Charting the probability of die values equal to or greater than a variable with N dice Math experts, I need your help with my probability modelling for a game.
What I have been trying to work out is how to check if a dice rolling situation progresses the probability in a reasonably linear manner.
(This paragraph is wrong but I'm keeping it because the answer is much easier to understand given my epic mistype) The game calls for a CL (Challenge level) which is the number at or above when rolled that counts as a success and the DR (difficulty rating) which is the number of successes needed to "win". I am attempting to model this using six-sided dice (d=6) in various numbers (n).
(What I should have said: The game calls for a CL (Challenge level)  which is the number of successes needed to "win" and the DR (difficulty rating) which is the number at or above when rolled that counts as a success. I am attempting to model this using six-sided dice (d=6) in various numbers (n).)
So I have put this together in a spreadsheet. I figured the best way to work out p(win) was to work out p(fail) and take that from 1.
For CL=1 I used =1-(((6-(7-DR))/6)^n) which I have checked here and I am fairly sure is right. For CL>1, I used =1-(((6-(7-DR))/6)^(n+1-CL)) 
as that seemed right at the time given that I was working out the probability of not rolling enough at or above CL. 
I calculated the probability for CL=1-5 and DR=2-6. The probability curve came out looking like this:

This is nothing like I was expecting and I have started to doubt my maths.
I tried reordering the CL/DR pairs to sorted by probability but that did not seem to have any pattern to it (here is charted for 5d6).

Can either (a) help me understand what I did wrong or (b) help me see why the CL/DR pairs progress so I can come up with a pattern that makes sense in a tabletop setting. They seem to be jumping all over the place?
I am hoping you are going to take a look at my work, enjoy a good laugh at my maths, and then show me what I did wrong because this does not seem like a very useful system the game in question unless I am very wrong somewhere.
 A: Below I have put two answers as I wasn't sure from the context which you are interested in.
In the first one, to win the game you must achieve the challenge level on successive throws, without a failure. Where the number of throws required to win is the difficulty rating.
In the second answer, I assume there is some fixed number of throws allowed, and that to win the number that exceed the challenge level must be at least the difficulty rating: but they do not have too happen successively.
Update: The original poster has clarified that Game 2 is the one of interest. Game 1 is left for posterity


Game 1.
In your game you have two parameters:
A roll of the dice is considered successful if it is equal to $k$ or greater (this is the challenge level), with $1 \leq k \leq 6$. Clearly if $k=1$ then every role is a success.
To win the game, you must have $n$ successful rolls in a row. i.e. you must roll at least a value of $k$ on $n$ rolls of the dice, without any failures. Let $p_{k,n}$ denote the probability of winning the game.
Due to the independence of each roll, the probability of winning when the difficulty rating is $n$ is equal to the probability of winning $n$ games with a difficulty rating of $1$. i.e.
$$p_{n,k} = p_{1,k}^n.$$
Then the probability of winning $1$ game is the probability of a dice roll being greater than or equal to $k$, which is
$$p_{1,k} = \frac{7-k}{6}.$$
Hence
$$p_{n,k} = \left(\frac{7-k}{6}\right)^n.$$


Game 2.
In this game you have three parameters:
A roll of the dice is considered successful if it is equal to $k$ or greater (this is the challenge level), with $1 \leq k \leq 6$. Clearly if $k=1$ then every role is a success.
In a given round we allow a total of $N$ rolls, and we say that the game was won if at least $n$ of these rolls are succesful (i.e. exceed $k$). $n$ therefore denotes the `difficulty rating' in your notation.
As I understand, you are asking: Given $k,\, N,\,n$ what is the probability that the player wins the game? 
As in the last game, the probability that a given roll is at least the challenge level $k$ is given by
$$p= \frac{7-k}{6}.$$
That is, the probability that a given roll is a success is distributed according to a $\text{Bernoulli}(p)$ distribution.
With this in mind, we now want to know the probability that in $N$ rolls, at least $n$ exceed $k$. We assume that each roll of the dice is independent, then this is the same as asking if a $\text{Binomial}(N,p)$ distribution is greater than or equal to $n$.
The above is a standard result in probability that relies on the fact that any given the number of successes of $N$ independent Bernoulli distributions is Binomially distributed.
Therefore the probability you are interested in is:
\begin{align*}
\mathbf{P}\big( \text{Bin}(N,p) \geq n \big) & =
\sum_{m = n}^N \binom{N}{m} p^m (1-p)^{N-m} \\
& = \sum_{m = n}^N \binom{N}{m} \left(\frac{7-k}{6}\right)^m \left(\frac{k-1}{6}\right)^{N-m}
\end{align*}
Unfortunately there is little more rearranging that is possible (save for pulling out the factors of $1/6$); this is known to be the case as there is no closed form expression for the partial sum of binomial coefficients.
In the below I plot the case where $N = 5$ dice are used, and we vary the challenge level ($x$-axis), and difficulty rating (different plot lines).

