ABC News
How About A Nice Game Of Chess?

Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here: math, logic and probability. These problems, puzzles and riddles come from many top-notch puzzle folks around the world — including you!

Recently, we started something new: Riddler Express problems. These are bite-size puzzles that don’t take as much fancy math or computational power to solve. For those of you in the slow-puzzle movement, worry not — we still feature our classic, more challenging Riddler.

You can mull both over on your commute, dissect them on your lunch break and argue about them with your friends and lovers. When you’re ready, submit your answer(s) using the links below. I’ll reveal the solutions next week, and a correct submission (chosen at random) will earn a shoutout in this column.1

Before we get to the new puzzles, let’s reveal the winners of last week’s. Congratulations to 👏 Steve Osborne 👏 of Frisco, Texas, and 👏 Matt Horrocks 👏 of London, our respective Express and Classic winners. You can find solutions to the previous Riddlers at the bottom of this post.

First up, this week’s Riddler Express, a rags-to-riches chess puzzle from Nik King, first posed to him by chess grandmaster Ben Finegold:

On a chessboard, under the standard rules, how many different paths can the king’s pawn (starting on square e2) take to make it to the other side of the board (e8), where it can become a queen?

If you need a hint, you can try asking me nicely. Want to submit a new Riddler Express? Email me.

And now, for Riddler Classic, it’s back to the board, with a couple of twists on a classic chess problem:

First, how long is the longest path a knight can travel on a standard 8-by-8 board without letting the path intersect itself?

Second, there are unorthodox chess pieces that don’t exist in the standard game, which are known as fairy chess pieces.2 What are the longest nonintersecting paths that can be taken by the camel (which moves like a knight, except 3 squares by 1 square), the zebra (3 by 2), and the giraffe (4 by 1)?

If you need a hint, you can try asking me nicely. Want to submit a new Riddler? Email me.

Here’s the solution to last week’s Riddler Express, which asked you to count the Cubs’ possible paths to a World Series victory. There are 12,250 unique strings of wins and losses that could deliver the Cubs through the playoffs and to the Commissioner’s Trophy.

Let’s break it down. In winning a five-game series, the Cubs could lose zero, one or two games. There is one way (WWW) to lose zero, three ways (LWWW, WLWW, WWLW) to lose one, and six ways (LLWWW, WLLWW, WWLLW, LWLWW, LWWLW, WLWLW) to lose two. That’s a total of 10 strings.

In a seven-game series, the calculations are similar, just longer. In these series, the Cubs could lose zero, one, two or three games. There is one way (WWWW) to lose zero, four ways (LWWWW, WLWWW, WWLWW, WWWLW) to lose one, 10 ways (suppressed here for space) to lose two and 20 ways (ditto) to lose three, for a total of 35 strings. The Cubs play two of these seven-game series.

Because series the Cubs win can never end in a lost game, and because the number of wins needed in each series is fixed, we can simply multiply these possibilities for each series together — 10*35*35 — arriving at 12,250.

And here’s the solution to last week’s Riddler Classic, concerning placing bets on a best-of-seven baseball series. The goal was to construct a series of bets on the individual games such that you’d win \$100 if the Cubs won the series and lose \$100 if the Red Sox won it. The challenge, of course, is that you don’t know ahead of time how many games the series will go. Regardless, you do know you should start the series off with a \$31.25 bet on the Cubs.

The rest of the solution is best expressed in a table. The number of Cubs wins is expressed along the left, the number of Sox wins along the top, and the entry between them is the bet you should place on the Cubs in that hypothetical game. So, for example, you should place a \$31.25 bet on the Cubs when the series is tied 0-0 in Game 1. If they win, you should place another \$31.25 bet. If they win again, bet \$25, and so on.

Here’s how to arrive at this plan, from our winner Matt Horrocks:

Work backwards from Game 7, in each case finding the two possible financial positions you desire at the end of that game, from a given starting series record. For Game 7, the teams start with the series tied at 3-3, and you want to finish either having won \$100 or lost \$100. Therefore, you need to start Game 7 at a balance of \$0 and bet \$100. For Game 6, if the Cubs are leading the series 3-2 you need to finish at \$100 in the case of a win and \$0 (carrying forward the answer from Game 7) in the case of a loss. Therefore you want to start from \$50 and bet \$50. The reverse is true if the Cubs are trailing 2-3: You want to finish at -\$100 or \$0, therefore you want to start from -\$50 and bet \$50. Continue this logic, working backwards from each desired pair of end states for each possible series outcome, calculating the midpoint of those two end states to identify the desired starting financial position.

Laurent Lessard calculated the bets you should place in a longer, best-of-41 series, illustrated below. He writes: “Roughly, you should bet more as you get closer to the end of the series, and you should also bet more when the score is closer.”

Elsewhere in the puzzling world:

Have a fantastic weekend, and enjoy the playoffs!

## Footnotes

1. Important small print: To be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday. Have a great weekend!

2. This is one of the greatest pages on Wikipedia, by the way.

Oliver Roeder was a senior writer for FiveThirtyEight. He holds a Ph.D. in economics from the University of Texas at Austin, where he studied game theory and political competition.