Can there be generalization of Monty Hall Problem?

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 $t(t-1)/(n(n-s-1)) + t(n-t)/(n(n-s-1))$ = $t(n-1)/(n(n-s-1))$


Following the notation of the previous posts in this thread, the contestant should switch if the following inequality is true: $$\frac{t(n-1)}{n(n-s-1)} > \frac{t}n$$ which reduces to: $$nst(n-s-1)>0$$ Ignoring solutions with negative values for $n$, $s$, and $t$, the above inequality is only true when $s>0$, $n-s>1$, and $t>0$.

The contestant should switch if Monty reveals at least one goat $(s>0)$, Monty leaves the contestant more than one door to choose from $(n-s>1)$, and there is at least one car $(t>0)$.

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Probability Theory

Monty Hall

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