So I was thinking of doing this recursively: $f(x,i)$ is equal to the probability of rolling greater than $x$ and landing on $i$ on the last roll. $f(0,i) = 1/6$ for $i = \{1,2,..,6\}$. $f(1,i) = 1/6 + 1/6f(0,i)$ for $i = \{2,...,6\}$ and $f(1,1) = 1/6f(0,1)$. Finally, we list out this recursion until we get $f(13,i)$ and see for what value of $i$ is $f$ the largest.
Is there a better way to approach this or an easy way to simplify this method?
On
The answer to your particular problem (which is the most probable last dice) is already given. Regarding the more general problem (computing $f(n,i)$), we can see that
$$f(n,i)= \frac{1}{6}\sum_{k=0}^{i-1} S(n-k)$$
where $S(m)$ is the probability that an unbounded running sum of dice hits the value $m$. Now, surely that problem must have already been analyzed in a fantastic math site I know... let me see... yes, here it is. We see that $S(m)$ has no simple closed formula, so your recursive approach is reasonable. And we can also intuitively see that $S(m)$ must asympotically tends to a constant (specifically, to 2/7), so $f(n,i) \approx i/21$ for large $n$.
On
Denote by $p_r(n)$ the probability that the last roll is $r$, when $n$ more points are needed to end the game. Then $$p_r(1)={1\over6}\quad(1\leq r\leq6)\ ,$$ and for convenience we put $p_r(n):=0$ when $n\leq0$. It is then easy to see that the $p_r(n)$ for each $r\in[6]$ satisfy the recursion $$p_r(n)={1\over6}\left(\sum_{k=1}^6 p_r(n-k) + 1_{1\leq n\leq r}\right)\qquad(n\geq1)\ .\tag{1}$$ For $n>r$ only the homogeneous part $$p_r(n)={1\over6}\sum_{k=1}^6 p_r(n-k)$$ remains. Its characteristic polynomial $\chi(\lambda)=6\lambda^7-7\lambda^6+1$ has a double zero at $1$ and five zeros of absolute value $<1$. It follows that there are constants $a$, $b$ with $$p_r(n)=a n + b +o(1)\qquad(n\to \infty)\ .$$ As $0\leq p_r(n)\leq 1$ for all $n$ we necessarily have $a=0$; but $b$ has to be determined from the initial conditions $\bigl(p_r(n)\bigr)_{r-5\leq n\leq r}$, which result after applying $(1)$ $r$ times.
Running $(1)$ on the computer corroborates the following conjecture: $$\lim_{n\to\infty} p_r(n)={r\over21}\ .$$ Heuristically this can be explained as follows: A roll of $r$ lets the sum jump over $r$ half integers. Since all rolls $r\in[6]$ are equiprobable and $\sum_{r=1}^6 r =21$, on average ${r\over21}$ of all half-integers are jumped over by a roll of $r$. When an $N\gg1$ is chosen then the probability that the mark at $N+{1\over2}$ is jumped over by a roll of $r$ is therefore equal to ${r\over21}$.
Professional random walkers should be able to prove this rigorously.
We are not asked the exact probabilities $f(13,i)$, but it is somehwat obvious that for $n\gg 0$ we have $f(n,i)\sim i$ (and hence $f(n,i)\approx \frac i{21}$). So even without calculation it is reasonable to assume that $f(13,6)>f(13,i)$ for all $i\ne 6$.
And indeed, any sequence of rolls that ends with $i$ exceeding $13$ can be mapped to a sequence of rolls that ends in a $6$ exceeding $13$ (and having the same rolls before). Therefore $f(13,6)\ge f(13,i)$ for all $i$. Now note that any sequence ending exactly at $13$ with an $i<6$ can be turned to a sequence exceeding $13$ and ending in $6$, we see that in fact $f(13,6)>f(13,i)$. (In this last step we used that $13\ge i$ for all $i<6$ so that the existence of a sequence summing to exactly $13$ with an $i$ as last roll is guaranteed).