At some point during a game, your opponent had at least a 90 percent chance of winning. Given this information, how likely was it that you pulled out the victory?
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. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.
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 recently saw a link to an ESPN graphic that got me thinking about comebacks in sports:
If you follow the NBA, then you probably know Boston Celtics had several improbable comebacks against the Indiana Pacers in this year’s playoffs. But given that the Pacers had a “90 percent chance or higher” to win at some point in a game, did that mean their probability of winning that game was actually 90 percent?
Let’s explore this with a toy (i.e., simplified) version of basketball.
Suppose you’re playing a game in which there are five “possessions.” For each possession, there’s a 50 percent chance that your team scores one point. If you don’t score, then your opponent instead scores one point.
After the game, ESPN reports that your opponent’s chances of winning were “75 percent chance or higher” at some point during the game (i.e., before the final possession is complete).
Given this information, what was the probability that your team actually won the game?
For Extra Credit, instead of five possessions, now suppose there are 101. Again, with each possession, there’s a 50 percent chance that your team scores one point. If you don’t score, then your opponent instead scores one point.
After the game, ESPN reports that your opponent’s chances of winning were “90 percent chance or higher” at some point during the game (i.e., before the final possession is complete).
Given this information, what was the probability that your team actually won the game?
There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. Last week, I was the guest lecturer at the American Regions Mathematics League (ARML), one of the largest in-person high school math competitions in the country.
After speaking to about 300 students the night before the competition, I stuck around and got a chance to relive some of my own ARML experiences from yesteryear. (I soon realized, however, that the last time I competed was before any of this year’s competitors had even been born.)
This week, I’m sharing Problem 6 from the individual round, which was as follows:
In the left figure below, the cell in the fourth row and third column of a 9 × 9 grid is deleted to obtain a set S of 80 cells. For each n = 1, 2, 3, …, 8, an L-shaped tile Ln consisting of 2n+1 cells is given, as shown in the figure at the right.
Compute the number of ways to tile S using these eight L-shaped tiles, each exactly once, with rotations of tiles allowed.
Pretty tricky, right? Now imagine that you have to solve this in 10 minutes or less, without the aid of a computer or calculator. Oh, and there’s another problem you also have to solve within those same 10 minutes that involves polynomial division. Yikes!
For even attempting this—let alone solving one or both problems in the time limit—these kids at ARML were pretty amazing!
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.
Last Week’s Fiddler
Congratulations to the (randomly selected) winner from last week: 🎻 Steve Curry 🎻 from Albuquerque New Mexico. I received 50 timely submissions, of which 41 were correct—good for an 82 percent solve rate.
Last week, you started with a line segment of length 1, and you randomly split it somewhere along its length into two parts. You computed the product of these two lengths. Then, you took each of the two resulting segments and repeated the process. That is, for each one, you randomly split it somewhere along its length into two parts and computed the product. Next, you did this for all four resulting segments, then the eight after that, and the 16 after that, and so on.
After doing this (forever), you added up all the products you computed throughout. On average, what value did you expect this sum to approach?
Before getting too rigorous, solver Robin Johannsen explored some different cuts to see what happened. For example, suppose each segment was always cut in half. That meant the first product was (1/2)2, or 1/4. The next two products were both (1/4)2, meaning their sum was 2·(1/16), or 1/8. The next four products were all (1/8)2, meaning their sum was 4·(1/64), or 1/16. As you can see, each successive group of products had half the combined sum of the previous group. So by cutting each segment in half, the total sum was 1/4 + 1/8 + 1/16 + 1/32 + …, a geometric series. A convergent geometric series with a first term of a and a common ratio of r converges to the value a/(1−r). Plugging in 1/4 for a and 1/2 for r meant this series converged to (1/4)/(1−1/2), which simplified to 1/2.
What if the cuts had instead been a third of the way along each segment, rather than halfway? This time, the first product was (1/3)·(2/3), or 2/9. The next two products were (1/9)·(2/9) and (2/9)·(4/9), the sum of which was 10/81. The next four products were (1/27)·(2/27), (2/27)·(4/27), (2/27)·(4/27), and (4/27)·(8/27), the sum of which was 50/729. This time around, each successive group of products had 5/9 the combined sum of the previous group. So by cutting each segment a third of the way along, the total sum was 2/9 + 10/81 + 50/729 + 250/6561 + …, another geometric series. Plugging in 2/9 for a and 5/9 for r meant this series converged to (2/9)/(1−5/9), which simplified to 1/2.
Wait a minute! Whether we split the segments at the halfway point or a third of the way, it made no difference. Either way, the sum of the products approached 1/2. Was it possible that all this talk of randomized cuts and computing averages was a red herring—that the sum always approached 1/2?
To figure this out, many solvers turned to calculus. (Note: If you want to skip the calculus entirely, you should still check out the geometric solution further down!) Some calculus arguments were more rigorous than others, but my personal favorite was a scaling argument put forth by solvers including Josh Silverman and Jim Gu.
For starters, it made good sense that, for a stick of length L, we could expect the sum of all the products to approach f(L) = kL2 on average, where k was some constant. In other words, multiplying pairs of lengths together should result in something proportional to L2.
Now, after the very first cut, the average product was:
Next, you could take the expected value of both sides:
\(k = \frac{1}{6} + ka^2 + k\left(1-a\right)^2\)
Integrating both sides of this equation with respect to a from 0 to 1 gave you the rather nice result that k = 1/2. And so, according to this line of reasoning, you expected the sum of the products to approach 1/2 (on average), which was in agreement with the trial results we found above. Meanwhile, solver Emily B Zhang showed you could get the same result even if you didn’t assume that f(L) was proportional to L2.
But what I loved most about this puzzle was that you could show the answer was always 1/2, no matter how you randomly split the segments, and that you could do this geometrically, without any calculus.
How? By drawing an infinitude of smaller and smaller rectangles. Consider the right isosceles triangular region below x + y = 1 with positive x and y, illustrated below by 🎬 Chris Connett 🎬.
The first time you split the segment, say, into lengths a and 1−a, the product of the resulting lengths is the area of the red rectangle in Chris’s diagram. On either side of this rectangle are two new right isosceles triangular regions, one with legs of length a and the other with legs of length 1−a. When each of the two new segments is split, the products of these newly generated lengths are the areas of the orange rectangles. The next four products are the four light green rectangles, followed by the eight neon green rectangles, and so on. As the segments multiply in number, the combined areas of all the rectangles is confined by (and approaches!) the area of the entire triangle, which is, of course, 1/2.
Several solvers generated similar diagrams, including 🎬 Sameer G 🎬 and 🎬 Michael Schubmehl 🎬, who animated the segments splitting and the resulting rectangles that progressively filled up the triangular region.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Sameer G 🎻 from Ithaca, New York. I received 18 timely submissions, of which 17 were correct—good for a 94 percent solve rate.
Another way to describe last week’s Fiddler was with a recursive function f, defined by f(L) = ab + f(a) + f(b). Here, a and b were random values between 0 and L, such that a + b = L. The question posed by last week’s Fiddler was this: On average, what value did f(1) approach?
For Extra Credit, you split segments into three parts rather than two. I also introduced the recursive function g, defined by g(L) = abc + g(a+b) + g(a+c) + g(b+c) − g(a) − g(b) − g(c). Here, a, b, and c were random values between 0 and L such that a + b + c = L.
On average, what value did g(1) approach?
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