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!

Last week, we started something new: Riddler Express problems. These will be 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 👏 **Katie Andrews **👏 of Great Falls, Virginia, and 👏 **Ben Sokolowsky **👏 of Stony Brook, New York, our respective Express and Classic winners. You can find solutions to the previous Riddlers at the bottom of this post.

First up, Riddler Express, inspired by **Eric Beck**:

Suppose the NCAA College Football Playoff (a single-elimination tournament) expanded to include not just four, but *all* 128 Division I Football Bowl Subdivision teams. How many individual games would be played in that playoff? (Hint: You probably don’t even need a piece of paper for this one.)

Submit your answer

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

And now, for Riddler Classic, a handy puzzle from **Eric Valpey: **

You’re on a DIY kick and want to build a circular dining table which can be split in half so leaves can be added when entertaining guests. As luck would have it, on your last trip to the lumber yard, you came across the most pristine piece of exotic wood that would be perfect for the circular table top. Trouble is, the piece is rectangular. You are happy to have the leaves fashioned from one of the slightly-less-than-perfect pieces underneath it, but there’s still the issue of the main circle. You devise a plan: cut two congruent semicircles from the perfect 4-by-8-foot piece and reassemble them to form the circular top of your table. What is the radius of the largest possible circular table you can make?

*Extra credit*: What is the largest circular table that can be made from *N* congruent pieces?

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 high Count Von Count can count on Twitter, given the 140-character limit. If he spells out the numbers, without commas or “and”s, plus an exclamation point, the highest he’ll get is the number 1,111,373,373,372. It fits nice and snug in a tweet!

There’s no silver bullet for arriving at this answer. Some trial and error is eventually required after you recognize that spelling out “one,” “two” and “ten” is relatively short compared to “three,” “seven” and “eight,” for example. Eventually, you’ll hit the character limit.

**Tim Supinie **graphed the length of the count’s tweets as the numbers being counted crept up:

But worry not, Count fans. At his current rate of about two tweets per day, he won’t hit the limit for another 1.5 billion years, at which point the oceans will have evaporated and only single-celled organisms and fictional nobles will have survived.

And here’s the solution to last week’s Riddler Classic, concerning an experimental sports league draft. Suppose a 30-team league switches from a system in which the previous year’s worst teams draft first to a system in which teams are randomly assigned to two groups via a series of coin flips, with all of the teams in one group drafting first, but the order within those groups still being sorted out by which teams were worst last season. If your team’s draft position was 10th in the old system, its expected draft position is **12.75 **under the new system.

Why? Half of the time, your team wins its coin flip, putting it in the first group. In this case, you expect to draft *after* 9/2 other teams. The ‘9’ represents the nine teams with worse records than yours, and the ‘2’ represents the 50-50 chance any of those teams ends up in your group. Therefore, in this case, you expect to draft in (9/2)+1, or 11/2, position.

The other half of the time, your team loses its coin flip, putting it in the second group. In this case, you expect to draft *before* 20/2, or 10, teams. The 20 is the 20 teams with better records than yours, and the 2 represents the chance that their coin flips place them in the first group. Therefore, in this case, you expect to draft in 30-10=20th position.

Combining these two cases, your team expects to draft in \(1/2\cdot (11/2)+1/2\cdot (20)=12.75\) position.

Turns out that the league’s rule change *does* help in alleviating the incentive for tanking at the end of a season.

For extra credit, I asked what a team’s expected draft position would be if a league of *N *teams was divided randomly into *T* groups. Per the puzzle’s submitter, Stephen Penrice, the expected draft position of the *i*th best team is

$$\frac{i}{t}+\frac{(t-1)(N+1)}{2T}$$

The math gets a little messy there, but the ideas are just the same as in the 30-team, 10th place case. For more, as ever, **Laurent Lessard** provides a lucid explanation.

And while the average pick becomes 12.75 rather than 10th under the new system, the *distribution* of that new pick is bimodal, and has a high variance, as this animated simulation from **Russell Maier** shows:

Elsewhere in the puzzling world:

- The world’s greatest detective [The New York Times]
- More mystery puzzles [Expii]
- A “September” puzzle [NPR]
- Slate now has a crossword puzzle [Slate]
- Better know a puzzle master: David Kwong [NBC News]

Have a wonderful weekend! May you win all of your coin tosses.