The first total past 100

Idea. The total jumps over 100 to some value $100+k$ with $k$ from 1 to 4. Far from the start, every number is visited with the same probability $\frac{2}{5}$, one over the mean roll. That gives the distribution of the overshoot $k$, and its mean.

Step 1: visiting a number. Let $u(m)$ be the probability that the running total ever equals $m$. A total of $m$ is reached from $m-1,\dots ,m-4$, so $$u(m)=\frac{1}{4}\sum _{j=1}^{4}u(m-j).$$ This average of the four previous values converges quickly to a constant, and since the total moves forward by $2.5$ per roll on average, that constant is $\frac{1}{2.5}=\frac{2}{5}$.

Step 2: the overshoot. The first total above 100 equals $100+k$ when the walk visits $100-j$ for some $j\ge 0$ and then rolls exactly $k+j$ (which must be at most 4). So $$\begin{array}{rl}\mathbb{P}(100+k)& =\sum _{j=0}^{4-k}u(100-j)\cdot \frac{1}{4}\\ & \approx \frac{2}{5}\cdot \frac{5-k}{4}=\frac{5-k}{10}.\end{array}$$ The overshoot is 1, 2, 3, 4 with probabilities $\frac{4}{10},\frac{3}{10},\frac{2}{10},\frac{1}{10}$.

Step 3: the mean. $$\begin{array}{rl}& 100+\frac{4\cdot 1+3\cdot 2+2\cdot 3+1\cdot 4}{10}\\ & =100+2=102.\end{array}$$

How exact is it? The only approximation is $u(m)\approx \frac{2}{5}$, and the error shrinks geometrically: at $m=100$ the exact answer is $102-1.9\times 10^{-20}$.

Why not 102.5? A uniform overshoot between 1 and 4 would give 102.5. But an overshoot of 1 can come from any of the four rolls (from 100 with a 1, from 99 with a 2, and so on), while an overshoot of 4 needs a 4 rolled from exactly 100. So small overshoots are more likely. In general, for a roll $X$, the mean overshoot is $$\frac{\mathbb{E}[X^2]+\mathbb{E}[X]}{2\mathbb{E}[X]}=\frac{7.5+2.5}{5}=2.$$