Martha has a fair die with the usual $6$ sides...What is the probability that she stops at a cumulative sum of $13$?

You may try to work backwards. If she stops at $13$, the previous total is $10,9,8$ or $7$. So the wanted probability is the probability of reaching $10$ and thowing a $3$, or reaching $9$ and throwing a $4$, or reaching $8$ and throwing a $5$, or reaching $7$ and throwing a $6$. I.e. $\frac{1}{6}$ of the probability of reaching $7,8,9$ or $10$. The probability of avoiding such states is the probability of reaching $6$ and throwing a $5$ or a $6$ or reaching $5$ and throwing a $6$. There are no other ways to avoid them. The probability of reaching $5$ (sooner or later) is $$ \frac{1}{6}+\frac{4}{6^2}+\frac{6}{6^3}+\frac{4}{6^4}+\frac{1}{6^5}= \frac{2401}{7776}$$ and similarly the probability of reaching $6$ (sooner or later) is $$ \frac{1}{6}+\frac{5}{6^2}+\frac{10}{6^3}+\frac{10}{6^4}+\frac{5}{6^5}+\frac{1}{6^6}= \frac{16807}{46656}$$ hence the probability of avoiding $7,8,9,10$ is $$ \frac{2401}{7776}\cdot\frac{1}{6}+\frac{16807}{46656}\cdot\frac{2}{6}$$ and the probability of reaching (sooner or later) $7,8,9,10$ is $\frac{57979}{69984}$.
The wanted probability of stopping at $13$ is so $$\frac{1}{6}\cdot \frac{57979}{69984}=\frac{57979}{419904}\approx{13.8\%}. $$

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Please take the above lines with caution, since a numerical simulation suggests the probabilities of stopping at $11-12-13-14-15-16$ are close to $29-24-18-14-9-5\%$. Indeed, I spotted the flaw: once we reach $7,8,9,10$, we may stay there for a few throws, then reach $13$. The above probability has to be corrected by considering this situation. The correct probability is the one mentioned in Christian Blatter's answer.

This MATLAB function builds the transition matrix of a Markov chain for this puzzle and then finds the probability mass function for the sum of the values after the die is rolled rolls times.

function [p, A] = stop13(rolls)
  if nargin < 1
    rolls = 11;
  elseif rolls > 11
    rolls = 11;
  end

  A = toeplitz(zeros(17,1),[0 ones(1,6)/6 zeros(1,10)]);
  A(1:6,1:6) = eye(6);

  p = A^rolls * [zeros(16,1); 1];
end

For the default input value of $11$ rolls, it returns $$\begin{align} p_{16} &= 0.0482 \\ p_{15} &= 0.0949 \\ p_{14} &= 0.1396 \\ p_{13} &= 0.1819 \\ p_{12} &= 0.2419 \\ p_{11} &= 0.2934 \\ p_{10} &= 0 \\ &\ldots \\ p_{0} &= 0\enspace, \end{align}$$ in excellent agreement with Jack's simulation.