Say you have two events which have a given chance to happen on each trial. I know that for any one of the events, the number of trials before a success is described by the geometric distribution.
What I want to know is how to calculate the chance that one of the events comes before the others.
Here is an example I have in mind: Imagine you are pulling pills randomly from a bag. Some of them are sugar pills which do nothing, some are poison, and some are antidotes. If you pull the antidote before the poison you survive, otherwise not. Assume the pills are replaced so there is always the same chance of pulling each type on any given trial.
The number of trials before poison or the number of trials before antidote is a geometric distribution. But I want to know the chance of getting the antidote before any poison pills.
I would like a general solution but as an example say 2% of the pills are poison and 4% are antidote. What is the chance you draw an antidote pill before any poison pills?
The best strategy in such cases with replacement is to toally ignore the object(s) that don't matter, here the sugar pills.
Secondly, since it is with replacement, all probabilities remain constant, and the probabilities ignoring sugar pills are $\Large\frac23$ for an antidote, and $\Large\frac13$ for a poison pill.
Thus P(antidote drawn before poison) $=\Large\frac23$