I've been having some rather morbid thoughts lately, so, naturally, I decided to share:
Suppose I have a deadly disease, which has a chance of killing me every day that it is left uncured. Also, suppose that there's this operation I can go through, but it is not without its risks: If it is successful, I won't have to worry about the disease killing me ever again. If it's a failure, I die there and then.
More formally, let $q : N \rightarrow [0,1]$ be the probability that the disease kills me at any given day $t \in N$. Also, let $p : N \rightarrow [0,1]$ be the probability the that the operation fails if I have it on day $t \in N$. Assume that both $p$ and $q$ are monotonically increasing. The question is: At what day $t_0$ do I have to go through the operation in order to maximise my expected lifespan?
Now, to solve this: It is obvious that at any day that $p(t) \leq q(t)$ you're better off having the operation because if you don't, you'd have the same chance of dying and no chance of being cured.
However, that can't possibly be the whole story. What if $p(t) > q(t)$ for every $t \in N$? You would still have to do the operation at some point, otherwise you'll eventually die because of the accumulated probability of all those days you're left uncured.
In that light, let $f(t) = \prod_{k=0}^{t}(1-q(k))$ be the probability that you survive $t$ days without going through the operation. Then, all you need to do is to solve the inequality $f(t_0) < f(t_0-1) \cdot (1-p(t_0))$. Is that correct? Or am I still missing something?