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 probabilityyou make your shot now? (Put another way, calculate the probability that |Y| < 1.)