Expected number of dice rolls require to make a sum greater than or equal to K? A 6 sided die is rolled iteratively. What is the expected number of rolls required to make a sum greater than or equal to K?
Before Edit
P(Sum>=1 in exactly 1 roll)=1
P(Sum>=2 in exactly 1 roll)=5/6
P(Sum>=2 in exactly 2 rolls)=1/6
P(Sum>=3 in exactly 1 roll)=5/6
P(Sum>=3 in exactly 2 rolls)=2/6
P(Sum>=3 in exactly 3 rolls)=1/36
P(Sum>=4 in exactly 1 roll)=3/6
P(Sum>=4 in exactly 2 rolls)=3/6
P(Sum>=4 in exactly 3 rolls)=2/36
P(Sum>=4 in exactly 4 rolls)=1/216

After Edit
P(Sum>=1 in atleast 1 roll)=1
P(Sum>=2 in atleast 1 roll)=5/6
P(Sum>=2 in atleast 2 rolls)=1
P(Sum>=3 in atleast 1 roll)=4/6
P(Sum>=3 in atleast 2 rolls)=35/36
P(Sum>=3 in atleast 3 rolls)=1
P(Sum>=4 in atleast 1 roll)=3/6
P(Sum>=4 in atleast 2 rolls)=33/36
P(Sum>=4 in atleast 3 rolls)=212/216
P(Sum>=4 in atleast 4 rolls)=1

I am not sure this is correct first of all and but I think this probability is related to the expected number of rolls?
But I don't know how to proceed further. Am I proceeding in the right direction? 
 A: This is so far only some ideas for another, more exact, approach, based on the same observation that my first answer. With time I will extend this ...
First, some notation. Let $K$ be some given, positive (large) integer. We want the distribution of $N$, which is the minimum number of throws of an ordinary dice to get sum at least $K$. So, first we define $X_i$ as the outcome of dice throw $i$, and $X^{(n)}=X_1+\dots+X_n$. If we can find the distribution of $X^{(n)}$ for all $n$ then we can find the distribution of $N$ by using
$$
   P(N \ge n)=  P(X_1+\dots+X_n \le K),
$$
and we are done. 
Now, the possible values for $X_1+\dots+X_n$ are $n,n+1,n+2,\dots,6n$, and for $k$ in that range, to find the probability $P(X_1+\dots+X_n=k)$, we need to find the total number of ways to write $k$ as a sum of exactly $n$ integers, all in the range $1,2,\dots,6$. But that is called an restricted integer composition, a problem well studied in combinatorics.  Some related questions on math SE is found by  https://math.stackexchange.com/search?q=integer+compositions
So searching and studying that combinatorics literature we can get quiet precise results. 
I will follow up on that, but later ...
A: there is no way to get exact expected number of rolls in general, but for a K.
Let N be the event of of expected rolling to get sum=>K.
for K=1, E(N)=1
for K=2, $E(N)=(\frac{5}{6}+2*1)/(\frac{5}{6}+1)=\frac{17}{11}$
and so on.
It will be going difficult to get E(N) for large K.
for example,for K=20 you'll need to expect from (4 rolls,20 rolls)
Central Limit Theorem will be more benefitiary with some % confidence.
as we know occurrence is uniformly distributed, for large values of K.
$$K(Sum)~follows~N(3.5N,\frac{35N}{12})$$(Normal Distribution)
Now you need "N" to get Sum at least K....
we convert it in standard normal distribution.$$\frac{K-3.5N}{\sqrt{\frac{35N}{12}}}=Z_\alpha$$ where $\alpha=1-confidence$%
You can get Z values from "Standard Normal Tables" or from here for example $Z_{0.01}=2.31,Z_{0.001}=2.98$
You know K,Z(at any error) ........ then you can get N=E(N) at some confidence % by solving equation.
A: I will give one method to find an approximate solution. First, let $X_i$ be the random variable, "result of throw $i$ with the dice" and let $N$ be the number of throws necessary to reach a sum at least $k$. Then we have that 
$$
  P(N \ge n) = P(X_1+X_2+\dots+X_n \le k)
$$
so to find the distribution of $N$ we need to find the convolutions of the distributions of the $X_i$ for $i=1,2,\dots,n$, for all $n$.  Those convolutions can be found numerically, but for large $n$ it might be much work, so we try instead to approximate the cumulative distribution function for the convolutions, using saddlepoint methods. For another example of saddlepoint methods, see my answer to Generic sum of Gamma random variables
We will use the Lugannini-Rice approximation for the discrete case, and follows R Butler: "Saddlepoint Approximations with Applications",  page 18 (second continuity correction).   First, we need the moment generating function of the $X_i$, which is
$$
M(T) = E e^{tX_i}= \frac16 (e^t+e^{2t}+e^{3t}+e^{4t}+e^{5t}+e^{6t})
$$
Then the cumulant generating function for the sum of $n$ independent dice becomes
$$K_n(t)=n \cdot log(\frac16\sum_{i=1}^6 e^{it})
$$
and we also need the first few derivatives of $K$, but we will find those symbolically using R. The code is the following:
 DD <- function(expr, name, order = 1) {
        if(order < 1) stop("'order' must be >= 1")
        if(order == 1) D(expr, name)
        else DD(D(expr, name), name, order - 1)
     }

make_cumgenfun  <-  function() {
    fun0  <-  function(n, t) n*log(mean(exp((1:6)*t)))
    fun1  <-  function(n, t) {}
    fun2  <-  function(n, t) {}
    fun3  <-  function(n, t) {}
    d1  <-  DD(expression(n*log((1/6)*(exp(t)+exp(2*t)+exp(3*t)+exp(4*t)+exp(5*t)+exp(6*t)))),  "t", 1)
    d2  <-  DD(expression(n*log((1/6)*(exp(t)+exp(2*t)+exp(3*t)+exp(4*t)+exp(5*t)+exp(6*t)))),  "t", 2)
    d3  <-  DD(expression(n*log((1/6)*(exp(t)+exp(2*t)+exp(3*t)+exp(4*t)+exp(5*t)+exp(6*t)))),  "t", 3)
    body(fun1)  <-  d1
    body(fun2)  <-  d2
    body(fun3)  <-  d3
    return(list(fun0,  fun1,  fun2,  fun3))
}

Next, we must solve the saddlepoint equation.  
That is done by the following code:
funlist  <-  make_cumgenfun()

# To solve the saddlepoint equation for n,  k:
solve_speq  <-   function(n, k)  {# note that n+1 <= k <= 6n is needed
    Kd  <-  function(t) funlist[[2]](n, t)
    k  <-  k-0.5
    uniroot(function(s) Kd(s)-k,  lower=-100,  upper=1,  extendInt="upX")$root
}

Note that the above code is not very robust, for values of $k$ far in either tail of the distribution it will not work. Then some code for actually calculating the tail probability function, approximately, by the Luganini-Rice approximation, following Butler, page 18, (second continuity correction):
Function for returning the tail probability:
#
Ghelp  <-  function(n, k) {
    stilde  <-  solve_speq(n, k)
    K  <-  function(t) funlist[[1]](n, t)
    Kd <-  function(t) funlist[[2]](n, t)
    Kdd <- function(t) funlist[[3]](n, t)
    Kddd <- function(t) funlist[[4]](n, t)
    w2tilde  <-  sign(stilde)*sqrt(2*(stilde*(k-0.5)-K(stilde)))  
    u2tilde  <-  2*sinh(stilde/2)*sqrt(Kdd(stilde))
    mu  <-  Kd(0)
    result  <- if (abs(mu-(k-0.5)) <= 0.001) 0.5-Kddd(0)/(6*sqrt(2*pi)*Kdd(0)^(3/2))  else
    1-pnorm(w2tilde)-dnorm(w2tilde)*(1/w2tilde - 1/u2tilde)
    return(result)
}
G  <- function(n, k) {
      fun  <- function(k) Ghelp(n, k)
      Vectorize(fun)(k)
  }

Then let us try to use this to calculate a table of the distribution, based on the formula
$$
  P(N \ge n) = P(X_1+X_2+\dots+X_n \le k) \\
          = 1-P(X_1+\dots+X_n \ge k+1) \\
          = 1-G(n,k+1)
$$
where $G$ is the function fron the R code above.
Now, let us answer the original question with $K=20$. Then the minimum number of rolls is 4 and the maximum number of rolls is 20. The probability that 20 rolls is needed is very small, and can be calculated exactly from the binomial formula, I leave that to the reader. (the approximation above will not work for $n=20$).
So the probability that $N \ge 19$ is approximated by
> 1-G(20, 21)
[1] 2.220446e-16

The probability that $N\ge 10$ is approximated by:
> 1-G(10, 21)
[1] 0.002880649

And so on.  Using all this, you can get an approximation for the expectation yourself.  This should be much better than the approximations based on the central limit theorem.
