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. The “Extra Credit” is where the analysis typically gets 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
From Bowen Kerins comes a chance at free money:
A casino offers you $55 worth of “free play vouchers.” You specifically receive three $10 vouchers and one $25 voucher.
You can play any or all vouchers on either side of an even-money game (think red vs. black in roulette, without those pesky green pockets) as many times as you want (or can). You keep the vouchers wagered on any winning bet and get a corresponding cash amount equal to the vouchers for the win. But you lose the vouchers wagered on any losing bet, with no cash award. Vouchers cannot be split into smaller amounts, and you can only wager vouchers (not cash).
What is the guaranteed minimum amount of money you can surely win, no matter how bad your luck? And what betting strategy always gets you at least that amount?
Hint: You can play vouchers on both sides of the even money game at the same time!
This Week’s Extra Credit
From Bowen Kerins also comes some Extra Credit:
You have the same $55 worth of vouchers from the casino in the same denominations. But this time, you’re not interested in guaranteed winnings. Instead, you set your betting strategy so that you will have at least a 50 percent chance of winning W dollars or more. As before, you cannot split vouchers and cannot wager cash.
What is the maximum possible value of W? In other words, what is the greatest amount of money you can have at least a 50 percent chance of winning from the outset, with an appropriate strategy? And what is that betting strategy?
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 fun read from Scientific American in which the following question is posed:
If I have some number of sticks with random lengths between 0 and 1, what are the chances that no three of those sticks can form a triangle?
As described in the article, the answer turns out to have a surprising connection to the Fibonacci sequence.
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 correct solutions to Fiddlers and Extra Credits are worth 1 point each. Solutions should be sent prior to 11:59 p.m. the Monday after puzzles are released. 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: 🎻 Kerry Brown 🎻 from Longmeadow, Massachusetts. I received 68 timely submissions, of which 57 were correct—good for an 84 percent solve rate.
Last week, all the many attendees at a particular Coldplay concert were couples. As the CEO of Astrometrics, Inc., you were in attendance with your romantic partner, who was definitely not the head of HR at Astrometrics, Inc. During the concert, the two of you spent half the time canoodling.
The camera operators loved to show people on the jumbotron during the concert, but time was limited and there were many attendees. As a result, the camera operators showed just 1 percent of couples during the concert. Couples were chosen randomly, but never repeated at any given concert.
You and your partner were shy when it came to public displays of affection. While you didn’t mind being shown on the jumbotron, you didn’t want to be shown canoodling on the jumbotron.
How many Coldplay shows could the two of you have expected to attend without having more than a 50 percent chance of ever being shown canoodling on the jumbotron?
First, let’s figure out the chances of your canoodling appearing on the jumbotron at one concert. There was a 1 percent chance you’d appear on the jumbotron at all, and, if you did, there was a 50 percent chance that you were canoodling at the time. So the probability of you being “caught” canoodling on the jumbotron at one concert was (1/100)·(1/2), or 1/200—that is, 0.5 percent.
How could you use this fact to determine your chances of being caught if you attended multiple concerts? If you attended N concerts, you could work out how likely it was for you to be caught once. But you could also be caught twice, three times, four times, all the way up to N times. There were many cases to consider, and so another strategy was called for. Instead, you could focus on the probability of never being caught across multiple concerts.
You had a 99.5 percent chance of not being caught at one concert. So the probability of not being caught at N concerts was (0.995)N. Thus, the probability of being caught (at least once) one minus this, or 1−(0.995)N. The puzzle was asking for the greatest value of N such that this probability didn’t exceed 50 percent. As an inequality, you wanted the greatest value of N such that you still had 1−(0.995)N ≤ 0.5.
You could plug in greater and greater values of N until the expression on the left ultimately exceeded 0.5. But here, let’s solve this inequality directly. With some rearrangement, you got (0.995)N ≥ 0.5, meaning your probability of not being caught remained at least 50 percent. Next, you could take the logarithm of both sides, which gave you N·log(0.995) ≥ log(0.5). To solve for N, you next had to divide both sides by log(0.995), which was a negative value. That meant you had to flip the inequality sign, giving you N ≤ log(0.5)/log(0.995), or N ≤ ~138.28.
Therefore, you could attend 138 concerts and still have only a 49.93 percent chance of being caught canoodling, just shy of 50 percent.
If you gave an answer of 138 or 139 (or anywhere in between), I still awarded full credit. Yes, the answer to the puzzle as worded was technically 138, but I wasn’t too picky when it came to finding the most concerts without a 50 percent chance of being caught versus the first concert where this probability finally broke 50 percent.
Several readers did miss the important detail that you only spent half the concert canoodling. They assumed you spent the entire time canoodling, resulting in an answer of 68 concerts. But that was not the puzzle, as no sensible CEO spends an entire concert canoodling with their head of HR—I mean, someone who is definitely not the head of HR.
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Mike Strong 🎻 from Mechanicsburg, Pennsylvania. I received 43 timely submissions, of which 38 were correct—good for an 88 percent solve rate.
Now, everyone at the concert spent at least some time canoodling. In particular, each member of a couple wanted to spend a particular fraction of the time canoodling, where this fraction was randomly and uniformly selected between 0 and 1. This value was chosen independently for the two members of each couple, and the actual time spent canoodling was the product of these values. For example, if you wanted to canoodle during half the concert and your partner wanted to canoodle during a third of the concert, you actually canoodled during a sixth of the concert.
Meanwhile, the camera operators loved to show canoodling couples. So instead of randomly picking couples to show on the jumbotron, they randomly picked from among the currently canoodling couples. (The time shown on the jumbotron was very short, so a couple’s probability of being selected was proportional to how much time they spent canoodling.)
Looking around the concert, you noticed that the kinds of couples who most frequently appeared on the jumbotron weren’t constantly canoodling, since there were very few such couples. Indeed, the couples who most frequently appear on the jumbotron spent a particular fraction C of the concert canoodling. What was the value of C?
