Square Roots and Great Comebacks

From the time I learned about them, I've been fascinated (probably to an unseemly amount) by the square root. I remember reading about a method for calculating square roots by long hand. There's no point, really; we have calculators to do that for us. (If you have some spare time and you enjoy this sort of thing, see if you can figure out the algorithm from the example at left.)

What use are square roots, anyway, aside from solving math problems about the diameters of circular lawns? (Have you ever seen any of those? They must encircle those conical swimming pools we dealt with in calculus class.) Here's one use: They can tell you when how big a lead your favorite basketball team needs to be secure in a win.

A few years ago, I derived a rule for determining when a lead was safe in a basketball game—specifically, an NBA game. (It matters, because the shot clock is different between an NBA game and a WNBA game and a NCAA men's game and a NCAA women's game.) You take the square root of the number of seconds left, and add three. For instance, if there's 3:45 left in the game, that's 225 seconds. Square root of 225 is 15, and you add 3, so an 18-point lead is pretty darned safe with 3:45 left. The "add 3" is for a trey at the buzzer. Go ask the Miami Heat about that 'round about now.

Pretty keen, huh? Although—not to put too fine a point on it—well-known sports statistician Bill James also came up with this very same rule. We'll call it independent discovery, at least on my part. I have no idea whether James stole it from me. Give him the benefit of the doubt, though.

But why? Why should this rule work? Why isn't it just the time remaining divided by some rate at which the team that's behind catches up? If a team can make up a 15 points in 225 seconds and then cap that with a trey to make up the 18, why can't it make up 33 points in 7:30? Or 63 points in 15:00?

And the sort-of answer to that is, it can. It's just terribly unlikely. Of course, it's already unlikely that a team can make up 15 points in 3:45, but it's still in the realm of possibility. Asking a team to do that twice in a row is just too much. If it was 100 to 1 against doing it once, doing it twice in a row would be 10,000 to 1 against. On the other hand, making up the same 15 points in twice the time is obviously easier. So in twice the time (7:30, natch), you should be able to make up some deficit in between. According to both me and Bill James—and honestly, are you going to go against both of us?—that deficit is 15 times the square root of 2. That's about 21, and if you add the 3 at the end it makes it 24.

Where on earth does this come from? One place is the drunkard's walk, otherwise known as the random walk (but I think "drunkard's walk" is more evocative). In this mathematical scenario, the eponymous drunkard starts off at some placemark—a lamppost, say. Each moment in time, he takes a step, but in a completely random direction. Might be in the same direction as the last step, might be in the opposite direction, might be anything. So after a bunch of steps, he might end up back at the lamppost where he started...or he might be home.

Odds are, though, he'll be at some intermediate distance. How far from the lamppost? Well, the first step is going to take him one step away for sure. We'll represent this by saying that d(1) = 1, where d(t) is the distance of the drunkard from the lamppost at time t. OK, now what about d(2)? Before that second step, he's one step away from the lamppost. His second step might take him two steps away, if he walks in the same direction, or zero steps away, if he walks in the opposite direction (back toward the lamppost). On average, though, he'll walk in some intermediate direction: let's say, perpendicular to his current progress from the lamppost. The Pythagorean theorem says then that

[d(2)]² = [d(1)]² + 1² = 1² + 1² = 1 + 1 = 2

or, in other words, d(2) = √2. We can go further. We've already got two examples where d(t) = √t and we'd like to get more. To do that, we'll use a process called induction. Suppose that you have a value of t for which d(t) = √t ; we'll now try to show that d(t) = √(t + 1) . Using the same argument as before—that the drunkard walks in some intermediate direction—we get

[d(t + 1)]² = [d(t)]² + 1² = [√t]² + 1² = t + 1

and then we directly get d(t + 1) = √(t + 1) . So as long as we can find a t where d(t) = √t , we're set; it's true for all greater values of t. But we already have such a value: t = 1! (And t = 2, for that matter.) It turns out, then, that the drunkard's walk, after time t, takes him a distance √t away from the lamppost.

Now, a couple of things. First, this isn't anything like a rigorous demonstration of the square root property of the drunkard's walk. You can look that up if you like. But if you work at it a little, it gives you an inkling of the intuition behind it. Secondly, though, and here we're back on track a bit: What has all this got to do with basketball games?

A basketball game is an alternating sequence of possessions. In each possession, the team with the ball is of course trying to score, and the other team is of course trying to prevent it from scoring. When the ball changes hands, the roles are reversed. In each individual possession, the effect on the score is biased: Only the team with the ball can score, usually. But in each pair of possessions, that bias cancels out, since both teams get a chance with the ball. The margin in the game can move in any direction—just like the drunkard's walk.

If the drunkard starts off 50 steps from home, he could conceivably get home in just 50 steps. But it's ridiculously unlikely: Each of those 50 steps would have to be in exactly the right direction. The square root property tells us he'll probably be just a bit over 7 steps from the lamppost; it would take 2500 steps to get him, on average, 50 steps from his starting point. After those 2500 steps, is he guaranteed to be home? Nope. He still has to be walking in the right direction. But it's at least plausible now.

In the same way, a basketball team that's down 18 points could conceivably make that up by scoring six three-pointers in a row while holding their opponents scoreless. If they did that by fouling and their opponents obliged by missing all of their free throws, the whole deficit could be made up in half a minute or so. But that's as unlikely as the drunkard walking 50 steps in exactly the right direction. Instead, a team will make up its deficit in halting fashion, sometimes making up three points, but other times giving up a point, or staying even, in any particular pair of possessions. The drunkard's walk, in other words, and that's why the square root rules great combacks.

I was going to follow this up with a discussion of sociology and mobilizing people, but this post is getting long (see, I do notice it!) and I'll defer that till next time.