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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?

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For any $s > 0$. –  Memming Mar 30 '13 at 16:47

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if you start with the usual Monty Hall conditions (that is, that the host knows where the goats are located and always reveals goats), even a single door opened implies that the chances of finding the car increase if you switch doors.

If there are $n$ doors with $t$ cars, the probability you choose a car is $t/n$, and changing door leaves a probability $(t-1)/(n-s-1)$ to eventually find the car. The probability you choose a goat is $(n-t)/n$, and changing door leaves a probability $t/(n-s-1)$ to eventually find the car. So the combined probability is $t(t-1)/(n(n-s-1)) + (n-t)(t-1)/(n(n-s-1))$ = $t(n-1)/(n(n-s-1))$

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