Can You Sweep the Series?

A famous sports personality recently commented that winning a best-of-seven series in five games is closer to a sweep than going to seven games. But what does the math say?

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

Sports personality Stephen A. Smith said something earlier this week that caught my attention—for mathematical reasons!

On the ESPN show, “First Take,” the discussion turned to the NBA’s New York Knicks, who would be facing the favored Boston Celtics in the best-of-seven Eastern Conference Semifinals. (Note that this segment aired prior to the beginning of the series.) 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 Stephen 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.

Let’s look at the first part of Stephen’s statement, that he believed the Knicks would lose to the Celtics in five games.

Let p represent the probability the Celtics win any given game in the series. You should assume that p is constant (which means there’s no home-court advantage) and that games are independent.

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

Determine the values of a and b.

This Week’s Extra Credit

Now that you’ve determined the values of a and b, let’s analyze the rest of Stephen’s statement. Is it true that losing in five games is “closer to a sweep than a seven-game series”?

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

Suppose p is randomly and uniformly selected from the interval (a, b), meaning we take it as a given that the most likely outcome is that the Knicks will lose the series in five games. How likely is it that p₄ is greater than p₇? In other words, how often will it be the case that probably losing in five games means a sweep is more likely than a seven-game series?

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 Sheetdoku, a game whose creators describe as “[combining] elements of Sudoku with spreadsheet formulas.” I would say it more closely resembles KenKen, but with a different set of operations. Either way, it’s pretty fun!

Here’s one of the daily puzzles from earlier this week:

Like a 5-by-5 KenKen, the whole numbers from 1 through 5 must appear exactly once in each row and column. Then, within each walled region, the corresponding spreadsheet factoid must also be true. For example:

• PROD(20) indicates that the product of cells C1, D1, and E1 must be 20. Since you are given that E1 = 4, that means C1 and D1 must be 1 and 5, in some order. Since D4 = 5, that means D1 = 1 and C1 = 5.

• MIN(1) indicates that the minimum of cells A1, B1, and A2 must be 1. Since D1 = 1, it must be A2 that equals 1. At this point, the first row has everything but a 2, so B1 = 2.

And so it goes. If you enjoy this puzzle, there’s a lot more where that came from, even though the game is still apparently in beta.

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: 🎻 Andrew Ford 🎻 from Madison, Wisconsin. I received 48 timely submissions, of which 42 were correct—good for an 87.5 percent solve rate.

I was recently a guest at Disney World, which had a new system called “Lightning Lane” for reserving rides in advance—for a fee, of course.

By purchasing “Lightning Lane Multi Pass,” you could reserve three of the many rides in a park, with each ride occurring at a different hourlong time slot. You happened to be visiting Magic Kingdom, which had 12 such time slots, from 9 a.m. to 9 p.m. So if you had the 3 p.m. time slot for a ride, then you can skip the “standby lane” and board the much shorter “Lightning Lane” at any point between 3 and 4 p.m. You could complete at most one ride within each hourlong time slot.

Once you completed the first ride on your Multi Pass, you could reserve a fourth ride at any time slot after your third ride. This way, you always had at most three reservations. Similarly, after you completed your second ride, you could reserve a fifth ride at any time slot after your fourth, and so on, up until you were assigned a ride at the 8 p.m. (to 9 p.m.) time slot. That was your final ride of the day.

Magic Kingdom happened to be very busy at the moment, so each ride was randomly assigned a valid time slot when you requested it. The first three rides of the day were equally likely to be in any of the 12 time slots, whereas subsequent rides were equally likely to occur in any slot after your currently latest scheduled ride.

On average, how many rides could you have expected to “Lightning Lane” your way through in one day at Magic Kingdom?

Thanks to the “magic” of Multi Pass (see what I did there?), you were guaranteed to go on at least three rides. And the maximum number of rides, of course, was 12. So the answer was somewhere between these extremes.

To get a handle on the situation and estimate the average number of rides, a few solvers simulated many, many days at Magic Kingdom. The histogram below shows the distribution for how many rides you went on in 1 million such simulations:

Most of the time you only got to go on three, four, or five rides. It was rather neat how the probability of going on three rides and the probability of going on five rides were each precisely 25 percent. Anyway, the average of these 1 million simulations was 4.2712 rides. Personally, I expected a greater figure. After all, you were guaranteed a minimum of three rides, so getting to go on an additional 1.2712 rides didn’t seem too impressive.

Let’s see if we can confirm this result analytically. To do that, we’ll split the problem into two parts: (1) reserving your first three rides, and then (2) iteratively reserving additional rides.

Let’s tackle the second part first. So you’ve already been on at least three rides, and now you’re booking another ride after what’s currently your latest scheduled ride. To keep things simple, let’s define the ride times as being 1 (9 a.m. to 10 a.m.) through 12 (8 p.m. to 9 p.m.).

• If your current last side was in slot 11, then you were guaranteed to go one final ride in slot 12. Huzzah!

• If your current last ride was in slot 10, then there was a 50 percent chance your next ride was in slot 12 (meaning one more ride) and a 50 percent chance your next ride was in slot 11 (meaning two more rides). On average, you went on 1 + 1/2, or 1.5 more rides.

• If your current last ride was in slot 9, then there was a 1-in-3 chance your next ride was in slot 12 (meaning one more ride), a 1-in-3 chance your next ride was in slot 11 (meaning two more rides), and a 1-in-3 chance your next ride was in slot 10 (meaning one more ride, plus an average of 1.5 more). On average, you went on 1 + 1/2 + 1/3, or about 1.83 more rides.

Continuing this trend, which you could prove using induction, if your current last ride was in slot n, then on average you could expect to go on 1/1 + 1/2 + 1/3 + … + 1/(12−n) more rides.

At this point, we’re ready to return to the original three rides. What mattered here was when the last of these rides occurred. If that third ride occurred in slot n, then your expected number of rides was 3 + 1/1 + 1/2 + 1/3 + … + 1/(12−n). The last step was working out how often each of the 12 slots was to be the last of the initial three slots.

In total, there were 12 choose 3 (or 220) equally likely ways to choose the first three slots. Of these, how many had a third ride in the 11th time slot? There were 10 remaining slots for the first two rides, so the answer here was 10 choose 2. And how many ways had a third ride in the 10th time slot? There were nine remaining slots for the first two rides, so now the answer was 9 choose 2. In general, the probability the third ride was in slot n was n−1 choose 2 divided by 12 choose 3.

Putting this all together, the average number of rides you could expect to go on was given by the following formula:

\(\sum_{n=3}^{12}\frac{\binom{n-1}{2}}{\binom{12}{3}}\cdot\left(3+\sum_{m=1}^{12-n}\frac{1}{m}\right)\)

Quite the expression, right? This was saying you had to consider when the last of the initial three rides was in each of the slots from 3 to 12, compute the resulting expected number of rides (three plus the sum of the unit fractions), and weight that value by the probability the third ride was in that slot. Adding it all up resulted in a sum of 118361/27720, or approximately 4.27. That was indeed the answer, and was in agreement with the simulations mentioned above.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Kerry Brown 🎻 from Longmeadow, Massachusetts. I received 35 timely submissions, of which 30 were correct—good for an 86 percent solve rate.

This time around, after you completed the first ride on your Multi Pass, you could reserve a fourth ride at any time slot after your first ride (rather than after your third ride). Similarly, after you completed your second ride, you could reserve a fifth ride at any time slot after your second, and so on, until there were no available time slots remaining.

As before, the first three rides of the day were equally likely to be in any of the 12 time slots, whereas subsequent rides were equally likely to occur in any remaining available slots for the day.

On average, how many rides could you expect to “Lightning Lane” your way through in one day at Magic Kingdom?