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.
Riddler Express
This week’s Express is for anyone who ever thought deeply about functions in high school algebra.
I have a mystery function that passes through the points (0, 1), (1, 2), (2, 4), (3, 8) and (4, 16). I also know the function is continuous and smooth. In other words, you can draw it in a single stroke without any sharp corners or cusps.
At this point, you might think this function has to be f(x) = 2x. But I wouldn’t be too sure of that.
Can you find a continuous, smooth function that passes through those five points, but is decreasing at x = 2? Your answer should be an algebraic expression for the function, rather than a graph or a sketch.
Extra credit: Now suppose I tell you that the function is always increasing. Can you find a function that is concave down at x = 2?
The solution to this Riddler Express can be found in the following column.
Riddler Classic
From Sean Sweeney, Chris Nho and Eli Luberoff comes a Classic that is for anyone who ever thought even more deeply about functions in high school algebra.
Suppose you have two distinct points on the x-axis of the coordinate plane. If I tell you a parabola passes through those two points, where on the plane could that parabola’s vertex be? Spoiler alert: The vertex can be anywhere on the perpendicular bisector of those two points. (Neat!)
Now, suppose the two distinct points are anywhere on the coordinate plane. If I tell you that a parabola with a vertical line of symmetry passes through those two points, where on the plane could that parabola’s vertex be?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to ÑÑâÐ Chris Sedlack ÑÑâÐ of Burlington, Iowa, winner of last week’s Riddler Express.
Last week, you and I were locked in a random number duel. We were both using random number generators, which should have given us both real numbers between 0 and 1. Whoever’s number was greater won the duel!
There was just one problem — I had hacked your random number generator. Instead of giving you a random number between 0 and 1, it gave you a random number between 0.1 and 0.8.
What were your chances of winning the duel?
Many readers solved this algebraically. If my random number happened to be anywhere between 0.1 and 0.8 — which happened 70 percent of the time — then both of our generators were effectively behaving the same way, meaning we both had a 50 percent chance of winning.
But 10 percent of the time my generator produced a number between 0 and 0.1, in which case you were guaranteed the victory. On the other hand, another 20 percent of the time my generator produced a number between 0.8 and 1, in which case I was guaranteed the victory.
Putting these results together, your chances of winning were 10 percent plus half of 70 percent, for a total of 45 percent.
A few solvers, like Paige Kester of Southlake, Texas, solved the puzzle geometrically. Paige represented the sample space of our two random number generators as a rectangle, as shown below:
The horizontal axis represented my random number, while the vertical axis represented yours. The area of the rectangle was 0.7. The region of the left that resulted in your victory was a trapezoid with height 0.7 and bases 0.1 and 0.8. That meant its area was 0.7·(0.1+0.8)/2, or 0.315. Finally, your chances of winning were the ratio of the area of the left trapezoid to the area of the rectangle, which was 0.315/0.7, or — once again — 0.45.
With the way I subtly rigged this game of chance, I should totally open a Riddler-themed casino.
Solution to last week’s Riddler Classic
Congratulations to ÑÑâÐ Tom Keith ÑÑâÐ of Toronto, Canada, winner of last week’s Riddler Classic.
Last week, I had in my possession 1 million fair coins. Among these, I want to find the “luckiest” coin.
I first flipped all 1 million coins simultaneously, discarding any coins that come up tails. I flipped all the coins that came up heads a second time, and I again discarded any of these coins that came up tails. I repeated this process, over and over again. If at any point I was left with one coin, I declared that to be the “luckiest” coin.
But getting to one coin was no sure thing. For example, I might have found myself with two coins, flipped both of them and had both come up tails. Then I would have had zero coins, never having had exactly one coin.
What was the probability that I would — at some point — have exactly one “luckiest” coin?
Instead of 1 million fair coins, suppose I had started with a much smaller number. If I had zero, the probability of getting the luckiest coin was, of course, zero. And if I had started with one coin, then I was guaranteed to have one coin at some point (even without making any flips!), which meant the probability was one. But that was it for the trivial cases.
If I had started with two coins, then there was a 25 percent chance they both came up heads, in which case I’d flip them both again as though nothing had ever happened. If we look only at the 75 percent of the time when something meaningful did happen, one-third of the time (25 percent overall) I wound up with two tails and no luckiest coin. But the remaining two-thirds of the time (50 percent overall), I wound up with one heads and one tails, and hence a luckiest coin. In short, when you started with two coins, the probability I found one luckiest coin was 2/3.
Had I started with three coins, one-eighth of the time I got three heads and started over, which means we only had to consider the remaining seven-eighths of the time. The probability of getting two heads was three-sevenths (from which my chances of finding the luckiest coin were 2/3, as we just calculated). Another three-sevenths of the time I got one head, in which case I had just found the luckiest coin. And one-seventh of the time I got three tails, with no luckiest coin. Putting these results together, the probability I found the luckiest coin was 3/7·2/3 + 3/7, or 5/7.
Continuing along these lines, you could find the probability of identifying a luckiest coin when you started with four coins, five coins, six coins and so on. Each time, you worked out the probabilities of the different numbers of heads you might get, multiplied those by the probabilities of finding the luckiest coin assuming you had started with those corresponding numbers of coins, and then adding up the results.
As the number of coins increased, the probability of finding the luckiest coin converged to a specific value — or at least it appeared to! Solvers David Lewis and David Ding graphed the results for up to 100 or so coins, finding the probability converged to a little more than 72 percent.
However, that convergence was merely an illusion. It turned out that as the number of coins increased, the probability oscillated, never quite converging to a specific value. In fact, the period of oscillation was not a constant, but rather a constant factor of 2. Here was the probability of finding the luckiest coin when you started with anywhere between 15 and 4,096 coins, courtesy of Laurent Lessard:
Solver Allen Gu took this a step further, figuring out the value around which the probability oscillated. By assuming there was a very large number of coins to begin with, thereby morphing summations into integrals, Allen found the center was 1/(2ln(2)), or about 0.7213475.
But we never actually answered the question: What was the probability of finding the luckiest coin when you started with 1 million coins? This week’s winner, Tom, had a computer churn away for more than two hours, ultimately arriving at a probability of 0.7213539633235218. (For the record, I accepted any answers that started off with 0.72.)
Finally, I am pleased to report that Walker Yane, son of this puzzle’s author, Gary Yane, was among this week’s solvers. Congratulations all around to the Yane Family!
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 riddlercolumn@gmail.com.