Just in time for March Madness, a(nother) basketball-themed Riddler:
Consider the following simplified model of free throws. Imagine the rim to be a circle (which we’ll call C) that has a radius of 1, and is centered at the origin (the point (0,0)). Let V be a random point in the plane, with coordinates X and Y, and where X and Y are independent normal random variables, with means equal to zero and each having equal variance — think of this as the point where your free throw winds up, in the rim’s plane. If V is in the circle, your shot goes in. Finally, suppose that the variance is chosen such that the probability that V is in C is exactly 75 percent (roughly the NBA free-throw average).
But suppose you switch it up, and go granny-style, which in this universe eliminates any possible left-right error in your free throws. What’s the probability you make your shot now? (Put another way, calculate the probability that |Y| < 1.)
The answer is about 90.4%.
Why no explanation? To be honest, I didn't actually solve this one analytically. This week's puzzle is, in my opinion, an example of what the Riddler (and casual-ish math puzzles in general) should not be. A good puzzle is one where you have to figure out what the math problem underlying it actually is; actually solving the math part is secondary and often trivial. There should be an "Aha!" moment when, after minutes or hours or days of thinking, you finally figure out the trick needed to solve the problem. And then you solve it.
This puzzle, on the other hand, is the opposite. There's no mystery here, the math problem is barely disguised by the basketball story written around it. This may as well just be pulled directly from a statistics textbook, which isn't all that fun. "Here's a math problem - do it."
I did have some fun simulating the problem, though. Here's the result of a Python simulation I did to calculate the result of a series of free throws. The ones on the left are unrestricted in the x-axis, and the ones on the right are the same shots thrown underhand: