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.
Here’s a puzzle that hopefully isn’t too complex:
Can you find three distinct numbers such that the second is the square of the first, the third is the square of the second, and the first is the square of the third? Assuming you can, what are three such numbers?
Extra credit: Can you find three other numbers with the same property?
The solution to this Riddler Express can be found in the following column.
From Fred Blundun comes a simple game of definitive diffidence:
Martina and Olivia each secretly generate their own random real number, selected uniformly between 0 and 1. Starting with Martina, they take turns declaring (so the other can hear) who they think probably has the greater number until the first moment they agree. Throughout this process, their respective numbers do not change. So for example, their dialogue might go as follows:
Martina: My number is probably bigger.
Olivia: My number is probably bigger.
Martina: My number is probably bigger.
Olivia: My number is probably bigger.
Martina: Olivia’s number is probably bigger.
They are playing as a team, hoping to maximize the chances they correctly predict who has the greater number.
For any given round with randomly generated numbers, what is the probability that the person they agree on really does have the bigger number?
Extra credit: Martina and Olivia change the rules so that they stop when Olivia first says that she agrees with Martina. That is, if Martina says on her turn that she agrees with Olivia, that is not a condition for stopping. Again, if they play to maximizing their chances, what is the probability that the person they agree on really does have the bigger number?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Last week, you were playing a game of cornhole with your friends, and it was your turn to toss the four bean bags. For every bean bag you tossed onto your opponents’ board, you got 1 point. For every bag that went through the hole on their board, you got 3 points. And for any bags that didn’t land on the board or through the hole, you got 0 points.
Your opponents had had a terrible round, missing the board with all their throws. Meanwhile, your team currently had 18 points — just 3 points away from victory at 21. You were also playing with a special house rule: To win, you had to score exactly 21 points, without going over. (Also, you were clearly playing a variation such that one team threw all four bean bags before the other, rather than alternating tosses.)
Based on your history, you knew there were three kinds of throws you could have made:
- An aggressive throw, which had a 40 percent chance of going in the hole, a 30 percent chance of landing on the board and a 30 percent chance of missing the board and hole.
- A conservative throw, which had a 10 percent chance of going in the hole, an 80 percent chance of landing on the board and a 10 percent chance of missing the board and hole.
- A wasted throw, which had a 100 percent chance of missing the board and hole.
For each bean bag, you could have chosen any of these three throws. Your goal was to maximize your chances of scoring exactly 3 points with your four tosses. What was the probability that your team would finish the round with exactly 21 points and declare victory?
Suppose you started with an aggressive throw. Then there was a 40 percent chance the bag went in the hole, in which case you could have wasted your remaining three throws. But there was also a 30 percent chance the bag landed on the board, in which case you needed to score 2 more points in your remaining three throws. Conservative throws were the best way to get those points. There was a 64 percent chance your first two throws landed on the board, a 6.4 percent chance your first and third throws landed on the board with a miss in between and another 6.4 percent chance your first throw missed while your last two landed on the board. So when your first throw landed on the board, that contributed 30 percent times 76.8 percent — or 23.04 percentage points — to your overall chances of winning.
But wait! There was a 30 percent chance that your first throw missed the board entirely. In this case, you had to score 3 points with your remaining three throws. If you were aggressive, you still had a 40 percent chance of winning on your second throw. But if that throw landed on the board (30 percent), you could still get 2 more points with conservative throws (64 percent). And if that second throw missed as well (30 percent), you could stay aggressive so that either your third or fourth throw went in the hole (40 percent plus 12 percent, or 52 percent). So with an aggressive second throw, you still had a 74.8 percent chance of winning.
Alternatively, your second throw could have been conservative. If so, it had a 10 percent chance of going on. If it landed on the board, that meant each of your final three throws had to land on the board for a (0.8)3, or 51.2 percent, chance of winning. But if your second throw missed (10 percent), you were then better off playing aggressively on your third throw, with a 52 percent chance of winning. So a conservative second throw gave you a 10 percent plus 51.2 percent plus 5.2 percent — or 66.4 percent chance — of winning. That was less than 74.8, so if you missed the board on your first throw, your second throw should also have been aggressive, contributing 30 percent times 74.8 percent — or 22.44 percentage points — to your overall chances of winning.
Putting all of this together, if your first throw was aggressive, then you would win with that throw 40 percent of the time, win by landing on the board 23.04 percent of the time and win despite missing the board 22.44 percent of the time. Overall, your chances of reaching exactly 21 points stood at 85.48 percent.
Now, what would have happened if you had kicked things off with a conservative throw, rather than an aggressive throw? First off, you would (accidentally) have scored 3 points 10 percent of the time. Meanwhile, for the 80 percent of the time your throw landed on the board, you needed 2 points in three throws, which we said had a 76.8 percent chance of happening. Finally, for the 10 percent of the time you missed the board, you needed 3 points in three throws, which we said had a 74.8 percent chance of happening. All together, this meant your probability of winning was 0.1 + (0.8)(0.768) + (0.1)(0.748), or 78.92 percent.
The last time I checked, 85.48 was greater than 78.92, which meant your best bet was to start with an aggressive throw. This gave you an 85.48 percent chance of scoring exactly 21 points.
That was a lot of casework. Several solvers, like Chris Smith and David Diamondstone, organized their calculations into a table with the number of points needed and throws remaining. By solving simpler versions of the problem with smaller numbers, they worked their way up (via a dynamic programming approach) to the given problem, which called for 3 points in four throws.
By the way, if you need a reminder of what it’s like being outside playing cornhole, here’s a photo from a neighborhood cornhole tournament, courtesy of solver Betts Slingluff.
Solution to last week’s Riddler Classic
Congratulations to 👏 Praveen Anumolu 👏 of Smithtown, New York, winner of last week’s Riddler Classic.
Last week, you were presented with a crossword puzzle title “Systematic Solving.” It had math:
Here was the solution, courtesy of the puzzle’s creator, Elise Corbin:
There were several themed clues, including 24-Across, which was the equation y = x−3, and 52-Across, which was the equation y = 2x+1. These two lines were visible in the puzzle as the circled coordinates, assuming the dead center of the grid was the origin of the coordinate plane.
Meanwhile, these two lines intersected in the southwest corner of the grid (also known as Quadrant III of the coordinate plane), at the point (−4, −7). These coordinates were part of the clues 37-Down and 70-Across.
Want more riddles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email Zach Wissner-Gross at email@example.com.