Monty Hall problem basically is: there are three doors, behind one of these doors is a brand new car, and behind the other two are goats, one picks a door that one thinks has a car behind, then the host opens one of the doors one didn't pick to reveal one of the goats, and the host asks the person whether one wishes to switch his choice. By the nature of probability, it is always better to switch.
I would like to generalize this case to the following variation:
Assume there are many finite doors (cardinality $n$). $t$ doors have cars behind. As in the original formulation, one picks a door. Then, the host reveals $s$ doors that have goats. Then, the host as in the original formulation asks whether one wishes to switch. For any possible $n$ bigger than 3, and given $t$, how is one able to find minimum $s$ that makes switching the choice always a better choice?