The sum up to the first six

Idea. Wald's identity: if $N$ is a stopping time with finite mean for i.i.d. steps $X_i$, then $\mathbb{E}[X_1+\cdots +X_N]=\mathbb{E}[N]\mathbb{E}[X_1]$.

Step 1: the stopping time. $N$, the time of the first $6$, is geometric with parameter $\frac{1}{6}$, so $\mathbb{E}[N]=6$. Whether we stop at roll $k$ depends only on rolls $1$ to $k$.

Step 2: why Wald applies. Write $S={\sum }_{k\ge 1}{X}_{k}1\{N\ge k\}$. The event $\{N\ge k\}$ means “no $6$ in the first $k-1$ rolls”, which is independent of $X_k$. So $$\mathbb{E}[S]=\sum _{k\ge 1}\mathbb{E}[X_k]\mathbb{P}(N\ge k)=3.5\cdot \mathbb{E}[N]=21.$$

Check another way. The rolls before the $6$ are uniform on $\{1,\dots ,5\}$ with mean $3$, and there are $N-1$ of them, $5$ on average. Then the final roll is $6$. Total: $5\cdot 3+6=21$.

The trap. Each roll before the stopping time is biased (it is never a $6$), and the last roll is always a $6$. Wald's identity says these two biases cancel exactly, provided the decision to stop never looks at future rolls.