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!

Some Riddler Nation news! Starting today, we’re going to offer up the occasional Riddler Express problem. 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’ll still feature our classic, more challenging Riddler. And for those of you who have found that Riddler to be a bit *too* challenging, hopefully Riddler Express is more your style.

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 (plural!), let’s reveal the winner of last week’s. Congratulations to 👏 **Mark Canning **👏 of Santa Clara, California, our big winner. You can find a solution to the previous Riddler at the bottom of this post.

But first, the inaugural Riddler Express! This one comes from **Kevin Huigens:**

Count Von Count, the counting count on “Sesame Street,” counts aloud on Twitter. If he counts up by one with each tweet — “One!” “Two!” “Three!” … “Five hundred thirty eight!” etc. — how high can he go before hitting the 140-character limit? Note: The count is enthusiastic and must end all of his tweets with an exclamation point.

Submit your answer

As always, 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 puzzle from **Stephen Penrice: **

You are one of 30 team owners in a professional sports league. In the past, your league set the order for its annual draft using the teams’ records from the previous season — the team with the worst record got the first draft pick, the team with the second-worst record got the next pick, and so on. However, due to concerns about teams intentionally losing games to improve their picks, the league adopts a modified system. This year, each team tosses a coin. All the teams that call their coin toss correctly go into Group A, and the teams that lost the toss go into Group B. All the Group A teams pick before all the Group B teams; within each group, picks are ordered in the traditional way, from worst record to best. If your team would have picked 10th in the old system, what is your expected draft position under the new system?

*Extra credit*: Suppose each team is randomly assigned to one of T groups where all the teams in Group 1 pick, then all the teams in Group 2, and so on. (The coin-flipping scenario above is the case where T = 2.) What is the expected draft position of the team with the Nth-best record?

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

And here’s the solution to last week’s Riddler, concerning some statisticians who find a $100 bill on the floor. If five of them play a game where, each turn, the bill has to do one of three things — move left, move right, or stay put (at which point the game ends) — and each thing is equally likely to happen, whoever starts with the bill has a **5/11 chance** of keeping it.

To get there, we can set up a system of equations. Let \(P_0\) be the probability you win when the bill starts in front of you, \(P_1\) be the probability if it starts one seat away, and \(P_2\) be the probability if it starts two seats away. If you start with the bill, for example, your chances of winning this turn are 1/3 plus your chances of winning if the bill is one seat away, which we can express mathematically like so:

$$P_0 = 1/3 + P_1/3 + P_1/3$$

Similarly for the other two possible positions:

$$P_1 = P_0/3 + P_2/3$$

$$P_2 = P_1/3 + P_2/3$$

Then we solve that system algebraically, which gives \(P_0=5/11\), \(P_1=2/11\) and \(P_2=1/11\).

For extra credit, I asked what the chances would be if there were *N* statisticians in the department, rather than five. The extra-credit solutions I received drew on all sorts of cool approaches, from simulations to Fourier transforms to contour integrals.

The full solution from the puzzle’s submitter, Bruce Torrence, will be published in a forthcoming issue of Mathematics Magazine. But simply put, all this fancy math winds up in a surprisingly beautiful place. For *N* players, the chances of winning if you start with the $100 bill are

$$ P_0= \begin{cases} F_N / L_N & \text{if $N$ is odd}\\ L_N / 5F_N & \text{if $N$ is even}\end{cases}$$

Where \(F_N\) is the *N*th Fibonacci number, and \(L_N\) is the *N*th Lucas number. The fifth Fibonacci number (not counting zero) is 5, and the fifth Lucas number is 11, recreating our 5/11 answer above. As the number of players increases to infinity, the chances of winning if you start with the bill converge to \(1/\sqrt{5}\).

Math!

Elsewhere in the puzzling world:

- A puzzle about walk signals [The New York Times]
- Four bookish brainteasers [The Guardian]
- Tryouts for the Varsity Math team [The Wall Street Journal]
- Some back-to-school puzzles [Expii]
- A race to the end of the alphabet [NPR]

Have a terrific weekend!