Every winter, when the deep snow comes, making going out uncomfortable and unsafe, we open a jigsaw puzzle to distract us from that beautiful white carpet outside. Our daughter drives over in her four-wheel-drive with the express purpose of putting together the border. This winter I was afraid the deep snow would never come. When finally enough snow arrived in February, it stayed around so long that we could have done two or three puzzles.
Meanwhile, I have come up with three entertaining jigsaw-puzzle puzzles for you. (Some may feel that the title of this column is entertaining enough.) In puzzles 1 and 2, we assume ordinary rectangular jigsaw puzzles, with rows and columns running cleanly from one edge to the opposite edge.
Puzzle 1. A jigsaw‑puzzle manufacturer wants to make puzzles with at least 1,000 pieces. Among those puzzles she wants the total number of pieces to be as small as possible. She also wants the ratio of the number of columns to the number of rows to be as close as possible to the Golden Ratio. She agrees that minimizing the sum of the percentage deviations from her two targets would be appropriate. Determine the number of rows and columns she should have in her puzzles.
Puzzle 2. A jigsaw puzzle has n edge pieces, where n is an even integer. How many pieces can the puzzle have altogether? For example, if n=14, then the puzzle could have 14, 18, or 20 pieces. (Your answers should be in terms of n.)
Puzzle 3. So far in this column, I have written “jigsaw” followed immediately by “puzzle” six times. Half the time I used a hyphen and half the time I didn’t. Provide the reason for hyphen use and non-use.
Solutions may be emailed to puzzles@actuary.org. In order to make the solver list, your solutions must be received by June 1, 2026. Providing the correct answer earns partial credit; full credit requires a proof, explanation, or argument, as appropriate.
Solutions to Last Issue’s Puzzles—Hole-In-One
By Josh Feldman
- What is the probability that Hoss does not make a single hole-in-one in a round? For the easiest six holes, Hoss’ probability of not making a hole-in one is (1- 1/3) = 0.666. For the toughest six holes the probability is (1 – 1/20) = 0.95. For the remaining six holes it is (1 – 1/6) = 0.8333. As there are six holes of each type, the overall probability is (0.666)6 * (0.95)6 * (0.8333)6 = 0.0216.
- What is the probability that Hoss gets seven or more holes-in-one during his first round? There are two ways to approach this problem. The first is to figure out the probability of making exactly six easy difficulty holes-in-one, one medium difficulty hole-in-one, and no hard difficulty hole-in-one. This is just (0.333)6 * 6 * (0.1666) * (0.8333)5 * (0.95)6 = 0.000405. Continue doing this for every combination, sum the results, and you will get a probability of 0.0264. Or, one can program in Excel the probability of getting a hole in one on each of the 18 holes, and run a quick simulation to figure out the number of holes in one for a round. Once you run the simulation once, running it multiple times is as easy as copying rows. To get the number of days in the challenge, just take the reciprocal of 0.0264 which equals approximately 37.8 days.
- What is the probability that Hoss gets eight or more holes-in-one on the first time he breaks the challenge? We could use Excel for this part as well, or we could build up our results. If we know the probability of getting exactly 0, 1, 2, … 7 holes-in-one during a round, we know the remaining probability leads to 8 or more holes-in-one. Doing the math (or simulation) leads to a 24.5% chance that this occurs.
Solvers: Anthony Salis, Al Spooner, David Promislow, Bob Conger, Bill Feldman, Jason Shaw, Cindy Hu, Michael Schachet, Sam Ellis, Jerry Miccolis, Daniel Wade, Douglas Levy, Rui Guo, Anna Quady, and Clive Keatinge.