Hark! Two Holiday Puzzles

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.

The holidays are upon us, and this week we’re offering up two games suitable to break out around the dinner table or in front of the fireplace. Please enjoy. The Riddler will be off next week. See you next year!

## Riddler Express

From Jeffrey Hope, open up your junk drawer and pull out those decks of cards:

Your challenge is to take any 50 cards from a standard 52-card deck and arrange them into 10 poker hands, one of each type from a royal flush down to a lowly high card.2 The hands you build always rank as the highest type of hand possible,3 and a card may not be reused across the hands. (This means, for example, that a four-of-a-kind better than four nines is impossible because it’d use a card you’d need to make a royal flush.) As in actual poker, it doesn’t matter what order the cards are arranged within a hand.

Sounds easy enough, and indeed there is more than one solution. But: Exactly how many solutions are there?

## Riddler Classic

From Bob Nasuti, grab your wallet and take a seat:

Consider a game of chance called Left, Right, Center. Everyone sits in a circle and begins with some \$1 bills. Taking turns, each person rolls three dice. For each die, if you roll a 1 or 2 you give a dollar to the person on your left, if you roll a 3 or 4 you give a dollar to the person on your right, and if you roll a 5 or 6 you put a dollar in the middle. The moment only a single person has any money left, the game ends and that person gets all the money in the center.

How long is the game expected to last for six players each starting with three \$1 bills? For X players each starting with Y \$1 bills?

## Solution to last week’s Riddler Express

Congratulations to 👏 Tom Loescher 👏 of Raleigh, North Carolina, winner of last week’s Riddler Express!

In our first-ever Riddler Express, I asked how high the “Sesame Street” character Count Von Count could count on Twitter, given its 140-character limit. The answer was 1,111,373,373,372, or, as the Count would write it, “One trillion one hundred eleven billion three hundred seventy three million three hundred seventy three thousand three hundred seventy two!” Since then, however, Twitter expanded its character limit to 280 characters, and last week I asked how high the Count could count now.

Now he can get to 101,373,373,373,373,373,373,372 or, in words, “One hundred one sextillion three hundred seventy three quintillion three hundred seventy three quadrillion three hundred seventy three trillion three hundred seventy three billion three hundred seventy three million three hundred seventy three thousand three hundred seventy two!” This number is nearly 100 billion times bigger than the answer with the shorter tweet length. If he tweets about two tweets per day, the Count won’t reach his maximum number until very long after every star in the universe has exhausted its fuel.

There’s no silver-bullet mathematical approach or magical equation that’ll get you to this answer — it’s more a problem of trial and error. One big clue: “Three” and “seven” are especially long digits to spell out, and “three hundred seventy three” is the longest three-digit number to spell out. (It’s tied with “seven hundred seventy seven,” but 373 is smaller, so we’ll have to contend with its length sooner in our counting.) Given that, we know that we need to bundle together those groups of 373s for as long as we can before hitting the character limit. Once we’ve beefed up the middle of our number with as many strings of 373 as possible, we begin to tick up those leading three digits — the “sextillions” place. We realize we can get that up to “101” and that we then get stuck at the last “372” in the ones place, and we’re done. Ah ah ah!

## Solution to last week’s Riddler Classic

Congratulations to 👏 Sam Bollier 👏 of Amherst, Massachusetts, winner of last week’s Riddler Classic!

Last week, I confessed my secret compulsion: trying to turn strings of digits I see in the world (license plates, ZIP codes) into true mathematical equations by inserting symbols such as addition, subtraction, multiplication, division and equals signs. It turned out I wasn’t alone, and many readers shared stories of playing their own, similar games. But how many strings that you see in the wild have a true mathematical equation lurking inside of them? And how does that answer change as the length of a string of digits increases? Is there some sufficiently long string such that every possible group of digits has a true equation inside?

This is a computational problem — there are too many possibilities and quirks in the combinations to handle it all by hand. You’ve got to tell your computer to grind through all the strings, inserting digits wherever possible and checking the validity of the resulting equation. Our winner, Sam, wrote a program in Python that iterated through every possible combination of digits, parentheses and mathematical operators. Those operators included the common +, -, *, /, ^, and =.

For two digits, for example, there are 100 total possibilities (the numbers 00 through 99) and only 10 percent of these have equations hidden within. These 10 percent are simply the 10 where you can insert an equals sign: 0=0, 1=1, 2=2 and so on. For three digits, things are a little more promising, and you can make equations with about a third of all possible strings. Things take off quickly from there, and you can make equations with about 80 percent of strings of length four (e.g. a PIN number) and 99 percent of strings of length five (e.g. a ZIP code). For strings of length six (e.g. a two-factor authentication code) you can find an equation 99.9852 percent of the time. For strings of length seven or longer (e.g. a phone number or a Social Security number) it appears you can always find an equation — with a single exception.

This chart got me wondering about those obnoxious strings of length six and seven that held no equations. What were they? There are just 148 out of the 1 million possibilities of length six that don’t work, and they range from a low of 295059 to a high of 980795. And just one string of length seven that doesn’t: 4870798. So that’s that: 4870798 is now officially my least favorite number.

## Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.

## Footnotes

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

2. The 10 types are royal flush, straight flush, four of a kind, full house, flush, straight, three of a kind, two pair, one pair and high card.

3. So, a 2-2-3-3-3 is a full house, not a pair, or two pairs, or a three of a kind.

Oliver Roeder is a senior writer for FiveThirtyEight.

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