Looking for an alternative solution to finding $E(N)$ for an experiment I've been given the following problem:

An infinite number of contestants play the following dice game: The contestants toss a die one by one, the first contestant to get a result either equal or higher than the result of the first dice toss by the first contestant wins the game. $N := $ The number of the winning contestant. $N \geq 2$. Find $E(N)$.


In the official solution the formula used to calculate the mean is $$E(X) =\sum_{x \in \mathbb{R}} P(X>x)$$
I'm aware of it and how to solve the problem using it, but up until then I've mainly used $$E(X) = \sum_{x \in \mathbb{R}} x \cdot \mathrm{p}_X(x)$$though my attempt has fallen short. How do I know to use the formula used in the official solution as opposed to the formula I'm used to? where is my solution wrong and is there an alternative way? 
My Attempt
 I started by trying to find the probability someone will roll higher than the first person, I did so using the 'Entire' probability formula:
$$K = \text{the first player's result} \quad X = \text{any following player's result}$$
$$P(\text{A toss wins}) = \sum_{k=1}^6 P(\text{A toss wins} |K=k) \cdot P(K=k) = \sum_{k=1}^6 \frac{7-k}{6} \cdot \frac{1}{6} = \frac{7}{12} $$
$$N \sim \text{Geom}(\tfrac{7}{12}) \implies E[\overbrace{N}^{ N=1 \quad \text{is the 1st player}}] = \frac{1}{p} +1= \frac{12}{7}+1=2.714 \neq 3.45 \quad (\text{which is the correct answer})$$
I've also tried computing the sum directly using the formula I'm familiar with $\sum_{n=0}^{\infty} n \cdot \frac{7}{12} \cdot \left(\frac{5}{12}\right)^{n-1}$ and got the same result. I'd love to see if there's a feasable way using these tools I'm familiar with, understanding where I got it wrong, and when is it better to switch up to $E(X) =\sum_{x \in \mathbb{R}} P(X>x)$ instead of using the other one.
 A: I'm just reiterating and adding a more detailed solution based on lulu's and drhab's advice:
$$X = \text{The first tosser},$$
$$P(\text{win against x}) = \frac{7-x}{6}.$$
why?
$x=1 \implies P(\{\text{win}\}) = P(\{1,2,3,4,5,6\})=\frac{6}{6},$
$x=2 \implies P(\{\text{win}\}) = P(\{2,3,4,5,6\})=\frac{5}{6},\dots$
so the pattern becomes apparent.
We know then that $$N-1\mid X \;\;\sim\;\; \text{Geom}\left(\frac{7-x}{6}\right).$$
We know that for geometric distribution $\mu = \frac{1}{p}.$ We know that for a discrete random variable $X$:
$$E(f(X)) = \sum_{x \in \mathbb{R}} f(x) \cdot \mathrm{p}_X(x).$$
So:
$$E(N-1)=E\big(E(N-1\mid X)\big)=E\left(\frac{6}{7-x}\right) = \sum_{x=1}^6 \left(\frac{6}{7-x}\right) \cdot \frac16 =2.45,$$
$$E[N]=E[N-1]+1=3.45.$$
A: First I try to solve the confusion about the difference in formulas for $E(X)$
The formulas are equivalent in this case, as can be seen as follows:
$$P(X>x)= P(X=x+1)+P(X=x+2)+\dots = p_X(x+1)+p_X(x+2)+\dots$$
And now
$$\sum_{x=0}^\infty P(X>x) =  (p_X(1)+p_X(2)+p_X(3)+p_X(4)+\dots) + (p_X(2)+p_X(3)+p_X(4)+\dots)+(p_X(3)+p_X(4)+\dots)+(p_X(4)+\dots)+\dots = p_X(1)+2p_X(2)+3p_X(3)+4p_X(4)+\dots.$$
Then about your attempted solution:
I think the problem is in your assumption that $N\sim\mathrm{Geom}(7/12)$. Instead, try to use $E(N)=E(N|K=1)P(K=1)+\dots+E(N|K=6)P(K=6)$.
Edit/extra hint: note that $N_{K=k} \sim \mathrm{Geom}(\frac{7-k}{6})$, maybe with a +1 somewhere
Edit 2:
The equation $E(N) = E(N|K=1)P(K=1)+\dots+E(N|K=6)P(K=6)$ comes from $P(N=n) = P(N=n|K=1)P(K=1)+\dots+P(N=n|K=6)P(K=6)$:
$$ E(N) = \sum_n n P(N=n) $$$$= \sum_n n \big(P(N=n|K=1)P(K=1) + \dots + P(N=n|K=6)P(K=6)\big) $$$$= \sum_n n P(N=n|K=1)P(K=1) + \dots + \sum_n n P(N=n|K=6)P(K=6) $$$$= E(N|K=1)P(K=1)+\dots+E(N|K=6)P(K=6)$$
