Skip to main content
ABC News
How Many Ways Can You Cast A Die?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the 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 the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter or send me an email.

Riddler Express

From Tim Curwick comes a sudoku-inspired puzzle:

A five-by-five Latin square is a grid with five rows and five columns, where each row and column contains each number from 1 to 5 exactly once. It turns out that there are quite a few five-by-five Latin squares.

If you pick one of these squares at random, what is the probability that no corner contains a 1?

Extra credit: If you pick a five-by-five Latin square at random, what is the probability that two of the four corners have the same number?

Submit your answer

Riddler Classic

I have an unlabeled, six-sided die. I make a single mark on the middle of each face that is parallel to one pair of that face’s four edges.

How many unique ways are there for me to mark the die? (Note: If two ways can be rotated so that they appear the same, then they are considered to be the same marking.)

Extra credit: Suppose you can also mark a face along one of its two diagonals. Now how many unique ways are there to mark the die?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to Ñб─Ñб÷âб─б≤Ðб▐ Paige Kester Ñб─Ñб÷âб─б≤Ðб▐ of Southlake, Texas, winner of last week’s Riddler Express.

Last week, you learned that Riddler Nation has many, many districts, each with the same population. Each district elected one representative (from one of the two major political parties) to the Riddler Nation Legislative Body by majority vote within the district.

However, after recent decisions by the highest court in the land, the redistricting process had resulted in districts that appeared to favor one party over the other.

What was the greatest percentage of Riddler Nation’s total population that could vote for one of the parties while resulting in that party not having a majority of seats in the Legislative Body?

This puzzle was an examination of gerrymandering, in which redistricting packed some districts (so the majority party represented the overwhelming majority) and cracked others (so the minority party had a slim majority). If you gerrymandered to the extreme, then 50 percent of the districts voted entirely for the majority party. In the other 50 percent, approximately half the population voted for the majority party. (Technically, it had to be a little under a half, but with large populations you could assume it was very close to a half.)

In summary, the majority party made up 50 percent of the population plus half of the other 50 percent, which was 75 percent of the total. (Of course, I accepted answers saying that the maximum was slightly less than 75 percent, depending on the population. This allowed the minority party to technically win a clear majority in half of the districts.)

This figure represented an upper bound in gerrymandering. Assuming free and fair elections (i.e., “fair” but still in the presence of gerrymandering), in a two-party system with winner-take-all district voting, it was possible for one party to receive as much as 75 percent of the vote — or very nearly that — and still not win a majority of seats in the Legislative Body. Meanwhile, a super-duper-majority of 80 percent was guaranteed to do the trick.

For more analysis on gerrymandering in the United States (rather than Riddler Nation), check out FiveThirtyEight’s Gerrymandering Project and redistricting overview.

Solution to last week’s Riddler Classic

Congratulations to Ñб─Ñб÷âб─б≤Ðб▐ Andrew J. Poker Ñб─Ñб÷âб─б≤Ðб▐ of Orange, California, winner of last week’s Riddler Classic.

Last week, the states of California, Colorado, New Mexico and New York had banded together to form the United States of Yellow (USY). Meanwhile, Florida, Texas, Utah and Wyoming were the first states to join the United States of Purple (USP). The remaining 40 contiguous states were joining either the USY or the USP, such that both new countries had 24 states; these sets of states themselves also had to be contiguous.

Virginia was leaning toward joining the United States of Yellow. But was there any way for it to do this?

For the purposes of this riddle, you were asked to consider Colorado as not bordering Arizona and Utah as not bordering New Mexico. Also, Michigan’s two peninsulas could be thought of as connected, so that Michigan bordered Ohio, Indiana and Wisconsin.

As it turned out, there was no way for Virginia to go yellow. (Okay, there was one way, if you made use of an ambiguity about states sharing a water border. But I’ll come back to that later.)

To see why that was, we can follow along with Peter Exterkate’s visual solution. First off, here were the eight states whose affiliation you were given in the problem:

Map of the Lower 48 states. In yellow: California, New Mexico, Colorado and New York. In purple: Utah, Wyoming, Texas and Florida.

Because the six New England states all connect to the rest of the nation through New York, that meant they had to be yellow as well, resulting in the following map:

Same as the previous map, but now the six New England states are also yellow.

At this point, for the USY to be contiguous, those seven yellow states in the northeast had to somehow connect to Colorado, New Mexico and California. If the USY cut through the middle to Colorado, then it would have separated Florida and Texas from Utah and Wyoming, meaning the USP wouldn’t have been contiguous. And so, the USY had to go north, connecting New York to California via Montana. If any state along the way had been purple, then one of the two groups wouldn’t have been contiguous. That left you with 20 yellow states:

Same as the previous map, but now all states along the northern border with Canada (plus Oregon) are now also yellow.

To connect Colorado and New Mexico to the rest of the USY, you still needed at least one more yellow state out west (e.g., Arizona). At the same time, for Virginia to be yellow, you also needed three more yellow states: Maryland, Delaware and New Jersey. If one or more of those states had been purple, then it (or Virginia) would have been cut off from the other states of its color.

Making all those connections would have meant the USY had 25 states, one more than its target of 24. In what amounted to a proof by contradiction, that meant Virginia could not have gone yellow and was instead destined for purple.

Some readers found it tricky to figure this out using a map, and instead turned the Lower 48 into a graph with each node representing a state and edges representing borders. Here was one such graph from solver Madeline Argent:

Graph showing connectedness of the Lower 48 states.

Finally, as I hinted at earlier, if you added one more edge to the graph above — between Minnesota and Michigan, whose Upper Peninsula borders Minnesota via Lake Superior — then it was possible for both the USY and the USP to each have 24 contiguous states. Given the ambiguity regarding water borders in the original puzzle, I also counted this line of reasoning as correct.

Want more puzzles?

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 Zach Wissner-Gross at


  1. Important small print: In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.


Related Interactives