every reader — er, coach — should submit two inputs: the area they want their shooter to target and the area they want their goalkeeper to defend. Like with the Battle for Riddler Nation, each participant’s inputs for shooting and for defending will be simulated against each another’s, and whoever accumulates the most net goals will be crowned ROFL Coach of the Year.

## Riddler Classic

From Liam Tucker, game theoretic shot selection:

Two players have taken to the basketball court for a friendly game of HORSE. The game is played according to its typical playground rules, but here’s how it works, if you’ve never had the pleasure: Alice goes first, taking a shot from wherever she’d like. If the shot goes in, Bob is obligated to try to make the exact same shot. If he misses, he gets the letter H and it’s Alice’s turn to shoot again from wherever she’d like. If he makes the first shot, he doesn’t get a letter but it’s again Alice’s turn to shoot from wherever she’d like. If Alice misses her first shot, she doesn’t get a letter but Bob gets to select any shot *he’d *like, in an effort to obligate Alice. Every missed obligated shot earns the player another letter in the sequence H-O-R-S-E, and the first player to spell HORSE loses.

Now, Alice and Bob are equally good shooters, *and* they are both perfectly aware of their skills. That is, they can each select fine-tuned shots such that they have any specific chance they’d like of going in. They could choose to take a 99 percent layup, for example, or a 50 percent midrange jumper, or a 2 percent half-court bomb.

If Alice and Bob are both perfect strategists, what type of shot should Alice take to begin the game?

What types of shots should each player take at each state of the game — a given set of letters and a given player’s turn?

What shot should Alice begin with if there were *N* players?

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ Jesi Lipp ÑÑâÐ of Kansas City, Missouri, winner of the previous Riddler Express!

Two weeks ago, you were presented with a classic problem of probability manipulation: You had two buckets and 100 ping-pong balls, 50 of which were red and 50 of which were blue. You were meant to arrange the balls into the two buckets however you liked, but each bucket needed at least one ball. Your friend would then blindly choose one of the two buckets and select a ball at random from the chosen bucket. How could you arrange the balls to maximize the probability that your friend chosen a red ball? What probability of success do you achieve?

Your friend achieves the greatest chance of drawing a red ball by placing a single red ball in one bucket and all the other balls in the other bucket. This gives a 50 percent chance he’ll draw a red ball for sure and a 50 percent chance he’ll have a 49/99 chance at a red ball. Summing that up: 0.5*(1) + 0.5*(49/99) = about a **74.7 percent chance** he’ll draw a red ball.

For extra credit, you were asked the same question but for 25 and 200 balls of each color. The solution idea is the same, and the calculations are similar.

For 25 balls of each color: 0.5 + 0.5*(24/49) = about a **74.5 percent chance**.

For 200 balls of each color: 0.5 + 0.5*(99/199) = about a **74.9 percent chance**.

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ Álvaro Begué Aguado ÑÑâÐ of Stony Brook, New York, and ÑÑâÐ Vince Vatter ÑÑâÐ of Gainesville, Florida, winners of the previous Riddler Classic, and Wordsmiths Extraordinaire of Riddler Nation!

The previous Riddler Classic brought the inaugural Superstring Scrabble Challenge. Using only the 100 tiles in a Scrabble bag, you laid them out into one long 100-letter string of your choosing. You looked through the string. For each word you found, you earned points equal to its score. Once you found a word, you don’t get any points for finding it again. The same tile could be used in multiple, overlapping words. So ‘“theater” includes “the,” “heat,” “heater,” “eat,” “eater,” “ate,” etc. The super challenge: What order of tiles gave you the biggest score? (The blank tiles are locked into the letter they represent once you’ve picked it.)

This is a challenging computational problem. As solver Michael Branicky observed, there are over \(10^{115}\) possible strings — more than the number of atoms in the universe — not to mention almost 200,000 different valid words in the word list.

Let’s begin with a visualization of the string submitted by solver Zach Wissner-Gross, which scored 1,076 points. Each stretch of blue marks out a word, and darker blue stretches represent more points.

This picture provides a helpful image of the intuition behind building a high scoring string: you want clusters with lots of overlapping words, and you want those clusters to be high-scoring, typically by containing Scrabble’s “power tiles” like the Z, Q, X and J. In Zach’s string, for example, there are valuable substrings such as ADOZERO, which contains the juicy words ADOZE, DOZE, DOZER and ZERO, and QUARTERNS, which contains QUA, QUART, QUARTE, QUARTER, QUARTERN and QUARTERNS.

And our inaugural Wordsmiths Extraordinaire, did just that in spades, to mix my game metaphors.

First, a quick note: Different solvers interpreted the scoring rules for this problem differently. Some assumed that if a word appeared more than once in a string, its score could be counted only once, while others assumed that a word could be counted whenever it appeared. Your humble Riddler editor originally intended the latter method, but he should have been more explicit, and so he will report both of those scores, in that respective order, for the high-scoring strings highlighted below. The solvers’ blanks are listed in brackets.

Vince Vatter, who also happens to be the reigning king of Riddler Nation and who said he “got out a Scrabble board and played around with the tiles until this pattern stood out to me,” managed a whopping 1,629/1,675 points with his superstring submission:

IOTTAVAUNTOYOUTACTINGAGABYEMBLAZONER[S]WITHEREDEVELOPERSHALLOWEDAMASKING[S]COFFERSUPEROXIDESQUIREDJINNII

Álvaro Begué Aguado explained that he “wrote a fairly straight-forward implementation of simulated annealing using C++ and let it run for about a day.” Here’s his juicy 1,629/1,657-point string:

OUTBLAZEDISCANTINGABYESQUIREDECODEVELOPER[S]IMPROVERSTUFF[S]WITHEREINTOKAYOURETAXINGAGAMASHALLOWEDJINNIO

Marc Broering generated “all the scores of each word in C, then chose the best using a metric based on score and word length” and arrived at 1,615/1,663 with:

TAUTOKAY[S]COFFINGUTATAMI[S]BEHAVERSETACODEVELOPERSUPEROXIDESQUIREDRAWINGYOGINITEMBLAZONERSHALLOWEDJINNI

And Jeremy Thorpe, who went with a “randomized computer search using lightweight constrained optimization techniques,” scored 1,595/1,651 with this beauty:

EMBLAZONER[S]WITHEREDEVELOPERSUPEROXIDESCANTINGABYE[S]QUIREDAMASKINGSHALLOWEDJINNIFACTORYOUTAVOFAGOUTITI

What do all of these top submissions have in common? BLAZE, SQUIRE, DEVELOPER and JINNI. I consider that a message from the universe which I will now spend the weekend decoding. Or perhaps it’s the prompt for a piece of flash fiction. In any case, well strung, solvers.

## 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 me at oliver.roeder@fivethirtyeight.com.

**CORRECTION (July 12, 2019, 4 p.m.):** A previous version of this column reported solvers’ scores for the Riddler Classic solution using two different scoring methods and without distinguishing between the two. It has been updated to include a description of the two methods and both scores for each submission mentioned, and to recognize an additional winner.