# What is the intuitive reason for this inequality?

I am facing the following question:

An archer is aiming at a circular target of radius 20 inches. Her arrows hit on average 5 inches away from the center, each shot being independent. Show that the next arrow will miss the target with probability at most $$\frac{1}{4}$$.

One can derive the answer if they let $$X$$ represent the he distance between the point hit and the center of the target, measured in inches. Because $$X \geq 0$$, and $$X \geq 20 \mathbb{I}_{\{X \geq 20\}}$$, we eventually derive that $$\frac{1}{4} \geq \mathbb{P}(X \geq 20)$$.

But what is the intuitive relationship that is limiting the probability of this tail event? Why can't we have more than 0.25 probability of the tail event?

• Markov inequality: $\mathbb P(X>a) \le \frac{\mathbb E[X]}{a}$ if $X$ is non-negative random variable and $a>0$ – Henry Sep 23 '20 at 12:40

This is the Markov inequality: $$\mathbb P(X>a) \le \frac{\mathbb E[X]}{a}$$ if $$X$$ is non-negative random variable and $$a>0$$
The intuitive argument (and a proof) is that to minimise the expectation, you cannot do better than concentrating the probability at or above $$a$$ down to the value of $$a$$ and concentrating the probability below $$a$$ down to the value $$0$$ so $$\mathbb E[X] \ge a\mathbb P(X>a) +0 \mathbb P(X\le a)$$
The intuition is that the expectation is a weighted average of the probabilities. If you put too much weight on one side, the expectation will move in the same direction and will not remain at 5. If you want the expectation to remain 5, every additional probability you put on $$\Pr(X>20)$$ must be somehow balanced by the probability of values below $$5$$. But since the total probability is $$1$$, you might not be able to have enough on the "short side" of $$5$$, to balance the excess probability on the longer side (think of a crane).