# Dart board probability example: general formula is independent of $\pi$ and $r$.

My textbook, Statistical Inference, Second Edition, by Casella and Berger, provides the following example:

Example 1.2.7 (Defining probabilities-II) The game of darts is played by throwing a dart a board and receiving a score corresponding to he humber assigned to the region in which the dart lands. For a novice player, it seems reasonable to assume that the probability of the dart hitting a particular region is proportional to the area o the region. This, a bigger region has a higher probability of being hit.

Referring to Figure 1.2.1, we see that the dart board has radius $$r$$ and the distance between rings is $$r/5$$. If we make the assumption that the board is always hit (see Exercise 1.7 for a variation on this), then we have

$$P(\text{scoring i points}) = \dfrac{\text{Area of region i}}{\text{Area of dart board}}.$$

For example,

$$P(\text{scoring 1 point}) = \dfrac{\pi r^2 - \pi(4r/5)^2}{\pi r^2} = 1 - \left( \dfrac{4}{5} \right)^2.$$

It is easy to derive the general formula, we find that

$$P(\text{scoring i points}) = \dfrac{(6 - i)^2 - (5 - i)^2}{5^2}, \ \ i=1, \dots, 5,$$

independent of $$\pi$$ and $$r$$.

I found it interesting that general formula is independent of $$\pi$$ and $$r$$. Since $$\pi$$ and $$r$$ are properties of the area of a circle, I'm wondering if I should be interpreting this to mean that the probability of scoring $$i$$ points on some dart board is independent of said dart board's shape? For instance, would this general formula be valid for a dart board of 5 square regions, rather than 5 circular regions?

I would greatly appreciate it if people could please take the time to review my reasoning here.

• You should probably add Figure 1.2.1, so we can see what’s going on. Commented Dec 25, 2019 at 8:30
• @URL Figure added. Commented Dec 25, 2019 at 8:33

This shouldn’t be too surprising given this very simple model. The underlying assumption is that the probability is uniformly distributed across the total area so that the probabilities are ratios of areas. In particular, any factor of $$\pi$$ in the expression for the area of one of the rings will cancel with the $$\pi$$ in the total area. Moreover, the shape and location of the subregions are irrelevant in this model, as is the point at which the player is aiming.

This isn’t a particularly realistic model, even assuming a novice player, but the point is to illustrate how a uniform distribution works, not to model a real darts player.