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 Eric Veneto, mathematical madmen are on the loose:
The year is 2000, and an arithmetical anarchist group has an idea. For the next 100 years, it will vandalize a famous landmark whenever the year (in two-digit form, for example this year is “18”) is the product of the month and date (i.e. month × date = year, in the MM/DD/YY format).
A few questions about the lawless ensuing century: How many attacks will happen between the beginning of 2001 and the end of 2099? What year will see the most vandalism? The least? What will be the longest gap between attacks?
Riddler Classic
From Matt Gold, a chance, perhaps, to redeem your busted bracket:
On Monday, Villanova won the NCAA men’s basketball national title. But I recently overheard some boisterous Butler fans calling themselves the “transitive national champions,” because Butler beat Villanova earlier in the season. Of course, other teams also beat Butler during the season and their fans could therefore make exactly the same claim.2
How many transitive national champions were there this season? Or, maybe more descriptively, how many teams weren’t transitive national champions?
(All of this season’s college basketball results are here. To get you started, Villanova lost to Butler, St. John’s, Providence and Creighton this season, all of whom can claim a transitive title. But remember, teams beat those teams, too.)
Solution to last week’s Riddler Express
Congratulations to 👏 Tom 👏 of Adams Run, South Carolina, and 👏 Brandon Mitchell 👏 of Lions Bay, British Columbia, winners of last week’s Riddler Express!
Last week brought a simple Riddler Express challenge: Submit a positive integer. Whoever submits the number closest to the median of all the submitted numbers wins!
Out of the 8,178 submissions we received, the median was 332. Our winners were closest, both submitting 333.
The submissions ranged from 1 to \(10^{200}-1\), the highest number that would fit in the submission form. Here is how most of the submissions broke down (for reasonability of scale, this chart excludes some of the extraordinarily large numbers that were submitted):
As always seems to be the case when we offer participatory Riddler challenges, there were solvers who prioritized subterfuge (or at least ballot-box stuffing) over prognostication. One man filled the submission box with hundreds of 9s hundreds of times. Another wrote a script to submit his favorite number, 213, scores of times. And yet another solver submitted dozens of copies of 32. I am a benevolent leader of Riddler Nation, but its laws must be enforced. Therefore, I considered only one submission from each submitter.
I’d love to offer a tidy lesson about how to game this kind of puzzle, but there really isn’t one. So much of it depends on the whims of the crowd. But in any case, 332 is a lovely number. It’s a New York City area code, it’s CCCXXXII in Roman numerals, it was a leap year, and, according to an unsourced Wikipedia article, it was the year Constantine I constructed “a bridge across the Danube in order to increase trade between the Visigoths and Rome.” Math!
Solution to last week’s Riddler Classic
Congratulations to 👏 Peter Ji 👏 of Madison, Wisconsin, winner of last week’s Riddler Classic!
Last week, while killing some time at your desk, you fired up a new game of Solitaire (the version where you deal out three cards from the deck at a time). But your boredom quickly turned to rage because your game was unplayable — you could flip through your deck, but you never had any legal moves! What are the odds?
They are about 0.25 percent, or about 1 in 400.
Let’s lay out the things we need to consider to tackle this problem. The first thing is the board of cards we’ve arranged on the table to begin the game. That board includes seven face-up cards (and 21 face-down cards). The second thing is the deck that we’ll deal from during the game. That deck includes the 24 remaining cards, but because we’re dealing three cards at a time, we only care about eight of them (the ones that will appear at the top of the three-card draws). These two sets — the seven face-up cards and eight in the deck — will determine whether we’ll play a nightmare game of Solitaire in which we have no legal moves.
What’s left to do is an intricate counting problem involving some very big numbers. In what follows, I’ve adapted the approach of solver Jacob Kes, who was kind enough to provide the code he used for counting card combinations.
Brute force is always the most elegant solution! The key criteria in this problem are a card’s rank (that is, whether it’s a number or a face card) and its color (red or black). You can use a computer script to find all the possible combinations for the seven face-up cards by rank and color only — for example, a red five, a black seven, a red jack, etc.3 But we can’t forget the suits!
For each set of seven number-and-color combinations, there’s a corresponding number M of possible combinations of seven cards (now taking suit into account). For each of these combinations, you can calculate how many of the remaining cards (of which there are 45 and might end up in the deck) wouldn’t allow us any legal Solitaire moves — let’s call this number N. That’s how many are neither aces, which can be moved to a designated area on the Solitaire table, nor are cards that can be moved onto a face-up card.
This means that, for a given seven cards, our chance of a nightmare deal is of P = (N choose 8)/(45 choose 8). Each face-up card combination has a 1/(52 choose 7) chance of occurring, so for a total probability, you just need to sum M*P/(52 choose 7) over all of the face-up number-and-color combinations. This winds up being 643,746,385,468/257,479,369,193,475, or about 0.0025, or about 1/400.
In addition to walking through the intricate math to arrive at this number, Laurent Lessard calculated the probability of a nightmare game in the version where you deal one card at a time. It’s significantly lower: about \(1.8\cdot 10^{-7}\).
Good luck avoiding a nightmare game, you brave, solitary warriors.
Want to submit a riddle?
Email me at oliver.roeder@fivethirtyeight.com.