Can there be generalization of Monty Hall Problem? [duplicate]

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?

Solution 1:

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

Solution 2:

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 $(t>0)$.

Solution 3:

It has been done here - The Monty Hall Paradox Probability Equations.

A quick summary below. It will greatly simplify the calculations if we start with these two ideas:

1. The probability values for $t$ doors having cars behind them are just $t$ times the original values (when there is only one door having a car). This is simply because the winning chances of all the strategies and subsets are simply inflated by $t$ by having $t$ winning doors. So let's start with $t=1$.

2. We assume that the host momentarily identified a set $y > s$ doors (just in his mind) and opened $s$ doors from this $y$ set. Later we'll just make $y=n-1$.

In the beginning, the winning probability of group $y$ as a whole is simply $y \over n$. And, if $y$ wins, the winning probability of each door inside it is $1 \over y$. With $s$ non-winning doors now opened from $y$, the winning probability of each door in $y$ becomes $1 \over y-s$, provided $y$ wins. But $y$ doesn't always win.

Hence, the absolute winning probability of each door within $y$ is simply ${y \over n} \times {1 \over y-s}$, or ${1 \over n} \times {y \over y-s}$. Since ${y \over y-s} >1$, switching to any door within $\boldsymbol{y}$ is always advantageous - even for $\boldsymbol{s=1}$.

Finally, on getting rid of $y$ by making it as large as $n-1$, i.e. it having every door except the player's original choice, and having $t$ doors instead of $1$, we get the winning probability of switching which is: $$ {t \over n} \times {y \over y-s} = {t \over n} \times {n-1 \over n-1-s} $$

Have a look at these two pages for comparing the real world results with the predictions from the equation - Monty Hall Paradox Test Tool and The Monty Hall Paradox Probability Equations.