Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-sized and 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 next week’s column. If you need a hint, or if you have a favorite puzzle collecting dust in your attic, find me on Twitter.
In standard American bingo, a card is a five-by-five grid of squares. The columns are labeled B, I, N, G and O, in that order. The five squares in the B column can be filled with the numbers 1 through 15, those in the I column with the numbers 16 through 30, those in the N column 31 through 45, and so on. The square in the very center of the grid is a “free space” on every card.
How many different possible bingo cards are there?
From Joe Vanderlans, a current constitutional problem:
Imagine that U.S. Supreme Court nominees are only confirmed if the same party holds the presidency and the Senate. What is the expected number of vacancies on the bench in the long run?
You can assume the following:
- You start with an empty, nine-person bench.
- There are two parties, and each has a 50 percent chance of winning the presidency and a 50 percent chance of winning the Senate in each election.
- The outcomes of Senate elections and presidential elections are independent.
- The length of time for which a justice serves is uniformly distributed between zero and 40 years.
Extra credit: It’s been ages, so let’s offer up a 🏆 Coolest Riddler Extension Award 🏆. Make more realistic assumptions, complicate the model, concoct an interactive simulation or add your own creative twist. Submit your extension via the form below. The winner gets a shiny emoji trophy next week.
Solution to last week’s Riddler Express
Congratulations to 👏 Patty Kokesh 👏 of Valparaiso, Indiana, winner of last week’s Express puzzle!
Two bright and honest math students are sitting together at lunch when their teacher hands them each a card with an integer on it. The teacher tells them that the product of their two numbers is 12, 15 or 18 and that whoever guesses the other’s number first wins. The first student looks at her card and says, “I don’t know what your number is.” The second student looks at her card and says, “I don’t know what your number is either.” The first student then says, “Now I know your number.” What number is on the loser’s card?
The number is 6. To begin with, the possible numbers the students might have are 1, 2, 3, 4, 5, 6, 9, 12, 15 and 18. These are the numbers that could multiply with another in the set to equal 12, 15 or 18. When the first student says she doesn’t know the other’s number, that means the first student does not hold the numbers 4, 5, 9, 12, 15 or 18. Why? Because those numbers combine with only one other number to form the announced products, and if her card held one, she could’ve deduced her friend’s number. That means the second student now knows that the first student’s card has a 1, 2, 3 or 6 on it. And, in turn, that means the second student must hold a 2, 3, 4, 5, 6, 9, 12, 15 or 18, because those are the numbers that combine with the first student’s possible numbers to make 12, 15 or 18.
When the second student announces that she doesn’t know the first student’s number, that means the second student can’t be holding a 2, 3, 4, 5, 9, 12, 15 or 18, because each of those multiplies with only one of the first student’s remaining possibilities to equal 12, 15 or 18. (And if the second student’s card held one of those, she would’ve known the first student’s number.) Therefore, the second student must be holding a 6, which could combine with either the first student’s 2 or 3. The first student now knows this and wins the game.
Solution to last week’s Riddler Classic
Congratulations to 👏 Matthew Lightman 👏 of Chicago, winner of last week’s Classic puzzle!
You were asked to consider the following game. In front of you is a shuffled stack of 10 cards, each containing a whole number from zero to nine. Sight unseen, you draw a card from the stack and place it somewhere in the multiplication equation below.
You then draw another card, place it and so on, until all four spaces are filled. Once a card is placed, it cannot be moved. Your goal is to create an equation with the lowest product. What is the optimal strategy, and how much better is that strategy than simply placing the cards randomly?
Let’s start by calculating the expected product with a thoughtless, random assignment of digits. To give us some sense of scope, there are 5,040 ways that the digits could be drawn into the equation above. (First there are 10 cards available, then nine, then eight, then seven. 10*9*8*7 =5,040 possible combinations.) Call the digits in the top row AB and those in the second row CD. Their product then equals (10A + B)*(10C + D) = 100AC + 10AD + 10BC + BD.2 Relying on the linearity of expectation, we can simplify the expected product to E(121AC) or 121*E(AC). The expected product of two different single digits — our randomly drawn A and C — is 19⅓, so the expected product of a random placement strategy is about 2,339.3.
But certainly Riddler Nation can do better than that! And indeed it did. The best strategy gives an expected product of about 1,056.8, or about a 55 percent reduction from the product under the random strategy. This game has a heavy component of skill.
Intuitively, you’ll want to place the lower numbers you draw in the tens-place boxes, on the left, as they’ll do less damage to your final product there. Similarly, you’ll want to place higher numbers in the ones-place boxes, where they’ll hurt you less. Specifically, for the first card you draw, you do best to place it on the left side if it’s a 4 or lower (the lower half of the cards) and on the right side if it’s a 5 or higher (the higher half of the cards).
Your next two decisions are a bit more complicated because you’ve already begun to fill your equation and learn about the cards remaining in the stack. So those are best made with the help of a computer. Solvers Nick Hansen and Hector Pefo were kind enough to share the code they wrote to solve the problem. Hector summarized the best strategy he found for placing the second card, which is the intricate crux of the plan, in the diagram below. (LL means place the card in the lower left, UR in the upper right, and so on.) This strategy defies easy summary, but again, in general, the lower cards should go on the left and the higher cards on the right.
When you go to place your third card, if it’s higher than the mean of the cards that remain in the stack, put it in the tens slot if you can, and if it’s lower, put it in the ones slot. (You’ll have to place a fourth card, too, but you won’t have any choice in the matter — there’ll only be a single slot remaining.)
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