This question came up during a D&D game I was playing in and I found that I was unable to come up with an answer. I believe this should be a summation question, but I may be entirely off:
You roll a 20 sided die. A success is defined as rolling 16 or higher and failure is defined as rolling 15 or lower. You roll the die until 60 trials have been completed or you accumulate 6 failures. What is the probability of accumulating 6 failures, and how would you find the probability of lasting an arbitrary number of trials before accumulating 6 failures?
I will answer the easiest first. You want the probability that in 60 rolls, you roll 16 or higher all but at most five times. That is a binomial probability. The probability of survival is given by this formula:
$$\sum_{k=0}^5 \dbinom{60}{k}\left(\dfrac{1}{4}\right)^{60-k}\left(\dfrac{3}{4}\right)^{k} \approx 1\times 10^{-25}\%$$
Now, the probability that you can survive a given number of trials is:
You have 100% probability of surviving up to 5 trials. Probability that you survive up to $n$ trials (for $n\ge 6$): $$\sum_{k=0}^5 \dbinom{n}{k}\left(\dfrac{1}{4}\right)^{n-k}\left(\dfrac{3}{4}\right)^{k}$$
Probability you lose on the $n$-th trial: $$\dbinom{n-1}{5}\left(\dfrac{1}{4}\right)^{n-6}\left(\dfrac{3}{4}\right)^6$$
Expected number of trials before finally succumbing to luck: $$\sum_{n\ge 6}n\dbinom{n-1}{5}\left(\dfrac{1}{4}\right)^{n-6}\left(\dfrac{3}{4}\right)^6 = 8$$
You can expect to fail by the eighth trial (on average).