Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here: math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world — including you! You’ll find this week’s puzzle below.
Mull it over on your commute, dissect it on your lunch break, and argue about it with your friends and lovers. When you’re ready, submit your answer using the link below. I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: To be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday — have a great weekend!
Before we get to the new puzzle, let’s return to last week’s. Congratulations to 👏 Mia Allende 👏 of Moorestown, New Jersey, our big winner. You can find a solution to the previous Riddler at the bottom of this post.
Now here’s this week’s Riddler, a summertime sports puzzle that comes to us from Nick Keenan of Washington, D.C.
Assume you have a sport (let’s call it “baseball”) in which each team plays 162 games in a season. Assume you have a division of five teams (call it the “AL East”) where each team is of exact equal ability. Specifically, each team has a 50 percent chance of winning each game. What is the expected value of the number of wins for the team that finishes in first place?
Extra credit: Let’s offer up a 🏆 Coolest Riddler Extension Award 🏆. Complicate the sport, change up the division, alter the teams, find out what happens if your team’s general manager walks around in a gorilla suit, or something even more creative. Submit your extension and its solution via the form below, or shoot me a link to your work on Twitter. The winner gets a shiny emoji trophy next week.
Submit your answer
Need a hint? You can try asking me nicely. Want to submit a new puzzle or problem? Email me.
And here’s the answer to last week’s Riddler, concerning the slaying of monsters and pilfering of their gems, which came to us from Brandon Hensley. I’ve adapted his solution here.
Given the rules of the game, you will collect an average of 3.65 common gems in the course of collecting at least one each of a common, uncommon and rare gem.
This problem can be viewed as having eight distinct states: You either have or do not have each of the three types of gem at any given time (that’s how we get to eight states: 2x2x2=8). To see all the permutations, denote each state by a three-digit binary number, e.g. 101, where the first digit indicates whether you’ve collected the common gem, the second the uncommon gem, and the third the rare gem. (“1” means you have the gem.) The goal is to get from state 000 to state 111. Let \(N_x\) denote the average number of common gems you would collect in going from state \(x\) to state 111. Since we’re wondering what happens when we start empty handed, the goal of this problem is to compute \(N_{000}\).
Since the probabilities of a monster dropping each of the three types of gems are in the ratio of 3:2:1, and must add up to 1, the probabilities are 1/2, 1/3 and 1/6. We are now equipped to consider each state case by case.
First, \(N_{111}=0\), as in you should expect zero average common gems going forward, since you are finished and won’t be collecting any more gems. In state \(N_011\), you are missing only the common gem. Upon slaying a monster, you have a 1/2 chance of collecting the missing gem and finishing the game and a 1/2 chance of returning back to \(N_{011}\). Thus:
$$N_{011}=\frac{1}{2}(1+N_{111})+\frac{1}{2}N_{011}=1$$
This makes sense — if you reach state \(N_{011}\), you haven’t collected any common gems and so will end the game by collecting your first one.
We can perform a similar analysis on \(N_{101}\) — there is a 1/2 chance of collecting a common gem and returning to the same state, a 1/3 chance of collecting the missing uncommon gem and ending the game, and a 1/6 chance of collecting the rare gem and returning to the same state:
$$N_{101}=\frac{1}{2}(1+N_{101})+\frac{1}{3}N_{111}+\frac{1}{6}N_{101}$$
$$=\frac{3}{2}$$
Finally, we can proceed similarly for \(N_{110}\):
$$N_{110}=\frac{1}{2}(1+N_{100})+\frac{1}{3}N_{110}+\frac{1}{6}N_{111}$$
$$=3$$
Now we can consider the cases in which only one gem has been collected:
$$N_{100}=\frac{1}{2}(1+N_{100})+\frac{1}{3}N_{110}+\frac{1}{6}N_{101}$$
$$=\frac{1}{2}(1+N+{100})+1+\frac{1}{4}$$
$$=\frac{7}{2}$$
$$N_{010}=\frac{1}{2}(1+N_{110})+\frac{1}{3}N_{010}+\frac{1}{6}N_{011}$$
$$=2+\frac{1}{3}N_{010}+\frac{1}{6}$$
$$=\frac{13}{4}$$
$$N_{001}=\frac{1}{2}(1+N_{101}+\frac{1}{3}N_{011}+\frac{1}{6}N_{001}$$
$$=\frac{5}{4}+\frac{1}{3}+\frac{1}{6}N_{001}$$
$$=\frac{19}{10}$$
And so finally:
$$N_{000}=\frac{1}{2}(1+N_{100})+\frac{1}{3}N_{010}+\frac{1}{6}N_{001}$$
$$=\frac{9}{4}\frac{13}{12}+\frac{19}{60}$$
$$=\frac{73}{20}=3.65$$
Thus, on average, you will end the game with 3.65 of the most common gem!
Reader Arthur Harris has this alternate, fanciful, anecdotal solution:
It is intuitively obvious that the answer must be greater than 3 and less than 4. A simple average of those two numbers would be 3.5, but that would be too easy. The geometric mean would be 3.46, but nobody learns that stuff anymore so that is an unlikely result. I then made a dartboard with numbers between 3.46 and 3.5. I blindfolded my son and he threw the dart with all of his strength. It flew out the open window and struck a car with the license plate TGX365. Using this divine inspiration, I concluded that the answer must be 3.65.
And, lest anyone still accuse us citizens of #RiddlerNation of being a band of philistine quants, I received this lovely Riddler poem (along with a very tidy solution) this week:
You were right, Andy, you were right. And I will ride into this weekend on the wings of your verse. May you all do the same. Happy weekend!
CORRECTION (May 27, 8:40 a.m.): An earlier version of this article misidentified a reader who submitted a solution to last week’s Riddler. It was Arthur Harris, not Matt Fox.