Can You Make The Fidget Spinner Go Backwards?
Illustration by Guillaume Kurkdjian
Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter or send me an email.
Riddler Express
My son noticed that when he held a fidget spinner in front of a television (with a 60 Hz refresh rate) and gave it a whirl, it appeared to suddenly spin backwards a few times before coming to a halt. While many fidget spinners have three lobes, this particular spinner had five lobes, as shown below.
After giving it a spin, we clearly saw it spin backwards three times before it stopped. How fast could it have been spinning at the beginning?
Extra credit: Upon closer examination, we also saw it spin backwards another three times, but in these cases it appeared to have twice as many lobes (i.e., 10). Now how fast could it have been spinning at the beginning?
Riddler Classic
From Roberto Linares comes a puzzle that will have you shouting βBingo!β:
A thousand people are playing LoterΓa, also known as Mexican bingo. The game consists of a deck of 54 cards, each with a unique picture. Each player has a board with 16 of the 54 pictures, arranged in a 4-by-4 grid. The boards are randomly generated, such that each board has 16 distinct pictures that are equally likely to be any of the 54.
During the game, one card from the deck is drawn at a time, and anyone whose board includes that cardβs picture marks it on their board. A player wins by marking four pictures that form one of four patterns, as exemplified below: any entire row, any entire column, the four corners of the grid and any 2-by-2 square.
After the fourth card has been drawn, there are no winners. What is the probability that there will be exactly one winner when the fifth card is drawn?
Solution to last weekβs Riddler Express
Congratulations to π Carl Schweppe π of Medford, Massachusetts, winner of last weekβs Riddler Express.
Last week, I had to cut a rug. After many years of using my favorite rug as a putting green, a narrow section in the middle had to be excised. The original rug was 12 feet long and 9 feet wide, while the middle strip was 8 feet long and 1 foot wide, as shown below.
Upon seeing the state of the rug, my neighbor suggested I cut it into two pieces and sew them back together to form a square rug, 10 feet by 10 feet, with no holes (shown below).
How was this possible?
To plug the hole in the middle, you could use 8 square feet from one of the adjacent columns. But that created a new hole, which could be filled by 6 square feet from the next column over. Then that was replaced by 4 square feet and finally 2 square feet. The end result called for two stair-shaped cuts on either side of the hole. Solver Glade Roper recorded a video of the rugβs rearrangement. Way to save that rug, Glade!
Thereβs not much more to say about this puzzle, other than the appreciation I feel for all the solvers who showed their work using ASCII art. Jenny Mitchell was smart to use square emojis with different colors:
πͺπͺπͺπͺπͺπͺπͺπͺπͺ
πͺπͺπͺπͺπͺπͺπͺπͺπͺ
π¨πͺπͺπͺβ¬πͺπͺπͺπͺ
π¨πͺπͺπͺβ¬πͺπͺπͺπͺ
π¨π¨πͺπͺβ¬π¨πͺπͺπͺ
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π¨π¨π¨π¨β¬π¨π¨π¨πͺ
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π¨π¨π¨π¨π¨π¨π¨π¨π¨
π¨π¨π¨π¨π¨π¨π¨π¨π¨
π¨πͺπͺπͺπͺπͺπͺπͺπͺπͺ
π¨πͺπͺπͺπͺπͺπͺπͺπͺπͺ
π¨π¨πͺπͺπͺπ¨πͺπͺπͺπͺ
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Solution to last weekβs Riddler Classic
Congratulations to π Michael Ringel π of Jacksonville, Florida, winner of last weekβs Riddler Classic.
Last week I was celebrating the birthday of a family member, which got me wondering about how likely it was for two people in a room to have the same birthday.
Suppose people were walking into a room, one at a time. Their birthdays happened to be randomly distributed throughout the 365 days of the year (and no one was born on a leap day). The moment two people in the room had the same birthday, no more people entered the room and everyone inside celebrated by eating cake, regardless of whenever that common birthday was.
On average, what was the expected number of people in the room when they ate cake?
This puzzle was a variation of the famed birthday problem, which asks how many people must be in a room for there to be at least a 50 percent chance that at least two of them have the same birthday. Here, the answer is paradoxically (to some, at least) small. With just 23 people in the room, thereβs a 50.7 percent chance that at least two of them have the same birthday. (Of course, this calculation makes the same assumptions as last weekβs puzzle β that there are 365 days in a year, and each day is equally likely to be someoneβs birthday.)
But rather than being asked for the number of people such that the probability exceeded 50 percent, you had to find the expected number of people such that two of them had the same birthday. While the answer was unlikely to again be exactly 23, it was surely close to 23.
Solver Adam Davitt found an exact expression for the expected number of people. First, the probability of eating cake when there was one person in the room was zero. Easy enough! The probability that two people in the room had the same birthday was 1/365 β the first person could have had any birthday, and the second person happened to have the same birthday. To eat cake with three people, the first person could again have any birthday, the second person had to have a different birthday, and the third person had to have either the first or second personβs birthday. The probability of this happening was (365/365) Γ (364/365) Γ (2/365).
In general, the probability of eating cake with N people was 365 Γ 364 Γ β¦ Γ (365βN+1) Γ (Nβ1)/365N. To find the expected number of people, you had to multiply each of these probabilities by N and then add them all up, for all values of N from 2 to 366. In the end, it turned out that the expected number of cake-eaters in the room was about 24.617, only slightly more than the 23 people from the birthday problem.
Solver Laurent Lessard noted that the famed mathematician Ramanujan previously found an excellent approximation for the sum in this problem. The answer turns out to be quite close to β(365π/2) + 2/3, or about 24.611. So you see, this puzzle really was about both cake and pi!
For extra credit, you assumed everyone ate cake the moment three people in the room had the same birthday. On average, what was this expected number of people?
This was closely related to a previous riddle that extended the classic birthday problem to three people. To have at least a 50 percent chance that three people in a room had the same birthday, you needed a total of 88 people in the room. So the answer to the extra credit should have been somewhere around 88. Sure enough, solver Ian Walker found the answer was very close: approximately 88.739 people.
Want more puzzles?
Well, arenβt you lucky? Thereβs a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. Itβs called βThe Riddler,β and itβs in stores now!
Want to submit a riddle?
Email Zach Wissner-Gross at riddlercolumn@gmail.com.
