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 if you have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

From Max Weinreich, a phantasmal puzzle:

Twenty ghostbusters are on their annual camping retreat. Two of them, Abe and Betty, have discovered that another pair, Candace and Dan, are in fact ghosts posing as ghostbusters. Abe and Betty hatch a plan: When all 20 campers are sitting in a circle around the campfire, Abe will fire his proton pack at Candace, and Betty will simultaneously fire her proton pack at Dan, annihilating the ghosts. However, if two proton streams cross, it means the end of all life on Earth.

If the ghostbusters are arranged randomly around the fire, what are the chances that Abe and Betty will cross streams?

## Riddler Classic

This week’s Classic, from Spreck Rosekrans, continues our camping theme. Here are four questions of increasing difficulty about finding sticks in the woods, breaking them and making shapes:

- If you break a stick in two places at random, forming three pieces, what is the probability of being able to form a triangle with the pieces?
- If you select three sticks, each of random length (between 0 and 1), what is the probability of being able to form a triangle with them?
- If you break a stick in two places at random, what is the probability of being able to form an acute triangle — where each angle is less than 90 degrees — with the pieces?
- If you select three sticks, each of random length (between 0 and 1), what is the probability of being able to form an acute triangle with the sticks?

## Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ Ian Leavitt ÑÑâÐ of Denver, winner of last week’s Express puzzle and the champion War-rior of Riddler Nation!

For a simplified game of War against a random opponent, you had to decide how to arrange your set of 13 cards: 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king and ace. In the one-round game, you go through all 13 of your cards once, with you and your opponent drawing a card off the top of your decks and comparing them. (Your opponent has the same cards.) If your card outranks your opponent’s, you get a point. After all 13 cards have been shown, the player with the most points wins. No points are awarded for ties. But there’s a catch: To enter the tournament, you must first beat me, the house. My cards are in this order: 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. I, being the house and all, get an advantage: I win ties. Plus, I can choose to play my deck forward or backward. Your deck must be able to beat both of my decks.

Over the weekend, 1,349 of you submitted decks. Of those, 1,278 were valid decks, and 1,163 beat my house deck both forward and backward, successfully earning entry into the tournament.^{2} (Twitter user @sl2c calculated that about 10 percent of all possible decks beat mine.) I matched each of these decks against every other — and the War results are in.

Our winner, Ian, described his approach this way: “Like an illusion, it just happened.” My best ideas come to me in the shower, so, Ian, I totally get it.

## Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ Keith HudsonÑÑâÐ of Nashville, winner of last week’s Classic puzzle!

In a standard, two-player, 52-card game of War, if I start with four aces and you start with the 48 other cards, randomly shuffled, your chances of winning are about **19 percent**. My chances, starting with the four aces, are about 81 percent.

At first, it might seem impossible to lose with all four aces — aces can’t be beat, after all. But later-in-the-game wars are what make this assumption problematic. If I take one of your fives, for example, with one of my aces early in the game, I will have to play that five later. At that point, it could match with one of your fives, leading to war. In that war, I have to place one of my cards^{3} face down, one of which could be an ace, and if I get unlucky, I could lose it to you. If that happens often enough, you could overtake me.

Ultimately, this is problem best tackled with computer simulations. Solvers turned to Perl, Python, R and C++. Julian Norton, Zack Segel, Jonathan Whitmore and Hector Pefo were kind enough to share their code.

Our winner, Keith, shared the following “pseudocode,” guiding us intuitively through the process that he used to create his simulations:

while neither deck is empty: { take the top card off of both decks and put them "on the table" if the cards are different: { shuffle the cards on the table put the table cards in the winner's desk } else { if neither deck is empty: { take the top card off of both decks and put them on the table } else { break } // end neither deck empty } // end cards are different } // end while calculate win-loss stats based on which deck is empty

How long can you expect all this warring to take? Hector shared a visual representation of his simulations, showing what share of the total number of games ended after a given number of turns flipping cards. He did this for a standard, fairly dealt game as well as our one-player-all-aces version:

There are slight variations in the rules of War that led to slightly varying results. Do you put one or two (or even three, as I did as a kid) cards face down during a war? Are the cards that are won in a war added randomly back to the bottom of a deck or can they be placed strategically? These led to small shifts in the probability, but regardless of the house rules, the starting ace-haver was always a heavy favorite.

In any event, get comfortable: You might be playing for a while.

## Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.