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Don’t Trust Anyone Older Than L

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.

Riddler Express

From Dominic van der Zypen, a shipwreck demands deduction:

We are marooned on an island that has the following curious property: Everyone over a certain age lies all the time. More specifically, there is an age limit L — a positive integer — and all islanders who are younger than age L only tell the truth, while islanders who are at least L years old only tell lies.

We are greeted by five islanders who make the following statements:

A: “B is more than 20 years old.”

B: “C is more than 18 years old.”

C: “D is less than 22 years old.”

D: “E is not 17 years old.”

E: “A is more than 21 years old.”

A: “D is more than 16 years old.”

B: “E is less than 20 years old.”

C: “A is 19 years old.”

D: “B is 20 years old.”

E: “C is less than 18 years old.”

What is L? And what did we just learn about the ages of the islanders?

Submit your answer

Riddler Classic

From Mark Whelan, shuffle up and deal. And count very, very high:

Bridge, that venerable card game, begins with an auction. There are four players around a table, each sitting across from his or her partner. At its simplest, the auction goes like this: Beginning with the dealer and orbiting around the table, players can place a bid or pass. Players who’ve passed can re-enter the bidding later. A bid is comprised of a number (one through seven) and a suit. Every legal bid must be higher than the one that came before, meaning either that its number is higher or that its number is the same and its suit is higher (from high to low, the suits go no-trump, spades, hearts, diamonds, clubs).

The auction ends if the other three players pass after a bid, or if all four players pass right away.

How many different legal bridge auctions are there?

Extra credit for you bridge nerds out there: What if we incorporate the possibility of doubling into this accounting exercise? Specifically, if the last bid was made by one of the opposing partners, a player may, while bidding, also double, which increases the stakes for the hand. If a player has been doubled before, they may redouble. Including these variations on bids, how many legal bridge auctions are there now?

Submit your answer

Solution to the last week’s Riddler Express

Congratulations to 👏 Alana Christie 👏 of Dallas, winner of last week’s Riddler Express!

Last week brought another maze that the enemies of Riddler Nation had crafted to entrap you. It looked like this:

Your goal was to get to the star in the middle, and you could start your journey in any square around the perimeter. You then moved through the squares — up, down, left or right — by following the squares’ arrows. If a square had a two-headed arrow, you could choose whichever of its directions that you liked. If you arrived in a square containing a number, you could exit in any direction you liked, but you had to add that number to your score. Your challenge was to solve the maze with the lowest possible score, which was the total of all numbered squares you crossed. What was the score?

The lowest possible score was 1. You can achieve it by entering via the green square on the middle of the maze’s right-hand edge, traveling left, then up to a “0,” then left some more, then up, left to another “0,” left, down, left through a “1,” left, down, and you’re there! That path looks like this, as illustrated by solver Matthew Modlin:

You could solve this maze simply by playing around a bit — by “following your nose,” as my high school calculus teacher used to say. (My nose, as it turned out, wasn’t especially good at calculus.) Or you could work backward, starting at the star and working toward the perimeter. Or you could employ the computational services of an algorithm. Solver Laurent Lessard, for example, used something called the Bellman-Ford algorithm, which calculates the shortest route and is used in applications such as routing data. His solution looks like this:

Solution to last week’s Riddler Classic

Congratulations to 👏 Shailesh Ingale 👏 of Chicago, winner of last week’s Riddler Classic!

Last week you came on down to play a game on “The Price Is Right” called Cover Up. You were trying to win a car by guessing its five-digit price. You had two numbers to choose from for the first digit, three numbers for the second digit, and so on, ending with six options for the fifth and final digit. To begin, you locked in a guess at the entire price of the car. If you got at least one digit correct on the first guess, the correct digit(s) were highlighted and you got to replace incorrect digits on a second guess. That process continued on subsequent guesses until the price was guessed correctly. However, if none of the new numbers you swapped in were correct, you lost. You could conceivably have won the car on the first guess or with up to five guesses.

What were the chances you won the car if all of your guesses were random selections, not based on any knowledge of cars? They were 231/720, or about 32.1 percent.

Why? In total, the car could have any one of 2*3*4*5*6 = 720 different prices, each of which is equally likely, as far as you’re concerned. There are also 720 guessing “paths” you could take, each of which is equally likely. The bulk of the solving comes in figuring out which of these paths lead you to driving home in a new car and which do not.

To figure that out, let’s denote your guess as ABCDE — or abCDe or ABcde, etc. — where the letters are capitalized if that particular digit is correct and lowercase if it is incorrect.

There is only one way to guess all the digits correctly on the first try: ABCDE.

There are five ways to go from four correct digits and one incorrect digit to winning the car: ABCDe → ABCDE, ABCdE → ABCDE, ABcDE → ABCDE, AbCDE → ABCDE, aBCDE → ABCDE.

And so on. Once we’ve counted up all these paths — solver Ryan Lafitte wrote a handy list of them — we arrive at a total of 231. We divide that by the total number of paths, 720, and we’re done!

Now suppose that you knew a little something about cars. Specifically, you knew the ten-thousands place of the car’s price for sure, therefore guaranteeing you at least one correct guess. How much does that knowledge improve your chances of winning the car? Not much, it turns out. Locking in the first digit means you have half as many possible prices and half as many routes for getting to the right price, so your chances of winning in this case are 116/360, or only about 32.2 percent. A little knowledge goes a little way.

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


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

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.