hitting a dart board probability

Question

You have a dart board which is split in half. If you hit the left half, you get $2$ points, if you hit the right half, you get $3$ points.

You have an 80% chance of hitting the dart board on any given throw and a 70% chance of hitting the side you're aiming for given that you hit the dartboard.

If you have an unlimited number of throws, what's the probability you get a score of exactly 7?

Answer 1

Only positive scores are relevant. So it is enough to look at them. Under condition that the score is positive by aiming on certain positive score ($2$ or $3$) there is probability of $\frac{56}{80}=\frac{7}{10}$ that this score is reached and a probability of $\frac{24}{80}=\frac{3}{10}$ that the other positive score is reached.

Let $p_{n}$ denote the probability that score $n$ is reached by applying a strategy that is optimal in the sense that it leads to a maximal probability to achieve score $n$.

Then $p_{1}=0$, $p_{2}=p_{3}=p_{5}=\frac{7}{10}$ and $p_{4}=p_{6}=\frac{7}{10}\frac{7}{10}=\frac{49}{100}$.

Choosing a strategy (at the start) to come to a score of $7$ we have the following options:

Aiming at $2$: there is a probability of $\frac{7}{10}p_{5}+\frac{3}{10}p_{4}=\frac{7}{10}\frac{7}{10}+\frac{3}{10}\frac{49}{100}=0.637$.

Aiming at $3$: we have an optimal probability of $\frac{7}{10}p_{4}+\frac{3}{10}p_{5}=\frac{7}{10}\frac{49}{100}+\frac{3}{10}\frac{7}{10}=0.553$.

Any 'other' strategy will give a probability $p\times0.637+\left(1-p\right)\times0.553\in\left[0.553,0.637\right]$ for some $p\in\left[0,1\right]$

So at the start we choose for aiming at $2$ and $p_{7}=0.637$ is the probability that score $7$ will be reached.

Answer 2

If you start at $3$, the only way to get to $7$ is to hit the left half twice. You eventually accomplish this with probability $0.7 \times 0.7 = 0.49$. If you start at $2$, the first hit does not matter, and the second hit has to be on the correct half. So you reach the target with probability $0.7$. Since reaching $2$ gives you a better chance, you always aim left at the start, so the final probability is $0.3 \times 0.49+0.7 \times 0.7 = 0.637$.