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

Consider the following figure, which shows 16 points arranged in a rhombus shape and connected to their neighbors by edges. How many distinct paths are there from the top point to the bottom along the edges such that:

• You never visit the same point twice.

• You only move downward or sideways—never upward.

Submit your answer

This Week’s Extra Credit

Consider the following figure, which shows 30 points arranged in a three-dimensional triangular bipyramid. As before, points are connected to their neighbors by edges. How many distinct paths are there from the top point to the bottom along the edges such that:

• You never visit the same point twice.

• You only move downward or sideways—never upward.

Submit your answer

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 a monthly challenge I made over at Amplify, called Contour Golf. It’s a mini-golf game (built inside the Desmos calculator and Activity Builder) with contour maps. The first hole is pretty easy, but the later ones are tough! Once you’re done, you can design your own hole and try ones others have made. (Here’s the link if you want to run the activity in your own classroom.)

By the way, there’s another monthly challenge from Amplify that’s a digital escape room—also built entirely in the Desmos calculator. Do you have what it takes to escape? (I think you do!)

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.)

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 David Slater 🎻 from Okemos, Michigan. I received 70 timely submissions, of which 66 were correct—good for a 94 percent solve rate.

Recently, on the ESPN show, “First Take,” the discussion turned to the NBA’s New York Knicks, who were facing the favored Boston Celtics in a best-of-seven Eastern Conference Semifinals. The question was whether the Knicks were more likely to be “swept” (i.e., lose the series in four games) or for the series to go to seven games. Here's what sports personality Stephen A. Smith had to say:

I got [the Knicks] losing this in five games, which means they’re closer to a sweep than a seven-game series. That’s how I’m looking at it right now.

Last week, you analyzed the first part of Stephen’s statement, that the Knicks would lose to the Celtics in five games.

Let p represent the probability the Celtics won any given game in the series. You assumed that p was constant (so there was no home-court advantage) and that games were independent.

For certain values of p, the likeliest outcome was indeed that the Celtics would win the series in exactly five games. While this probability was always less than 50 percent, this outcome was more likely than the Celtics winning or losing in some other specific number of games. In particular, this range could be specified as a < p < b.

What were the values of a and b?

A great way to start was to work out the probabilities for winning the series in various numbers of games. Yes, this was a whole bunch of work, but it provided insight into the puzzle and paid off down the road.

First off, what was the probability that the Celtics would win in four straight games (a sweep)? That was p⁴. Not too tricky, right?

Next, what was the probability that the Celtics would win the series in five games? Winning in five meant they won four games, with probability p⁴, and lost one game, with probability (1−p). Also, there were five (i.e., 5 choose 1) ways these games could have been ordered (such as WWLWW, LWWWW, etc.). So the probability the Celtics won four out the first five games was 5·p⁴·(1−p). However, to win the series in five games, they couldn’t have won the first four, which was one of the five orderings we just considered. So the probability they won the series in five games (and not four) was in fact 4·p⁴·(1−p).

Moving on, what was the probability that the Celtics would win the series in six games? In this case, they again won four games, with probability p⁴, and lost two games, with probability (1−p)². This time, there were 15 (6 choose 2) ways these games could have been ordered. However, to win the series in six games, they couldn’t have won four out of the first five games. Another way to say this was that they had to win the sixth game. There were 10 (5 choose 2) ways to order two losses among the first five games, so the probability they won the series in six games was 10·p⁴·(1−p)².

Finally, what was the probability that the Celtics would win the series in seven games? Here they won four games, with probability p⁴, and lost three games, with probability (1−p)³. Remember, they had to win the final game, and there were 20 (6 choose 3) ways to order three losses among the first six games. So the probability they won the series in seven games was 20·p⁴·(1−p)³.

As for the Knicks, the formulas for their winning in four, five, six, or seven games were the same as the formulas we found for the Celtics, but with the probabilities swapped. That is, if you replaced p with 1−p and vice versa, you had expressions for the Knicks’ probabilities.

Now that we’ve worked out all these cases, let’s see what they looked like. Solver 🎬 Tom Keith 🎬 graphed them:

It’s tough to tell from the graph because of the dimensionless vertical axis (it’s just “probability”), but the chances the Celtics (or the Knicks, for that matter) won in exactly five games were always less than 50 percent. In other words, 4·p⁴·(1−p) was never greater than 0.5 for values of p between 0 and 1. (That explained some of the careful wording in the original puzzle.) The maximum occurred when p was 0.8, in which case the Celtics would win in five games 32.768 percent of the time.

Getting back to the puzzle, you were charged with finding when the Celtics winning in five games was the most likely outcome. Based on the graph above, as p increased, at some point the Celtics winning in five games overtook winning in six games as the most likely outcome. They were equal when the expression for winning in six games, 10·p⁴·(1−p)², was equal to the expression for winning in five games, 4·p⁴·(1−p). Setting these equal and dividing both sides by p⁴·(1−p) gave you 10·(1−p) = 4. Solving gave you p = 0.6. This was the value of a, the lower bound where winning in five games was the likeliest outcome.

As p increased further, winning in five games was eventually overtaken by winning in four as the likeliest outcome. This occurred when the expression for winning in four games, p⁴, was equal to the expression for winning in five games, 4·p⁴·(1−p). Setting these equal and dividing both sides by p⁴ gave you 1 = 4·(1−p). Solving gave you p = 0.75. This was the value of b, the upper bound where winning in five games was the likeliest outcome.

Putting these results together, winning in five games was the likeliest outcome when 0.6 < p < 0.75. A few solvers misread the problem and interpreted p as the probability the Knicks (rather than the Celtics) won each game. This resulted in the range 0.25 < p < 0.4, for which I still awarded full credit.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Jason Weisman 🎻 from Wilton, Connecticut. I received 60 timely submissions, of which 43 were correct—good for a 72 percent solve rate.

For Extra Credit, you analyzed the rest of Stephen A. Smith’s statement. Was it true that losing in five games was “closer to a sweep than a seven-game series”?

Let p₄ represent the probability that the Celtics would sweep the Knicks in four games. And let p₇ represent the probability that the series would go to seven games (with either team winning).

Suppose p was randomly and uniformly selected from the interval (a, b) from last week’s Fiddler, meaning we took it as a given that the most likely outcome was that the Knicks would lose the series in five games. How likely was it that p₄ was greater than p₇? In other words, how often was it the case that probably losing in five games meant a sweep was more likely than a seven-game series?