You keep on throwing a dice and add the digit that appears to a sum. You stop when sum $\ge 100$. What’s the most frequently appearing digit in all such cases? $1$ or $6$?
I believe the probability of $1$ and $6$ should be equal as the whatever the number of rolls, the probability of getting a number should not be affected. However I don't have a formal proof for it and am not sure if this is right.
On
The expected number of $1$'s is the same as the expected number of $6$'s. Let $n_j(k)$ denote the expected number of digit $j\in \{1,\ldots,6\}$ appearing in the sequence until the running sum reaches $k$ (in your case $k=100$). Then $$ n_j(k)=\frac{1}{6}(1+n_j(k-j))+\frac{1}{6}\sum_{i\ne j} n_j(k-i)=\frac{1}{6}+\frac{1}{6}\sum_{i\in \{1,\ldots,6\}} n_j(k-i) $$ with $n_j(1-i)=0$ for $i=1,\ldots,6$. This recurrence relation is the same for all $j$ and so its solution, $n_j(k)$, is the same for all $j$.
Basic approach. Imagine drawing a tree, with a root labelled $0$. The running count of each node is the label on that node, plus the sum of the labels of all of its direct ancestors. We build on the tree as follows: Under any node whose running count is not yet $100$, we add six more nodes, labelled $1$ through $6$. We repeat until there are no nodes left whose running count is less than $100$.
At the end of this process, we obviously have a finite tree. How many $1$s are there? How many $6$s? Was there any time when we added a $1$ but not a $6$, or vice versa?