Rolling Until the Sum Is Not Seven

Key idea. The final roll has the law of a single roll conditioned on the sum not being $7$ , and that law is still symmetric about $7$ .

Let $S_1,S_2,\dots$ be the i.i.d. sums of successive rolls. Write $p=\mathbb{P}(S_i=7)=\frac{6}{36}=\frac16$ , and let $N$ be the index of the first roll whose sum is not $7$ .

For $k\neq 7$ , split on the value of $N$ :

\mathbb{P}(S_N=k)=\sum_{n\ge1}\mathbb{P}(S_1=\dots=S_{n-1}=7,\ S_n=k)=\sum_{n\ge1}p^{\,n-1}\,\mathbb{P}(S_1=k)=\frac{\mathbb{P}(S_1=k)}{1-p}.

This equals $\mathbb{P}(S_1=k\mid S_1\neq 7)$ . So the payout $S_N$ has the law of one sum conditioned on not being $7$ .

Next use symmetry. Replacing each die face $d$ by $7-d$ shows that $S_1$ and $14-S_1$ have the same distribution. The event $\{S_1\neq 7\}$ is unchanged by this replacement. Hence

\mathbb{E}[S_1\mid S_1\neq7]=\mathbb{E}[14-S_1\mid S_1\neq7]=14-\mathbb{E}[S_1\mid S_1\neq7],

which gives $\mathbb{E}[S_1\mid S_1\neq 7]=7$ . Therefore

\boxed{\mathbb{E}[S_N]=7}

Alternative. The law of total expectation gives the same result without using symmetry:

7=\mathbb{E}[S_1]=p\cdot 7+(1-p)\,\mathbb{E}[S_1\mid S_1\neq7].

Rearranging gives $\mathbb{E}[S_1\mid S_1\neq 7]=\frac{7-7p}{1-p}=7$ . More generally, discarding the outcome equal to the mean never changes the conditional mean.