Can You Throw the Card Game?

If you cheat in your opponent’s favor, how likely are you to win? (In which case, I guess you're really losing.)

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “extra credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

Every so often, I play the card game Slap with someone who really doesn’t like to lose. (You know who you are.) But before we get into all that, here are the rules of the game:

• Two players are both dealt 26 cards from a standard deck.

• From there, they take turns putting down one card at a time. Let’s interchangeably designate the players as “A” and “B.”

• If A puts down a jack, then B puts down one card. If it’s a face card, the game continues with A. Otherwise, A wins the round and adds those cards to the bottom of their pile so that the last card played is at the very bottom of the pile.

• If A puts down a queen, B has two chances to put down a face card, at which point the game would continue with A. Otherwise, A wins the round and collects the cards.

• If A puts down a king, B has three chances to put down a face card, at which point the game would continue with A. Otherwise, A wins the round and collects the cards.

• If A puts down an ace, B has four chances to put down a face card, at which point the game would continue with A. Otherwise, A wins the round and collects the cards.

• If, at any point, two cards have been played in succession (by different players or the same player) that have the same number or face, then both players can slap the pile of played cards to win the round and collect the cards.

• Whoever won the previous round plays the first card in the next round.

Now look. I really like this opponent and I don’t want to hurt their feelings. And so, while shuffling, I rig the deck so that they get a majority of the jacks, a majority of the queens, a majority of the kings, and a majority of the aces.

Nobody likes a cheater, and so my opponent rightfully accuses me of rigging the game. In my defense, I say that the probability of one player randomly receiving the majority of each kind of face card is reasonable. What is this probability?

Extra Credit

It turns out that rigging the deck while shuffling is very hard to do. Instead, I decide to cheat in my opponent’s favor by ensuring my reaction time is slower than theirs at every single slapping opportunity. In other words, I lose every slap.

Assuming I shuffle and distribute the cards randomly, and that my opponent plays the first card in the game, what is my probability of winning?

Making the Rounds

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing a fun statistics-related puzzle I saw on social media earlier this week. Here’s the link if you want to see the original statement, but to avoid spoilers, I’m repeating the puzzle below, with minor edits:

Toss a coin five times and look at the flips that immediately follow a heads. If there are none, repeat the experiment with another five tosses.

If you repeat this experiment many, many times, then among the flips immediately following a heads, what is the average proportion of these flips that are also heads? In particular, is this proportion less than 0.5, equal to 0.5, or greater than 0.5? 

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Eric Snyder 🎻, from Everett, Washington. I received 72 timely submissions, of which 55 were correct—good for a 76 percent solve rate!

Last week, you looked at the factorial number system, a “mixed-radix” system in which the Nth digit from the right has a maximum allowable digit of N−1. As a result, the rightmost digit is always 0, the digit to its left can be 0 or 1, the next digit can be 0, 1, or 2, and so on. This is called the factorial number system because the place value of the N^th digit is (N−1)!.

I also provided an example that was incorrect in my original posting, and was thereafter corrected (sorry about that!). Suppose a number was written 103210 in the factorial number system. In base 10, this number was 1(5!) + 0(4!) + 3(3!) + 2(2!) + 1(1!) + 0(0!), or 143. (Note that the numbers in bold were exactly the digits from the factorial number representation.)

What was the smallest number in the factorial number system that was divisible by the whole numbers from 1 through 5, while also containing all of the digits 0 through 5? (Note that this number was allowed to have more than one copy of any given digit, and it could contain additional unspecified digits, like 0.)

Before digging in, I wanted to highlight one question that several readers asked: Did the number have to be divisible by 1 through 5 in base 10, or in the factorial base? The answer was that it doesn’t matter. If a number A is divisible by a number B, then it doesn’t matter what base you use—A will always be divisible by B. For example, the quantity 15 is divisible by the quantity 5. That’s in base 10, of course. In base 2, you’d say that the quantity 1111 is divisible by the quantity 101. But they mean the same thing: You can always evenly split 15 things among 5 groups (or groups of 5, if you prefer).

Now, back to the puzzle. It was tempting to start with the smallest number in the factorial number system that included all six digits, which was 543210. In base 10, this was 5(5!) + 4(4!) + 3(3!) + 2(2!) + 1(1!) + 0(0!), or 719. Alas, this was not divisible by 2, 3, 4 or 5. Instead of trying out small numbers with all six digits, it was more efficient to look for numbers that were divisible by 1, 2, 3, 4, and 5, and then ensure all six digits were present.

All whole numbers are divisible by 1, so that didn’t require any extra thought. To be divisible by 2 (i.e., to be even), the number had to end with the digits 00, since the place values of all digits to the left of them were also even. Similarly, to be divisible by 3, the number had to end with the digits 000, since the place values of all the digits to their left were also divisible by 3.

From there, things got a little trickier. To be divisible by 4, the number certainly could have ended with the digits 0000. Then, to be divisible by 5, the number would have had to end with the digits 00000. So, a reasonably small number with the desired properties was 1234500000 in the factorial number system, or 462,120 in base 10. However, there was a smaller number.

Let’s return to divisibility by 4. Sure enough, all factorials starting with 4! are multiples of 4, whereas 3! is not a multiple of 4. Nevertheless, 3! has a factor of 2, so placing a 2 in the corresponding place value will still result in a multiple of 4. More explicitly, the factorial number 2000 is 2(3!) + 0(2!) + 0(1!) + 0(0!), or 12, in base 10, which is indeed divisible by 4.

From there, you still needed the rightmost five digits to give you a number that was divisible by 5. If the fifth digit from the right was A, then you had A2000, or 24A+12 in base 10. This was divisible by 5 when A was 2 (i.e., 22000 is 60 in base 10).

And that was it! As long as your number ended in 22000, it was guaranteed to be divisible by 1, 2, 3, 4, and 5. The smallest such number that included all the digits from 0 to 5 was 134522000 in the factorial number system. (If you answered with “13452200” due to my errant initial posting of the puzzle, I still marked you as correct.) As promised, I allowed you to submit your answer in base 10 as well, which was 1(8!) + 3(7!) + 4(6!) + 5(5!) + 2(4!) + 2(3!) + 0(2!) + 0(1!) + 0(0!), or 58,980.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Tom Keith 🎻, from Toronto, Canada. I received 58 timely submissions, of which only 33 were correct—good for a 57 percent solve rate, the second lowest ever for an Extra Credit around here.

For Extra Credit, you had to find the smallest number in the factorial number system that contained each of the digits 1 through 9 exactly once, and which was also divisible by all of the whole numbers from 1 through 9. (Note that this number was allowed to contain any amount of zeros.)

As I noted in last week’s Fiddler, to have the number be divisible by 1 through 5 the last five digits had to be 00000 or 22000. Here, because no duplication of the digits from 1 through 9 was permitted, that ruled out 22000, which meant the last five digits had to be 00000.

Since 5! is divisible by 6, that meant the sixth rightmost digit could have been anything. But for the seven rightmost digits to give you a number that was divisible by 7, things again got a little tricky. Sure, the last seven digits could have been 0000000. But if these digits were the more generic AB00000, then you wanted 720A + 120B to be divisible by 7. In other words, you wanted 120 · (6A + B) to be divisible by 7, which was true if and only if 6A + B was divisible by 7. And that was only true if A and B were the same digit. (In particular, 3300000 was divisible by all the numbers from 1 to 9.) However, since repeated nonzero digits were prohibited, that meant the last seven digits really were 0000000.

Because 7!, 8!, and all higher factorials are already divisible by the numbers from 1 through 9, that meant all the preceding digits could have been anything, or at least anything allowable under the factorial number system. While it might have been tempting to submit an answer of 1234567890000000, that 9 wasn’t allowed in the 7! place. The smallest number with the desired properties turned out to be 1234569870000000. I asked for the number in factorial base only, since writing it out as approximately 1.5 trillion in base 10 wasn’t particularly interesting in my opinion. Again, due to the error in the original posting of the puzzle, I also accepted answers of “1234567000000” (i.e., with six trailing zeros rather than seven).

However, I did not accept answers like 12459876330000 or 359876142200, since they contained duplicate digits. (You’ve got to read the directions, people!)

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.