How Many Rabbits Can You Pull out of a Hat?

A hat contains a known number of orange, green, and purple rabbits. Rabbits are randomly pulled out one at a time, without replacement. For how many rabbits can you correctly guess the color?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “Extra Credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

I have a hat with six small toy rabbits: two are orange, two are green, and two purple. I shuffle the rabbits around and randomly draw them out one at a time without replacement (i.e., once I draw a rabbit out, I never put it back in again).

Your job is to guess the color of each rabbit I draw out. For each guess, you know the history of the rabbits I’ve already drawn. So if we’re down to the final rabbit in the hat, you should be able to predict its color with certainty.

Every time you correctly predict the color of the rabbit I draw, you earn a point. If you play optimally (i.e., to maximize how many points you get), how many points can you expect to earn on average?

This Week’s Extra Credit

Now, instead of two rabbits of each of the three colors, my hat contains 10. That is, it contains 10 orange rabbits, 10 green rabbits, and 10 purple rabbits. As before, every time you correctly predict the color of the rabbit I draw, you earn a point.

With optimal play, how many points can you expect to earn on average?

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing another Substack I came across, called A Piece of the Pi. There, Professor Richard Green discusses interesting mathematical topics—most recently dimer tilings—in casual longform.

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

I’m tracking submissions from paid subscribers and compiling a leaderboard, which I’ll reset every quarter. All timely correct solutions to Fiddlers and Extra Credits are worth 1 point each. At the end of each quarter, I’ll 👑 crown 👑 the finest of Fiddlers. (If you think you see a mistake in the standings, kindly let me know.)

Two months in, Michael Schubmehl sits alone at the top, with a perfect score. But will it stay that way?

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Jeffrey Ling 🎻 from Redwood City, California. I received 57 timely submissions, of which 53 were correct—good for a 93 percent solve rate.

A new season of LearnedLeague recently kicked off! Here’s how it works: Every day, you and your opponent for the day are presented with the same six trivia questions. You each do your best to answer these, and you assign point values for your opponent, without knowing (until the following day) which questions your opponent answered correctly. You must assign point values of 0, 1, 1, 2, 2, and 3 to the six questions.

For example, suppose I assigned values of 1, 2, 0, 3, 2, and 1 to the six questions, in order. Unbeknownst to me, my opponent answered the first, third, and fourth questions correctly. That meant they got 1 + 0 + 3, or 4 points for the match. My own score would depend on which questions I got right, and how these were scored by my opponent. If I got more than 4 points, I won the match.

Now, when someone answered three questions correctly, like my opponent just hypothetically did, the fewest points they could earn was 0 + 1 + 1 = 2, while the most points they could earn was 2 + 2 + 3 = 7. The fact that they got 4 points wasn’t great (from my perspective), but wasn’t terrible. In LearnedLeague, my defensive efficiency is defined as the maximum possible points allowed minus actual points allowed, divided by the maximum possible points allowed minus the minimum possible points allowed. Here, that was (7−4)/(7−2), which simplified to 3/5, or 60 percent.

By this definition, defensive efficiency ranges somewhere between 0 and 100 percent, provided it’s even defined (i.e., your opponent gets more than zero and less than six questions right).

Suppose you knew for a fact that your opponent would get two questions right. However, you had absolutely no idea which two questions these were, and so you randomly applied the six point values to the six questions.

What was the probability that your defensive efficiency for the day would be greater than 50 percent?

The two questions your opponent answered correctly were effectively random, and the point values you assigned were also random. To keep these independent, it was only necessary to randomize one set or the other. So, for the sake of simplicity, solver (and fellow LearnedLeaguer!) Austin Shapiro assumed you assigned point values to the six questions in ascending order: 0, 1, 1, 2, 2, 3. Meanwhile, the two questions your opponent answered correctly remained random. In total, there were 6 choose 2, or 15, ways your opponent could have done this.

Next, what did it mean to have a defensive efficiency greater than 50 percent? To figure this out, you needed to know the most and fewest points your opponent could have scored when answering exactly two questions correctly. The most points was 3 + 2, or 5. The fewest points was 0 + 1, or 1. So if they earned P points, then your defensive efficiency was (5−P)/(5−1), or (5−P)/4. For this to be greater than 50 percent, you needed (5−P)/4 > 1/2. Solving for P gave you the inequality P < 3. In short, as long as your opponent scored less than 3 points (i.e., they scored 1 or 2 points), your defensive efficiency exceeded 50 percent.

Of the 15 cases we mentioned above, in how many of them did your opponent score 1 or 2 points? What made this tricky was that two questions were worth 1 point and another two questions were worth two points. To better keep track of these point values, you could assign them unique labels: 0, 1_A, 1_B, 2_A, 2_B, 3.

There were two ways your opponent could earn 1 point: 0 + 1_A and 0 + 1_B. There were three ways your opponent could earn 2 points: 0 + 2_A, 0 + 2_B, 1_A + 1_B. Of the 15 total cases, your opponent earned fewer than three points is five of them. That meant the answer was 5/15, or 1/3.

With a little more work, you could show that the probability of achieving a defensive efficiency that was exactly 50 percent was also 1/3, and the probability of achieving an efficiency less than 50 percent was the remaining 1/3.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Brad Slavens 🎻 from Atlanta, Georgia. I received 48 timely submissions, of which 43 were correct—good for a 90 percent solve rate.

This time around, your opponent was equally likely to get one, two, three, four, or five questions correct.

As before, you randomly applied the six point values (0, 1, 1, 2, 2, 3) to the six questions.

What was the probability that your defensive efficiency would be greater than 50 percent?