0
$\begingroup$

I am doing a problem that states: If you are throwing a dart at a circular board with radius $R$, what is the expected distance from the centre?

If $x$ is the expected radius, then it would be the distance from the centre such that the area created from the radius $x$ is half that of the total area. Then we have:

$$\frac{\pi x^2}{\pi R^2} = \frac{1}{2}$$ $$\therefore x = \sqrt{\frac{1}{2}} R$$

So $x=\sqrt{\frac{1}{2}} R$ creates a circle where its area is half the size of the board from the centre, then that’s the expected radius of the dart throw.

The solution instead says:

We first assume that the probability of landing in a subset of the board is equal to its area divided by the area of the board. This means that we (almost) always hit the dartboard but beyond that all points are equally likely. The dart board has radius $R$ so the probability of landing inside a circle of radius $s$ is $$ \left(\frac{s}{R}\right)^2 . $$

The density of the radius is therefore the derivative of this with respect to $s$ which is $$ \frac{2 s}{R^2} . $$

The expected distance from the centre is then $$ \int_0^R \frac{2 s}{R^2} s d s=\frac{2}{3} \frac{R^3}{R^2}=\frac{2 R}{3} $$

Why is the solution $\frac{2R}{3}$ and not my solution?

$\endgroup$
4
  • 2
    $\begingroup$ Short version: You gave the median distance, not the mean distance. If you don't get why those are different, please say so, and I'll probably write out a fuller answer. $\endgroup$
    – Brian Tung
    Commented Feb 26 at 23:52
  • $\begingroup$ @BrianTung Not quite sure what you mean - I thought what I wrote is the mean distance... $\endgroup$
    – Xerium
    Commented Feb 26 at 23:59
  • $\begingroup$ No, what you wrote is the median distance: the value which is exactly in the middle of the possible distances. By way of analogy, if the possible distances were discrete and were $\{2, 6, 8, 9, 10\}$ with probability $1/5$ each, then the median would be $8$—the middle distance. But the mean would actually be $(2+6+8+9+10)/5 = 7$. In the continuous case, as we see here, we need an integral rather than a summation to compute the mean, but the principle is the same. $\endgroup$
    – Brian Tung
    Commented Feb 27 at 0:32
  • $\begingroup$ I'll add an answer to explain more fully. $\endgroup$
    – Brian Tung
    Commented Feb 27 at 0:34

1 Answer 1

1
$\begingroup$

Your answer doesn't agree with the official solution because you produce the median of the distribution, rather than its mean. In general, they are computed differently, and they are not guaranteed to be equal.

It is easier to see this in the discrete case. Suppose that the distances are equally likely to be $2$, $6$, $8$, $9$, or $10$, with probability $1/5$ each. The median value is $8$, because it most precisely divides the possible values into equal halves; there are two values less than $8$ and two values greater than $8$. For the median, we don't care how much less or greater they are, only that they are. We would get the same median if the collection of numbers were instead $2, 6, 8, 9, 1000$.

In contrast, the mean does "care" about what the values are, and not just their ordering, because for a discrete set of equally probable values, it is the sum of the values divided by how many there are—in this case, $\frac{2+6+8+9+10}{5} = \frac{35}{5} = 7$. If the $10$ were replaced by $1000$, we would instead get a mean of $\frac{2+6+8+9+1000}{5} = \frac{1025}{5} = 405$.

In the present problem, we are asked to find the mean distance. Because the distribution of distance is continuous rather than discrete, we need an integral rather than a summation, but the principle is the same: We add up all the possible distances (the $s$ from $0$ to $R$), weighted by how likely each distances is (the $\frac{2s\,ds}{R^2}$). The result is the mean.

In contrast, what you determined is the value that falls in the middle of the possible values—that is, the value for which the distance is equally likely to be more or less. As you correctly ascertained, that is the distance within which the portion of the dartboard has half the total area—but that is the median, not the mean.


Two further points:

  • This question implicitly assumes that all points are equally likely to be struck; for example, you're not more likely to strike the bullseye than some other area of the dartboard equal in size to the bullseye. In real physical processes, this isn't always the case. By all rights, the question should have been worded to make this explicit.

  • There are cases where the median is equal to the mean, of course. Sometimes this happens "by accident"—that is, there isn't anything special about the problem that explains why they're the same; they just work out to the same value. Sometimes, though, there's a good reason, such as that the distribution of values around the median is symmetric. That isn't the case with either my discrete example or the dartboard; in each situation, it's possible to be less than the median by more than it is possible to be more than it.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .