Cubs World Series Puzzles, For Fun And Profit

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 рџ‘Џ Michael Cortina рџ‘Џ of Baltimore and рџ‘Џ Steve Schaefer рџ‘Џ of Carlsbad, California, 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, just in time for October:

The best team in baseball this year, the Chicago Cubs, have clinched their playoff spot and will play their first playoff game a week from today. The Cubs’ road to the World Series title consists of a best-of-five series followed by two best-of-seven series. How many unique strings of wins and losses could the Cubs assemble if they make their way through the playoffs and win their first championship title since 1908? (For example, one possible string would be WWWWWWWWWWW — three straight sweeps. Another would be WWWWWWWLLLWWWW — two sweeps plus a dramatic World Series comeback.)Eamus Catuli.

Submit your answer

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, let’s make it interesting with a puzzle adapted from Keith Wynroe:

You are a gambler and a Cubs fan. The Cubs are competing in a seven-game series against the Red Sox — first to four games wins. Your bookie agrees to take any even-odds bets on any of the individual games. Can you construct a series of bets such that the guaranteed outcomes are: You win $100 if the Cubs wins the series and lose $100 if the Red Sox win it?

Submit your answer
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 how many games would be played if the NCAA College Football Playoff expanded to include all 128 Division I Football Bowl Subdivision teams.

Sure, you could add up the number of games played in each round: 64+32+16+8+4+2+1. But what are you, an abacus? You could also cut through the arithmetic and recognize that each game eliminates exactly one team, and exactly one team is left victorious at the end. Therefore, there must be 127 games played in this hypothetical tournament. That’s a lot of football.

And here’s the solution to last week’s Riddler Classic, concerning making the biggest possible circular table out of a 4-foot-by-8-foot rectangular piece of wood. You wanted the option of adding leaves to your new table, so you were tasked with finding the largest radius you could achieve by cutting out two semicircular pieces. It’s about 2.70545 feet.

There are two good candidates for the arrangement of the table halves to consider here, each trying to squeeze the largest possible semicircles into that rectangle, as illustrated by the puzzle’s submitter, Eric Valpey. One is this:

And the other is this:

Which of these two produces the largest table turns out to depend on the specific dimensions of our original rectangular piece of wood. In our 4-by-8 example, though, it’s the latter. Let’s do a little math to figure out just how big our new table’s radius will be, again courtesy of Eric. We can break that second arrangement down geometrically like this:

Per the diagram above, we can divide the top (the 8-foot side) into three sections of length \(r-r \cos \alpha\), \(r \cot \alpha\), and \(r\). (Remember SOHCAHTOA?) After that, it’s just a matter of algebra.

$$8 = r – r \cos \alpha + r \cot \alpha + r$$

$$8 – 2r = -r \cos \alpha + r \cot \alpha = 2 r \sin \alpha$$

$$\cot \alpha = 2 \sin \alpha + \cos \alpha$$

$$\alpha = 0.498945$$

We also know

$$r = 4 / (1 + \sin \alpha)$$

So \(r = 2.70545\).

Laurent Lessard crafted the following illustration, showing when each arrangement scheme is optimal. When the table gets wide enough, we should switch carpentry strategies.

The extra credit last week — trying to cut the optimal table using N congruent pieces rather than just the two — was, admittedly, ridiculously hard, if not impossible. (Sorry.) Neither I nor the citizens of Riddler Nation could develop a generalized solution. But, Zach Wissner-Gross, delegator extraordinaire, decided to just let the table try to figure it out for itself:

Elsewhere in the puzzling world:

• Puzzles about lines and squares [The New York Times]
• Puzzles about dominoes and nuts [The Wall Street Journal]
• Playing around with digits [The Players’ Tribune]
• Are you smarter than the Gogglebox brainbox? (Don’t ask.) [The Guardian]
• Words, backward and forward [NPR]
• 42! [Expii]

Have an excellent weekend!