I'm trying to reason about the probability of a random outcome that can't be measured based on the existence of other random outcomes that can be measured (silent vs detected data corruption, for example). I boiled down the problem to this abstraction:
Suppose I roll a fair 20-sided die an unknown amount of times, and I truthfully reveal I rolled 11-20 five times. What's the probability I never rolled a 1?
My intuition tells me to break the problem down and sum the probabilities: $$\sum_{n=0}^\infty P(A|r=n+5)P(r=n+5)$$
i.e. If I rolled $n+5$ times and $5$ times were 11-20, the number of outcomes where I don't roll a 1 is $9^n*10^5*{n+5\choose 5}$, and the number of outcomes where I roll $n+5$ times and $5$ times are 11-20 is $10^n*10^5*{n+5\choose 5}$, which produces a probability of $(\frac9{10})^n$ that I don't roll a 1 given that I rolled $5$ 11-20s... But I can't simply sum this from $n=0\to\infty$ (which produces $10$ as the answer) because it's missing a multiplicand—the probability that I rolled $n+5$ times. That metric doesn't make sense to me as an independent variable, so this is clearly the wrong approach, but I can't think of a better one.
For $~k \in \{5,6,7,\cdots\},~$ let:
$~f(k)~$ denote the probability that you rolled 11-20 exactly $~5~$ times out of the $~k~$ dice rolls.
Let $~W~$ denote $~\displaystyle \sum_{k=5}^\infty f(k).$
Let $~g(k)~$ denote the probability of not rolling a 1 in any of the $~k~$ dice rolls, given that you rolled 11-20 exactly $~5~$ times [e.g. $~g(5) = 1~$].
Then, the probability that you never rolled a $~1~$ is
$$\sum_{k=5}^\infty \left[ ~\frac{f(k)}{W} \times g(k) ~\right]. \tag1 $$
In (1) above, the fraction $~\frac{f(k)}{W}~$ represents normalizing $~f(k).~$ That is, the fraction represents the relative probability that there were exactly $~k~$ dice rolls.
So, the problem reduces to :
Crafting the formula for $~f(k)$
Computing $~W.$
Crafting the formula for $~g(k).$
$$f(k) = \frac{\binom{k}{5}}{2^k}.$$
$$W = \frac{\binom{5}{5}}{2^5} + \frac{\binom{6}{5}}{2^6} + \frac{\binom{7}{5}}{2^7} + \cdots. \tag2 $$
Personally, while I suspect that there is a closed form expression for $~W,~$ I don't know how to obtain it.
Notice that for $~k \geq 6,~$
$$\frac{\binom{k+1}{5}}{2^{k+1}} \div \frac{\binom{k}{5}}{2^k} = \frac{k+1}{2(k-5)}.$$
So, by the Ratio Test, $~W~$ will be finite.
$$g(k) = \left[\frac{9}{10}\right]^{k-5}. \tag3 $$
In (3) above, you know that exactly $~(k-5)~$ dice rolls were in the range 1 through 10. The probability of an individual dice roll, from these $~(k-5)~$ dice rolls, not being $~1~$ is $~\displaystyle \frac{9}{10}.$