I see it's a very old question, but let me add my two cents. It's only an approximate solution and at times it involves some guesswork, but it turns out to be quite good. (Also, it doesn't need a computer and the math is pretty elementary.)
Let $a_n$ denote the probability of rolling total of $n$ in any number of rolls (now without the "stop when > 100" condition). After some thinking, we get a recurrence relation for these:
$$ a_n = (a_{n-1} + a_{n-2} + \ldots + a_{n-6})/6,\quad a_{-5} = a_{-4} = \ldots = a_{-1} = 0, a_0 = 1. $$
This is a linear recurrence, which can be solved "easily" by forming the characteristic equation $6\lambda^6 - \lambda^5 - \lambda^4 - \lambda^3 - \lambda^2 - \lambda - 1 = 0$. If we denote its roots by $l_1$ to $l_6$, the explicit formula for the recurrence has the form of
$$ a_n = \sum_{0 < i < 7} C_i l_i^n. $$
(The $C$'s are obtained from the boundary conditions.) Since the $a_n$'s represent probabilities, they should be in the interval of $<0;1>$. From this it seems to be reasonable that $|l_i| \leq 1$. If there were any that don't satisfy this condition, the $a_n$'s would be unbounded (since the powers, and even differences of two of them with different bases, would be unbounded).
Now we see it has a root of $\lambda = 1$. So, $a_n$'s eventually converge to $C_1$ (which we don't know, but we don't care), since all other $C_i l_i^n$ converge to 0 (because of the $|l_i|<1$ condition).
So probability of getting 101 is $a_{101}$. Getting 102 has a probability of $a_{102} - a_{101}/6$, since we can't obtain it by rolling 101+1. Similarly, rolling 103 has a probability of $a_{103} - (a_{102} + a_{101})/6$ (no 101+2 nor 102+1), etc. Now we guess that the $a_n$'s converge so well that $a_{101}$ through $a_{106}$ are essentially equal. That gives the 6:5:4:3:2:1 ratio, and furthermore, the (correct) limit of $a_n$'s, 2/7.
I know this is somewhat estimatory (some may say, "physicist's") approach, but even with that, I hope it could have some value.