# What is the probability that the total score after throwing darts is divisible by $3$.

To play a game of darts Michael throws three darts at the dart board shown. The number of points $$(1,$$ $$5$$ or $$10)$$ for each of the three regions is indicated. His score is the sum of the points for the three darts. If the radii of the three concentric circles are $$1,$$ $$2$$ and $$3$$ units, and each dart Michael throws hits this dart board at random, what is the probability that his score is evenly divisible by $$3?$$ Express your answer as a common fraction. After taking the values modulo $$3$$, we have $$1, 2, 1$$. I am pretty sure that the only way we can get divisible by $$3$$ in this problem is if we have modulos $$1, 1, 1$$ or $$2, 2, 2$$ for the darts. This means that the probability is $${\left(\dfrac23\right)}^3+{\left(\dfrac13\right)}^3=\dfrac13$$.

I feel as if I am missing something, or am I correct?

Thanks!

EDIT: "At random" means that the likelihood of a dart landing in a region is the total area of that region divided by the total area of the dart-board.

• "Each dart hits the board 'at random'" This is not clear. Are we meant to assume that each of the three regions are equally likely to have been struck? Or are we meant to assume that the regions are as likely according to the ratio of their area compared to the overall circle? Or is there some other distribution that the darts might hit? This problem is ambiguous. Depending on the distribution you will get different answers like you see in Bertrand's Paradox. Sep 1, 2020 at 2:32
• An expert dart player still "hits the board randomly"... just that those random points all happen to be tightly packed assuming they have a high enough accuracy (low spread) and good enough aim (center of spread). An expert dart player might then hit a bullseye effectively every time and as such have a 100% chance of having a total score divisible by $3$ since they will always get a score of $30$. Sep 1, 2020 at 2:36
• @MikeSmith As I explained in a deleted comment to Teresa Lisbon, your answer is correct if you assume each region is as likely based on its area compared to the overall circle. However, as pointed out in JMoravitz's comments, there are multiple ways to interpret the question's statement in terms of the distribution of probabilities of the dart hitting different parts of the board. Is this what you were wondering about or is there something else? Sep 1, 2020 at 2:44
• hmm... paying a bit closer attention, the question seems to have been crafted to give the same result whether we were to treat regions as equally likely (region for 5 occurring $\frac{1}{3}$ as there are three regions) or if we were to treat regions as occurring relative to their relative area ($\frac{\pi\cdot 2^2 - \pi\cdot 1^2}{\pi\cdot 3^2}$)... Still, it is crucial to be aware of the ambiguity of the problem and the different interpretations. Had the $5$ and the $1$ switched places or the values otherwise changed slightly the differences in the interpretations will come into play here. Sep 1, 2020 at 2:52
• @JMoravitz: that means it is a bad problem, as many problems depend on knowing which distribution is intended. The problem should be clear and the wrong answer should be wrong. Sep 1, 2020 at 3:06

Suppose probability corresponding to modulo $$1$$ is $$p$$, then probability correponding to modulo $$2$$ is $$1-p$$.

Hence it should be $$p^3+(1-p)^3=p^3 + (1-p)^3.$$

The radius are $$r_1=1, r_2=2, r_3=3$$ respectively and the probability is proportional to the area, then

$$1-p=\frac{\pi r_2^2 - \pi r_1^2}{\pi r_3^2}=\frac{r_2^2-r_1^2}{r_3^2}.$$

Since $$r_2=2r_1$$ and $$r_3=3r_1$$, then $$1-p=\frac{4-1}{9}=\frac13$$

While the numerical value coincides, you should illustrates that your probability is computed based on the area.

The areas $$A_1,A_5,A_{10}$$ areas are

$$\begin{cases} A_{10}: &\pi r^2 = pi\\ A_5: &4\pi-\pi=3\pi\\ A_1: &9\pi-4\pi=5\pi \end{cases}$$

Out of $$(1+3+5)\pi=9\pi$$, the probabilities to get:

$$\begin{cases} P(10) &= 1/9\\ P(5) &= 3/9\\ P(1) &= 5/9 \end{cases}$$

Any of the $$27$$ possible configurations $$(a,b,c)$$ has the probability $$\dfrac{p}{27\cdot 3^3 = 3^6}$$

To reach a divisible by 3 score, there is

• only one way to make either $$(10,10,10)$$, $$(5,5,5)$$ or $$(1,1,1)$$ of respective probabilities $$\dfrac1{3^6}, \dfrac{27}{3^6}, \dfrac{125}{3^6}$$which sum is $$\dfrac{153}{3^6}$$
• three ways to make each of $$(1,1,10)$$ or $$(1,10,10)$$, $$3\dfrac{25+5}{3^6}$$

Summing the whole gives $$\bbox[5px,border:2px solid #ca9]{\dfrac{243}{3^6}=\dfrac{3^5}{3^6}=\dfrac13}$$