Number of times to roll a dice to get 4 or 2 How many times do we need to roll a dice until 4 or 2 turns up?
I guess the probability for each is $1/6$. Since we have a "or" I guess it is $1/6+1/6$ and $4/6$ unwanted numbers. But I don't know how to compute the number of times we need to roll.   
 A: Your success event is: we have 2 or 4.
The probability of this event (as you noted) is: 
$p = 2/6 = 1/3$.   
So the expected number of times to perform the experiment
in order to get your first success is: $1/p = 1/(1/3) = 3$.   
A: Yes 2/6 or 1/3 represents that on a given roll, it ends up being a 4 or a 2.
Theoretically, there is a chance that you may have to roll $99999^{99999}$ times until you get a 4 or a 2 for the first time, or it could happen on your first roll. What I think you are looking for, is that given you have $n$ rolls, what the chance that by the nth roll you will get a 2 or a 4.
Note: chance of not getting 2 or 4, as you pointed out, is: $4/6$
Assume you don't a 2 or a 4 on your first roll (this will 4/6 times). Then it may happen again that on you second roll, you still don't get a 2 or a 4. This will occur 4/6 out of the times on the second roll.
So the chance that you don't roll a 4 or a 6 on your first roll, and then fail again on your second row is: $(4/6)*(4/6)$
Clearly, the chance that you succeed is $1 - P($failure$)$ so: $1-(4/6)*(4/6)$ 
As you can work out, in general, the chance that you will roll a 4 or a 2 by your nth role (succeed) is:
$1-(2/3)^n$ 
The probability exponentially approaches 1, which makes intuitive sense, as getting no 4s or 2s after something like 50 rolls is extremely low, just as the formula suggests.
A: For any number of rolls, we can't be sure that a 4 or a 2 shows up. Namely, we still have cases without a 4 or a 2, e.g. only ones. 
A: You have a variable $X\sim Ber(\frac{1}{2},n)$ that says the number of successes after $n$ rolls.
$\mu = \frac{n}{3} \ \ \ \sigma =\sqrt{\frac{1}{3}\times \frac{2}{3}}\times \frac{1}{\sqrt{n}}=\sqrt{\frac{2}{9n}}$
We want to be somehow sure of $X\geq 1$. We should begin but what is for you sure enough, but let's use for that $95\%$, which is usually used.
From that, it follows:
$P(X\geq 1)\sim P(Z\geq \frac{1-\mu}{\sigma})=P(Z\geq \frac{(2-n)\sqrt{2n}}{2})$
If we want to be $95\%$ sure, we are looking when $Z\geq -1.645$ ($Z$ follows a normal distribution)
All together:
$\frac{(2-n)\sqrt{2n}}{2}=-1,645\\
n\approx 3.2838$
So, with 3 rolls, you can't be yet $95\%$ sure. For that level of certainty you'd need 4 rolls.
