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Come On Down And Escape The Maze

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 Andy Reigle, once more into the maze, dear friends:

Those pesky enemies of Riddler Nation are at it again! A couple of weeks ago, they trapped you in a maze without walls. Most of you escaped, but the enemies remain undeterred. They’ve been hard at work building a new maze without walls, shown below.

Before banishing you to the maze, they hand you a list of rules.

  1. You can enter via any perimeter square. The goal is to reach the yellow star in the center with the lowest possible score, which is calculated by adding up all the numbers that appeared in any squares you crossed.
  2. You can only move horizontally or vertically (not diagonally) to bordering squares.
  3. If you enter a square with an arrow (↑), you have to exit that square in the direction the arrow indicates. Some squares have double arrows (↔), giving you a choice of two directions.
  4. If you enter a square with a number, you must add it to your score, but you can exit in the direction of your choice.
  5. If an arrow leads you to a square with a skull, you die. But you knew that already.

What is the lowest score you can achieve while solving the maze and saving Riddler Nation?

Submit your answer

Riddler Classic

From Josh Berry, come on down!

You’re playing a “Price Is Right” game called Cover Up, which has contestants try to guess all five digits of the price of a brand new car. You have two numbers to choose from for the first digit, three numbers to choose from for the second digit, and so on, ending with six options for the fifth and final digit. You’re not winning any $100,000 cars in this game.

First, you lock in a guess at the entire price of the car. If you get at least one digit correct on the first guess, the correct digit(s) are highlighted and you get to replace incorrect digits on a second guess. This continues on subsequent guesses until the price is guessed correctly. But if none of the new numbers you swapped in are correct, you lose. A contestant could conceivably win the car on the first guess or with five guesses, getting one additional correct digit highlighted on each guess.

First question: If you’re guessing entirely by chance, what’s the likelihood of winning the car?

Second question: Suppose you know a little bit about cars. Specifically, you are 100 percent certain about the digit in the ten-thousands place, but have to guess the remaining four digits by chance. What’s the best strategy, and what’s the likelihood of winning the car now?

Submit your answer

Solution to the last week’s Riddler Express

Congratulations to рџ‘Џ Derek Truesdale рџ‘Џ of San Jose, California, winner of last week’s Riddler Express!

Last week took us to Broadway. The song “Seasons of Love” from the musical “Rent” states that a year has 525,600 minutes and, indeed, 365×24×60 = 525,600. (Leap years need not apply.) That, naturally, raised a question: Given any three random integers — X, Y and Z — what are the chances that their product is divisible by 100?

The chances are exactly 12.43 percent.

To get there, we want to figure out the chances that XYZ contains at least two factors of 2 and at least two factors of 5 — 2×2×5×5 = 100.

To get started with this calculation, brush off that elementary school lesson on factors. As solver Hector Pefo explained, note that the probability that this number has at least two factors of 2 is 1 minus the probability that it has zero factors of 2 minus the probability that it has one factor of 2. Half of all integers are divisible by 2. So the probability that XYZ has zero factors of 2 equals \((1/2)^3 = 1/8\). For XYZ to have exactly one factor of 2, one of its elements has to have exactly one factor of 2 and the other two have to be odd. Half of all even numbers, or a quarter of all numbers, have exactly one factor of 2 — that is, they are not divisible by 4. So the probability that XYZ has one factor of 2 equals \(3\cdot (1/4)(1/2)(1/2)\). So the probability that we’ll successfully get our factors of 2 into our mystery number is \(1 – 1/8 – 3/16 = 11/16\).

Now for the factors of 5. The logic is the same: The probability that it has at least two factors of 5 is 1 minus the probability that it has zero factors of 5 minus the probability that it has one factor of 5. That probability is given by \(1 – (4/5)^3 – 3\cdot (1/5 – 1/25)(4/5)^2 = 113/625\).

We can multiply those two probabilities together to find the chance that XYZ is divisible by 4 and 25, or, as we’ve wanted all along, 100. \((11/16)(113/625) = 0.1243\), or 12.43 percent.

“Seasons of Love,” you got lucky.

Solution to last week’s Riddler Classic

Congratulations to рџ‘Џ Andrew Petersen рџ‘Џ of Cedar Rapids, Iowa, winner of last week’s Riddler Classic!

Last week, you and I were playing a game. Spread out on a table in front of us, face up, were nine index cards with the numbers 1 through 9 on them. We took turns picking up cards and putting them in our hands. (There was no discarding.) The game ended in one of two ways. If we ran out of cards to pick up, the game was a draw. But if one player had a set of three cards in his or her hand that added up to exactly 15 before we ran out of cards, that player won. (For example, if you had 2, 4, 6 and 7, you would win with the 2, 6 and 7. However, if you had 1, 2, 3, 7 and 8, you hadn’t won because no set of three cards added up to 15.) Let’s say you went first. With perfect play, who would win and why?

No one will win — the game will be a draw. Why? This is really tic-tac-toe in disguise.

Specifically, it’s tic-tac-toe played on top of a magic square. Consider arranging the nine index cards in the following way:

\begin{matrix}\nonumber 6&1&8\\7& 5 &3\\2 &9& 4\end{matrix}

The cards in every row, column and diagonal add to 15 — exactly what one of us needs to win. If picking up a card and putting it in one’s hand is akin to marking that card with an X or an O, then this game is just tic-tac-toe on this specific sort of board. With perfect play, it’s well known that tic-tac-toe is a guaranteed draw. So this game is, too! (But it’s still fun, I promise.)

If you’re in the mood to take it to the next level, here’s an article from 2018 about how something called “ultimate tic-tac-toe” has plenty to tell us about modern politics.

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.