Dice-Coin Paradigm

Step 1: Probability of seeing a 1, for a fixed $n$

Suppose we already knew the first tails landed on flip $n$, so we roll the die exactly $n$ times. Rolling "no $1$ at all" in $n$ independent rolls has probability $(5/6{)}^n$, so

$$P(\text{at least one }1\mid n\text{ rolls})=1-{\left(\frac{5}{6}\right)}^n$$

That handles the die, for a fixed $n$. The remaining question is: how likely is each $n$ in the first place?

Step 2: Probability that the first tails lands on flip $n$

For the first tails to occur exactly on flip $n$, the first $n-1$ flips must all be heads, and the $n$th must be tails:

$$P(\text{first tails on flip }n)={\left(\frac{1}{2}\right)}^{n-1}\cdot \frac{1}{2}={\left(\frac{1}{2}\right)}^n$$

Step 3: Combine and sum over every possible $n$

The event "we see a $1$" can happen via any $n=1,2,3,\dots$, and these are mutually exclusive (the coin gives exactly one value of $n$). So we weight each $n$'s die-probability by how likely that $n$ was, and sum over all of them:

$$P(\text{see a }1)=\sum _{n=1}^{\infty }{\left(\frac{1}{2}\right)}^n\left[1-{\left(\frac{5}{6}\right)}^n\right]$$

It's convenient to instead sum from $n=0$: the $n=0$ term is $1\cdot \left[1-1\right]=0$, so throwing it in changes nothing, but it lets us use the clean closed form ${\sum }_{n=0}^{\infty }{r}^n=\frac{1}{1-r}$ directly instead of the shifted version. Splitting the sum in two:

$$P(\text{see a }1)=\sum _{n=0}^{\infty }{\left(\frac{1}{2}\right)}^n-\sum _{n=0}^{\infty }{\left(\frac{1}{2}\cdot \frac{5}{6}\right)}^n=\sum _{n=0}^{\infty }{\left(\frac{1}{2}\right)}^n-\sum _{n=0}^{\infty }{\left(\frac{5}{12}\right)}^n$$

Step 4: Evaluate both geometric series

$$\sum _{n=0}^{\infty }{\left(\frac{1}{2}\right)}^n=\frac{1}{1-\frac{1}{2}}=2,\sum _{n=0}^{\infty }{\left(\frac{5}{12}\right)}^n=\frac{1}{1-\frac{5}{12}}=\frac{12}{7}$$

so

$$P(\text{see a }1)=2-\frac{12}{7}=\frac{14-12}{7}=\frac{2}{7}$$

A quick sanity check: $2/7\approx 0.286$. Since $n$ is at least $1$ with probability $1$, and $P(\text{at least one }1\text{ in a single roll})=1/6\approx 0.167$, while more rolls make it more likely, a number moderately above $1/6$ is exactly what we'd expect.

Thus, the probability of observing a die value of $1$ is $\frac{2}{7}$