Welcome to The Riddler. Most weeks, I offer up two problems related to the things we hold dear around here: math, logic and probability.

But this week is special. The time has flown, and somehow it’s been a year since I (peacefully) took over this column from my predecessor, Oliver Roeder.

It is only fitting that we continue one of Ollie’s (and my) favorite traditions here at The Riddler: the Battle for Riddler Nation. In order to have a chance at 👑 winning 👑 and becoming the next ruler, I need to receive your battle plans before 11:59 p.m. Eastern time on Monday. Have a great weekend!

## This week’s Riddler

Some readers may be familiar with the first, second, third and fourth Battles for Riddler Nation. If you missed out, you may want to consult the thousands of attack distributions from these previous contests.

I am pleased to say that this week marks the *fifth* such competition — but this time, the number of castles has changed!

In a distant, war-torn land, there are 13 castles. There are two warlords: you and your archenemy. Each castle has its own strategic value for a would-be conqueror. Specifically, the castles are worth 1, 2, 3, …, 12, and 13 victory points. You and your enemy each have 100 soldiers to distribute, any way you like, to fight at any of the 13 castles. Whoever sends more soldiers to a given castle conquers that castle and wins its victory points. If you each send the same number of troops, you split the points. You don’t know what distribution of forces your enemy has chosen until the battles begin. Whoever wins the most points wins the war.

Submit a plan distributing your 100 soldiers among the 13 castles. Once we receive all your battle plans, we’ll adjudicate all the possible one-on-one matchups. Whoever wins the most wars wins the battle royale and is crowned ruler of Riddler Nation!

Who can steal the crown from 👑 David Love 👑 of Ambler, Pennsylvania, who currently sits atop the throne? Also, don’t count out Vince Vatter of Gainesville, Florida, a two-time champion eager to take back what was rightfully his.

Will *you* defeat them?

Do you have the cunning and logic to be the next ruler of Riddler Nation?

The results to this Riddler can be found in the following week’s column.

## Solution to last week’s Riddler Express

Congratulations to 👏 Dan Speirs 👏 of Newtown Square, Pennsylvania, winner of last week’s Riddler Express.

Last week, you had a giant sheet that lay flat on the ground, covering the Earth (assumed to be a perfect sphere with a radius of 6,378 kilometers).

You wanted to raise the sheet so it was instead always 1 meter off the ground. To make it so, how much did you have to increase the area of the sheet?

You were essentially being asked to find the difference in surface area between two spheres — one with the radius of the Earth, or 6,378,000 meters, and another whose radius was one meter longer, or 6,378,001 meters.

Most readers knew the formula: A sphere with radius *r* has a surface area of 4𝜋*r*^{2}. So one approach was to precisely calculate the areas of the two spheres and subtract them.

You could also have worked through a little algebra before plugging anything in. If *R* is the radius of the Earth, then the difference in surface area was 4𝜋(*R*+1)^{2}−4𝜋*R*^{2}. After canceling out the squared terms, this difference became 8𝜋*R*+4𝜋 square meters, or approximately 160.3 million square meters. Since there are a million (not a thousand!) square meters in one square kilometer, this was equivalent to **160.3 square kilometers**.

In the grand scheme of things, the answer was a pretty tiny fraction of the Earth’s surface — a shade over 0.00003 percent, to be precise.

For extra credit, you had to identify a city, country, land mass or body of water whose area was very close to the answer. As far as countries went, Liechtenstein, with an area of 160 square kilometers, was the closest.

So when it comes time to grow our Earth-covering quilt, I think the most prudent thing to do would be to head to Liechtenstein and host a quilting bee. I’ll see you there!

## Solution to last week’s Riddler Classic

Congratulations to 👏 Richard Guidry Jr. 👏 of Baton Rouge, Louisiana, winner of last week’s Riddler Classic.

Last week, you wanted to play a very special game of War.

War is a two-player game in which a standard deck of cards is first shuffled and then divided into two piles with 26 cards each — one pile for each player. In every turn of the game, both players flip over and reveal the top card of their deck. The player whose card has a higher rank wins the turn and places both cards on the bottom of their pile. In the event that both cards have the same rank, the rules get a little more complicated, with each player flipping over additional cards to compare in a mini “War” showdown.

Assuming a deck was randomly shuffled before every game, how many games of War would you expect to play until you had a game that lasted exactly 26 turns, with no mini “Wars?”

As you might have guessed, such a “perfect” game of war is very rare. So rare, in fact, that trying to simulate it was a fruitless exercise. Instead, your best bet was to work out an analytical solution.

A good first step was to determine the probability *p* of having a perfect game of War (either for you or for your opponent — there was no such distinction in the problem). If we knew the value of *p*, then the probability of achieving one perfect game in exactly one attempt would be *p*, in exactly two attempts it would be (1−*p*)*p*, in exactly three attempts it would be (1−*p*)^{2}*p*, and so on. By combining these probabilities, the expected number of games until having a perfect one was then *p* + 2(1−*p*)*p* + 3(1−*p*)^{2}*p* + 4(1−*p*)^{3}*p* + …, an infinite arithmetico-geometric series whose sum was simply 1/*p*.

So all you had to do was find the probability of having a perfect game of War, and then compute the reciprocal of that number. Alas, determining this probability was easier said than done.

You could have tried a back-of-the-envelope calculation. If we put aside mini “Wars” for a moment, pretending that each player has a 50 percent chance of winning each turn outright, what’s the probability of your winning a perfect game? You would have to win all 26 turns, meaning your chances stood at 1/2^{26}, or approximately 1.49×10^{−8}. Pretty unlikely.

But that was still an estimate. For each turn, you actually had *less than* a 50 percent chance of winning outright, since there was a nonzero chance of a mini “War” in which their card had a matching rank, leaving you and your opponent to split the leftover probability. That meant 1.49×10^{−8} was an overestimate.

Finding the exact probability, it turned out, was an advanced exercise in combinatorics. Solver Laurent Lessard defined the probability as a function of the remaining number of cards of each rank. He then set up a recurrence relation and had his computer crunch the numbers via memoization.

Meanwhile, Peter Norvig similarly “abstracted” what a deck was, considering only how many cards there were of relative ranks (rather than their specific ranks), and then tracked the probability trees in which one player won all the rounds outright.

Both Laurent and Peter found that the probability of sweeping your opponent in 26 turns was approximately 3.1324×10^{−9} — not too far off from our back-of-the-envelope calculation. which meant you would expect to play about **319 million** games before winning in such a fashion. Solver Angela Zhou calculated this value more precisely, finding it was 29,908,397,871,631,390,876,014,250,000 divided by 93,686,147,409,122,073,121. Wow!

This was the expected number of games until *you* won in a sweep. The expected number of games until either you or your opponent won in a sweep was *half* as many (since the probability of either of you achieving this was twice as high), or about **159,620,171** games. The question was ambiguous as stated, so I gave credit for both answers.

Finally, if you enjoyed this riddle, you might try your hand at a few similar combinatorial challenges, courtesy of solver Eric Farmer. You’ll find boring snaps, Frustration Solitaire and prohibited subwords. Sounds like just another day at The Riddler.

## 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 Zach Wissner-Gross at riddlercolumn@gmail.com