How to find the Number of tries to get a value greater than 4 tossing a fair die You've decided to repeatedly toss a fair six-sided die. Let's make the reasonable assumption that tosses are independent. 
On average, how many tosses does it take until you see a number larger than four?
I have seen a few solution of the kind here, but not able to understand why do I need a probability distribution curve or average.
Let's say I get a value greater than 4 in nth attempt, will the probability not be 2/6n.
Since E is 2 and S 6xn.
Can I not equate this with sum of a gp with n terms?
 A: I will give it a try. There are several ways to approach this problem. You indicate something with series, so I will use that.  You need Expectation for this.If you get more than $4$ in one go, than "contributes" in the Expectation as $(1)(\frac{2}{3})^0\frac{1}{3}$. First throw 4 or below, second throw more than 4 contributes as $(2)(\frac{2}{3})^1\frac{1}{3}$. First two throws 4 or below, third throw more than 4 contributes as $(3)(\frac{2}{3})^2\frac{1}{3}$. So if you factor out $\frac{1}{3}$ and use $x=\frac{2}{3}$, then essentially we are interested in the series $\frac{1}{3}(1+2x+3x^2+4x^3+...)$. We recognize the derivative of the geometric series here. The geometric series is $\frac{1}{1-x}=1+x+x^2+x^3+...$ and thus derivative is $\frac{1}{(1-x)^2}=1+2x+3x^3+...$. For $x=\frac{2}{3}$ this yields $9$, subject to $\frac{1}{3}$ gives $3$. So the expectation would be $3$ throws. If this is not what you are looking for, I can take it off.
A: So you have a sequence of trials which independent and identical success rate $2/6$, and you wish to find the expected count of trials until the first success.
Thus the count has a Geometric$_1$ Distribution.   There are several ways to derive the expected value of a Geometric$_1$ Distribution, but the result is well known.
The easiest way is to use the Law of Total Expectation, and the fact that if we make it to the second trial the "experiment" resets with one trial done. [Called the "memmoryless" property.]
$$\begin{align}\mathsf E(X)~=~&\tfrac 46~\mathsf E(X\mid X>1) + \tfrac 26~\mathsf E(X\mid X=1) \\ =~& \tfrac 23(1+\mathsf E(X))+\tfrac 13 \\~\\ \therefore \mathsf E(X)~=~&3\end{align}$$
A: Here is another perhaps a bit "unusual" approach: Recursion. Let $n$ be the expected number of throws for success (throwing greater than 4). On any given moment, the probability having greater than 4 is of course $1/3$. But if you fail $(2/3)$, you are back to square one and you have to throw again. So if you have already failed, say $n$ times, to fail another time, you will throw $n+1$ times so that is $(2/3)(n+1)$. Now the important step comes: Recursion: The expected number of throws for success is then $n=(1/3)(1)+(2/3)(n+1)$. First component is for success, second component for fail. From here $n$ can be solved. Verify that it follows $n=3$
