A dice is rolled and summed over and over , What is the probability That the sum will be "m" , "m" is a positive integer

My problem starts after the numer 6 , as I start to loose options , as I can't use 1 dice for higher numbers then 7 , same goes for 2 dices after 13 and so on.

Any1 got a soultion ?

Here are 2 possibilities (there are many others):
First possibility: A die is rolled k times and the numbers produced are summed. What is the probability that the sum is m?
For a particular k:
Let f(m) be the number of sequences of length k with all terms in {1, 2, 3, 4, 5, 6} and with sum k.
I will denote the number of ways to choose b elements of a set of size a by aCb.
I will denote the integer part of a number r by <r>
Let F be the polynomial in X with the coefficient of X^m being f(m).
Let G = 1 + X + X^2 + X^3 + X^4 + X^5 + X^6 = X(1-X^6)/(1-X).
So we have:
F = G^k
= X^k * (1-X^6)^k * (Sum(i = 0, infinity, X^i))^k
= X^k * sum(i = 0, k, (-1)^i kCi X^6i) * sum(i = 0, infinity, ((i + k - 1)C(k-1)) X^i)
= X^k * sum(i = 0, infinity, (sum(j = 0, min(k, <i/6>), (-1)^j kCj ((i - 6j + k - 1)C(k-1))) X^i)
giving f(m) = sum(j = 0, min(k, <(m-k)/6>), (-1)^j kCj ((m - 6j - 1)C(k-1))).

The probability is then f(m)/6^k.

For example, if k = 4 and m = 10 we have:

f(m) = sum(j = 0, 1, (-1^j) 4Cj ((10 - 6j - 1)C3))
= 1 4C0 9C3 + (-1) 4C1 3C3
= 84 - 4 = 80.
So the probability of this outcome is 80/6^4 = 5/81.

Second possibility: A die is rolled repeatedly and after each roll the number produced is added to a running total. The process terminates as soon as the running total is greater than or equal to m. What is the probability that the final total is m?
Of course, this could be computed by summing the results obtained in the first version over all values of k. But that is quite ugly. Here is another solution:

The problem may be restated as follows:
A die is rolled repeatedly and after each roll the number produced is added to a running total. What is the probability (denoted p(m)) that at some point the running total equals m?
p(m) is also the probability that, for some i, at some point the running total is n-i and on the next throw after that point, the number on the die is i.
That is, p(m) = (p(m-1) + p(m-2) + p(m-3) + p(m-4) + p(m-5) + p(m-6))/6. This recurrence does have a closed-form solution, but it is pretty disgusting. It's probably easier to just use the recurrence, so for example:

p(0) = 1 = 1.000...
p(1) = 1/6 = 0.166...
p(2) = 7/36 = 0.194...
p(3) = 49/216 = 0.227...
p(4) = 343/1296 = 0.265...
p(5) = 2401/7776 = 0.309...
p(6) = 16807/46656 = 0.360...
p(7) = 70993/279936 = 0.254...
p(8) = 450295/1679616 = 0.268...
p(9) = 2825473/10077696 = 0.280...
p(10) = 17492167/60466176 = 0.289...

It is not too hard to show that p(n) tends to a limit as n tends to infinity.

Thankyou

sfwc
<><