14

Darts and Ballistic Missiles

14.1 THE PROBLEM

Two darts players, A and B, stand in front of a square dartboard with dimensions 2R by 2R. On the board is painted a circular target area of radius R with its center at the center of the board. We imagine this center point is the origin of an x, y coordinate system with axes parallel to the edges of the board. A and B each throw a dart at the board, and although each dart does indeed land somewhere on the board, the players have distinctly different throwing techniques. When A throws her dart it lands at a point with coordinates (X, Y), where X and Y are independent random variables, each uniformly distributed from −R to R. When B throws his dart it lands at a point with coordinates (X, X), where X is, as before, a random variable uniformly distributed from −R to R.

As a warm-up for the real puzzler in this problem (soon to come), calculate the probability that A’s dart lands in the circular target area (call the result P_A), and then do the same for B’s dart (call that probability P_B). These are not difficult calculations to do, and you should be able to verify that P_A > P_B independent of R.

Once you’ve done that, here’s the real problem for you. Instead of X and Y being uniformly distributed, take them as still identically distributed and independent but now normally distributed with zero means. That is, their probability density functions are

where σ is a positive parameter (with units of distance) called the standard deviation. These are the equations for the famous bell-shaped Gaussian probability curves named after the great German mathematician, Carl Friedrich Gauss (1777–1855).

Now, of course, the dartboard is to be imagined as arbitrarily large, as either dart could potentially land arbitrarily far from the origin in the infinite x, y plane of the board. But our questions remain as before: what are P_A and P_B, the probabilities, respectively, that the darts of A and B land within distance R of the origin of the coordinate system? These new questions are, admittedly, probably now better associated with the impact points of ballistic missiles on distant targets (σ is then a measure of the miss distance of the missile) rather than with a darts game, hence the second part of this puzzler’s name.

Is it still true that P_A > P_B for all R, as it is in the first case of darts?

14.2 THEORETICAL ANALYSIS

First, the warm-up exercises. When A throws her dart it can land anywhere on the board, with each tiny patch of area as likely to receive the dart as any other tiny patch of area of the same size. So the probability of the dart landing inside the circular target area is simply the ratio of the circular target area to the area of the board. That is,

When B throws his dart it lands (by the Pythagorean theorem) a radial distance of from the origin. To be in the circular target area, this must be less than or equal to R. Thus,

or, since X is uniform from −R to R,

These two results have no dependency on R, and so A always has the greater probability hitting the circular target area. Let’s now see if this remains true when we go to the case when X and Y are normally distributed.

For A we have

Since X and Y are independent, we know their joint probability density function is

and so, integrating over a circle of radius R centered on the origin of coordinates,

where dA = dxdy is the differential area in the x, y coordinate system. Using the well-known trick of converting to polar coordinates, we have x² + y² = r² and the differential area is now dA = rdr dθ. To cover the circular area, r and θ vary from 0 to R and from 0 to 2π, respectively. So,

This last integral can be done (and we eventually will), but as you’ll soon see, we don’t have to do it quite yet.

For B we have

where I’ve taken advantage of the symmetry of the integrand around x = 0. If we change the dummy variable of integration from x to r, with the transformation , then

Now, let’s define the function f(R) to be the difference between P_B and P_A. That is,

Next, recall how to differentiate an integral (for a freshman calculus derivation, see my book The Science of Radio [Springer 2001], pp. 415–418):

Since the integrals on the right-hand side in the equation for f(R) have integrands with no dependency on R, we can write df/dR by inspection:

where

For any σ > 0 it is easy to show that g(R) = 0 has two positive solutions (I’ll prove this in a technical note at the end of this analysis), that is, there are two values of R (call them R1 and R2) such that g(R1) = g(R2) = 0, which means that at those values of R we have df/dR = 0. Here’s why this is useful to know. When R = 0 we see that f = 0 because both integrals in the definition of f are zero when R = 0, and that when R = 0. So, for 0 ≤ R < R1, the function f increases and then, at R = R1, f reaches a maximum, and so, for R > R1, the function f then decreases until R = R2, when f must again reach an extrema, which must now be a minimum. For R > R2, f then again increases. Since we have

and since the first integral is well known (look in any math tables book, or see my book Mrs. Perkins’s Electric Quilt [Princeton 2009], pp. 282–283), and the second integral is easy to do, we have

and this means, since f is increasing for R > R2, that f is negative (that’s the only way an increasing function can approach zero).

To summarize what we’ve learned: starting at R = 0, we see that f(R) = P_B − P_A increases from zero until R = R1, then in the interval R1 < R < R2 we see that f decreases and passes through zero to go negative until it begins to increase again toward zero when R > R2. This means there is some value of R between R1 and R2—call it R_C—such that

Therefore, it is not true, as it was in darts, that P_A > P_B, always. Rather, if R < R_C, then B’s missile is the more likely to be in the circular target area, while if R > R_C, then it is A’s missile that is the more likely to be in the circular target area.

To find the actual value of R_C, let’s change variables in the f(R) equation using the transformation r = uσ. Then dr = σ du, and so

or, if we write w = R/σ,

We wish to find the value of w that gives f = 0. This is done fairly easily by using one of any number of algorithms; I used a particularly popular number-cruncher called the Newton-Raphson method (see my book When Least Is Best [Princeton 2004], pp. 120–123, for its derivation and historical background). It is an iterative procedure that, given an estimate for w (written as w_n), generates the next (and presumably better) estimate of w_{n + 1}:

where f(w_n) is as given above and f′(w_n) is df/dw, which is given (recall our integral differentiation discussion) by

The calculation of f(w) requires good integration software, and I used MATLAB^®’s powerful quad (for quadrature) command. Starting with an initial guess of w₁ = 1 (see the technical note below for where this guess comes from), the Newton-Raphson method quickly converges to w = 1.75294. That is, R_C = 1.75294σ.

Technical Note

We now have one final task, that of showing g(R) = 0 always has two positive solutions. In Figure 14.2.1 you see plots of 1/σ R e^−R2/4σ2 for three values of σ (0.5, 1, and 2), and you can see that all three curves peak above the constant (as σ increases, the peak shifts to the right) with the result always being two intersections with the horizontal line. This figure is just for illustration purposes, and it is easy to prove that the peak value is greater than for all σ > 0 and so 1/σ R e^−R2/4σ2 = twice, always.

The maximum of 1/σ R e^−R2/4σ2 occurs when its derivative equals zero. So,

Figure 14.2.1. The maximum of 1/σ R e^−R2/4σ2 is greater than .

when R = σ. This gives a maximum value of

a result independent of σ. Now, because 2/e > 1/π because 2π > e because 2π > 6 while e < 3.

If you look at the middle curve (σ = 1, and so w = R), you see it crosses the line at about R ≈ 0.5 and R ≈ 2.5. These are the values of R1 and R2, and so R_C is somewhere in between. That’s why I picked w₁ = 1 (but any initial value in the approximate interval 0.8 to 2.2 would work just as well).