Skip to main content
Menu
How Many Soldiers Do You Need To Beat The Night King?

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 Tom Hanrahan, three colorful journeys:

In grade school, you may have learned about the three primary colors — blue, yellow and red — and the three secondary colors — green (blue + yellow), purple (red + blue) and orange (yellow + red).

And now it’s time to put that knowledge to use. Try to get through the maze below, a nine-by-nine grid of lines, three times: once as blue, once as yellow, and once as red.

If you are blue, you may only travel on lines that include the color blue. So you may travel on lines that are blue, green, purple or white (which contains all colors). You may not travel on orange, yellow, red or black (which contains no colors). The analogous rules hold for your trips as yellow and red.

In all three cases, you are attempting to travel between the same two points on the maze’s edge. Send me links, pictures or descriptions of your strategy!

Submit your answer

Riddler Classic

From Greg Burnham, it had to happen eventually, at long last and not a moment too soon, The Riddler meets “Game of Thrones”:

At a pivotal moment in an epic battle between the living and the dead, the Night King, head of the army of the dead, raises all the fallen (formerly) living soldiers to join his ranks. This ability obviously presents a huge military advantage, but how big an advantage exactly?

Forget the Battle of Winterfell and model our battle as follows. Each army lines up single file, facing the other army. One soldier steps forward from each line and the pair duels — half the time the living soldier wins, half the time the dead soldier wins. If the living soldier wins, he goes to the back of his army’s line, and the dead soldier is out (the living army uses dragonglass weapons, so the dead soldier is dead forever this time). If the dead soldier wins, he goes to the back of their army’s line, but this time the (formerly) living soldier joins him there. (Reanimation is instantaneous for this Night King.) The battle continues until one army is entirely eliminated.

What starting sizes of the armies, living and dead, give each army a 50-50 chance of winning?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Stuart Tooley 👏 of Edinburgh, United Kingdom, winner of last week’s Riddler Express!

Last week I gave you the following sequence of numbers and asked you what number came next?

2
6
10
3
8
9
4
7
?

The correct missing number is 8.

The numerical solution actually had more to do with letters, and with my favorite game, Scrabble. Starting at the top of the list: The number 2 is spelled “two,” which is worth six points if spelled with Scrabble tiles, so 6 becomes the next number in the list. The number 6 is spelled “six” which is worth 10 points in Scrabble, so 10 comes next in the list, and so on. To find the missing number, we write 7 as “seven” and tally that it’s worth eight points, so 8 is our answer.

Solution to last week’s Riddler Classic

Congratulations to 👏 Brian Hare 👏 of Raleigh, North Carolina, winner of last week’s Riddler Classic!

Last week we were introduced to five brothers who had joined the Riddler Baseball Independent Society, or RBIs. Each of them enjoyed a career of 20 seasons, with 160 games per season and four plate appearances per game.2 Given that their batting averages were .200, .250, .300, .350 and .400, what were each brother’s chances of beating DiMaggio’s 56-game hitting streak at some point in his career? (Streaks could span across seasons.)

Their chances of besting DiMaggio were, respectively, approximately 0, 0, 0.01, 0.8 and 13.9 percent. Put another way, their chances were roughly 1-in-9,000,000,000, 1-in-3,000,000, 1-in-8,000, 1-in-130 and 1-in-7.

To get there, first we need to compute the chances that each brother get at least one hit in any given game. Suppose a brother’s batting average is A. The chances that that brother does not get a hit in a given game is (1-A)^4, because he gets four at bats. So the chances a brother does get a hit in any given game is 1 minus that. That gives the brothers a chance of 0.5904, 0.683594, 0.7599, 0.821494 and 0.8704.

We then need to turn those individual game chances into streak chances. To put a bit of mathematical structure on this problem, we are looking for the probability \(P\) that at least \(r\) consecutive games with a hit appears in a sequence of total length \(n\) games given a probability \(p\) of a hit in each game. In our case, \(r=57\), \(n=3,200\) (20 seasons of 160 games), and we calculated \(p\) in the paragraph above.

As solver Michael Branicky explained, this probability, assuming the number of games played is large enough to contain a sufficiently long streak, must satisfy the following recursion:

\begin{equation*} P_{n+1} = P_n + ( 1 – P_{n-r} ) (1-p) p^r \end{equation*}

This is because a streak of length \(r\) in \(n+1\) games either occurred in \(n\) games, or occurs for the first time in game \(n + 1\). From there, the calculation can be done with a bit of computer code. Michael also illustrated the chances of breaking the streak by a player’s batting average:

As it happened, the brothers also had a cousin with a whopping .500 average, but he would get banned from the league after 10 seasons after testing positive for performance enhancers. What were his chances of beating the streak? Despite the shortened career, they were about 93.3 percent, an answer we can arrive at with the same approach described above.

As solver Chris Jones concluded, “Moral of the story: Doping pays off for record-breaking, bros.”

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 me at oliver.roeder@fivethirtyeight.com.

CORRECTION (May 17, 2019, 10:48 a.m.): An earlier version of the Riddler Express displayed the wrong color for one line. The leftmost horizontal bar in the fourth row from the top should be white, not black. The graphic has been updated.

Footnotes

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

  2. To make this simple, we assumed that each plate appearance resulted in a hit or an out, so there were no sacrifices or walks to complicate the math.

Oliver Roeder is a senior writer for FiveThirtyEight.

Comments