Hector's Puzzled

Now that Halloween has come and gone, your chances of getting free candy have similarly disappeared. Desperate for sugar, you wander into the candy store, where three kinds of candy are being sold: Almond Soys (yummy, sounds vegan!), Butterflingers and Candy Kernels. You’d like to buy at least one candy and at most 100, but you don’t care precisely how many you get of each or how many you get overall. So you might buy one of each, or you might buy 30 Almond Soys, six Butterflingers and 64 Candy Kernels. As long as you have somewhere between one and 100 candies, you’ll leave the store happy. But as a member of Riddler Nation, you can’t help but wonder: How many distinct ways are there for you to make your candy purchase? [Solution]

I want to carve the perfect eye for my pumpkin this Halloween, but I can’t seem to make it the right size. Since symmetry is the key to beauty, the triangular eye should be equilateral and equiangular, which is easier said than done on the surface of a spherical pumpkin! To be a proper triangle, the three corners should be connected by arcs that are “straight,” meaning they go directly from one corner to the next via the shortest possible path. (Think about air travel: When you flying from the West Coast of the U.S. to Europe, you’ll fly north of the Arctic Circle along the way, since that’s the shortest path over the Earth’s curved surface.) I want an eye that’s one-sixteenth of the pumpkin’s surface. For such an ideal pumpkin eye, at what angle should each of the sides meet? [Solution]

The classic birthday problem asks about how many people need to be in a room together before you have better-than-even odds that at least two of them have the same birthday. Ignoring leap years, the answer is, paradoxically, only 23 people — fewer than you might intuitively think. But Joel noticed something interesting about a well-known group of 100 people: In the U.S. Senate, three senators happen to share the same birthday of October 20: Kamala Harris, Brian Schatz and Sheldon Whitehouse. And so Joel has thrown a new wrinkle into the classic birthday problem. How many people do you need to have better-than-even odds that at least three of them have the same birthday? (Again, ignore leap years.) [Solution]

You’ve made it to the $1 million question, but it’s a tough one. Out of the four choices, A, B, C and D, you’re 70 percent sure the answer is B, and none of the remaining choices looks more plausible than another. You decide to use your final lifeline, the 50:50, which leaves you with two possible answers, one of them correct. Lo and behold, B remains an option! How confident are you now that B is the correct answer? [Solution]

You are the coach for Team Riddler at the Tour de FiveThirtyEight, where there are 20 teams (including yours). Your objective is to win the Team Time Trial race, which has the following rules: Each team rides as a group throughout the course at some fixed pace, specified by that team’s coach. Teams that can’t maintain their pace are said to have “cracked,” and don’t finish the course. The team that finishes the course with the fastest pace is declared the winner. Teams ride the course one at a time. After each team completes its attempt, the next team quickly consults with its coach (who assigns a pace) and then begins its ride. Coaches are aware of the results of all previous teams when choosing their own team’s pace. Assume that all teams are of equal ability: At any given pace, they have the exact same probability of cracking, and the faster the pace, the greater the probability of cracking. Teams’ chances of cracking are independent, and each team’s coach knows exactly what a team’s chances of cracking are for each pace. Team Riddler is the first team to attempt the course. To maximize your chances of winning, what’s the probability that your team will finish the course? What’s the probability you’ll ultimately win? Extra Credit: If Team Riddler is the last team to attempt the course (rather than the first), what are its chances of victory? [Solution]