# Why the chance of getting single real number in [0,1] is 0?

I have just started learning probability course and my professor said something really strange.

1. If we take a random real number between 0 and 1 (including) then the chance of getting 1/2 is 0

2. The chance of getting a number in [0,1/2] is 1/2

While I totally agree with the second point I don't agree with the first one.

Why?

I know that the chance is really really small since there are non countable real numbers between 0 and 1 but saying that the probability is 0 means that you will never get 1/2 but that is not true since 1/2 is a possible chance.

The key point of your problem is that your conclusion is wrong. If an event has probability zero, it doesn't mean that it can't happen. That is the mind-blowing thing when starting to think about probability theory. That is also the reason why we say that events with probability zero do not occur almost surely. Now that we have corrected that misconception, I hope it will be easier to accept the computations in the other answers.

If a randomly chosen number falls at $$\frac 1 2$$ it falls within $$(\frac 1 2 -\epsilon, \frac 1 2 +\epsilon)$$ for any $$\epsilon$$. So the probability of this event cannot exceed $$2\epsilon$$ for any $$\epsilon$$. Hence it must be $$0$$.

• But again 0 means it is impossible to get 1/2 which isn't true – john Oct 25 '20 at 9:05
• @john There is a difference between empty set and event of prabability $0$. The event of gettting $\frac 12$ has probability $0$ but it is not the empty set. – Kavi Rama Murthy Oct 25 '20 at 9:09
• @john You should leave your "intuitive idea" that the probability of selecting $1/2$ is the same than throwing a dice. It is not. In particular because a dice has a finite number of faces. Here you're dealing with infinite sets. – mathcounterexamples.net Oct 25 '20 at 9:10

One key property of a probability function $$p$$ is that it is countably additive, i.e. for a countable family of disjoint subsets $$\{S_n\}$$, you have

$$p\left( \cup_{n \in \mathbb N} S_n\right) = \sum_{n \in \mathbb N} p(S_n)$$

Now, you know that $$[0,1] \cap \mathbb Q$$ is countable. If $$p(\{1/2\})$$ would be positive, you'll get the contradiction

$$p\left([0,1] \cap \mathbb Q\right) = \infty$$

The following is a very inexact way of explaining your problem, but perhaps it will help.

Think about it like this: the probability of an event is the number of ways your event can happen divided by the number of ways any event can happen.

In the case of selecting the number $$\frac{1}{2}$$ from the interval $$[0,1]$$, there is exactly 1 way to select $$\frac{1}{2}$$ from the infinitely many ways to select any number from $$[0,1]$$.

So $$\Pr(\textrm{selecting }\frac{1}{2}\textrm{ from }[0,1]) = \frac{1}{\infty}=0$$.

• But again 0 means it is impossible to get 1/2 which isn't true – john Oct 25 '20 at 10:17
• Other people have already told you this, but a probability of 0 does not mean it is impossible. The only way it would be impossible would be if $\frac{1}{2}$ wasn’t available to be chosen. Measure theory is what is being glossed over here, but it turns out that the probability of selecting any number of finite or countable infinite possibilities from and uncountable set will always be 0. – Laars Helenius Oct 25 '20 at 10:42
• Maybe this will help: youtu.be/ZA4JkHKZM50 – Laars Helenius Oct 25 '20 at 10:47