0
$\begingroup$

n arrows are shot to n targets and each arrow will hit one of the targets, each target having the same probabilty (1/n) to be hit by the arrow, independent of the previous shots. We divide the n targets into the ones hit (by 1 or more arrows) and the ones not hit. What is the expected percentage of the targets hit? It is easy to see that it is more than 50% and less than 75%. Is the golden ratio 61.8% the expected percentage? (in the limit for growing n)
Additional question: If one considers only the hit targets, what is the expected percentage of them to be hit by exactly 1 arrow?

$\endgroup$
0
$\begingroup$

Let us define the probability $p$ that a given target is hit by a given arrow : $p = 1/n$

After $n$ tries, the probability that a target in not hit is $(1-p)^n$.

The probability that a target is hit is therefore $$P_h = 1 - (1-p)^n = 1 - \left(1 - \frac{1}{n}\right)^n$$

This correspond to the pourcentage of hit targets.

As $n \to \infty$, this probability converge to $1-e^{-1}$, about 63.2%.

The probabilty that one target is hit by exactly one arrow is equal to $$np(1-p)^{n-1} = \left(1-\frac{1}{n}\right)^{n-1}$$

If we consider the hit targets only, the ratio is equal to

$$\frac{\left(1-\frac{1}{n}\right)^{n-1}}{1 - \left(1 - \frac{1}{n}\right)^n} $$

As $n \to \infty$, this probability converges to

$$\frac{e^{-1}}{1-e^{-1}} = \frac{1}{e-1} $$

About 58%

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy