You have 3 doors. There are two goats and one super car in the doors.
If you select one door, the show host will open one of the remaining two doors. In that door is a goat.
Will you change your decision??
There are three events.
First, the car is in \(i^{th}\) door. Let this be \(C_i\).
Two, you choose \(i^{th}\) door out of three doors. Let this be \(D_i\).
Three, host opens \(i^{th}\) door with a goat. Let this be \(G_i\).
Naturally \(C_i\)s are disjoint, \(D_i\)s are disjoint, \(G_i\)s are disjoint.
Let’s say that you chose the door #1. Then the host showed the door #2 with a goat.
If you have a more rational argument than to decide to change the decision by throwing a coin, it is natural to follow it.
And yeah, it does exist!!
Let the probability of the car not being in the door you’ve chosen be \(P(C_3\ |\ D_1,G_2)\).
If \(\theta\)s are disjoint, \[P(\theta\ |\ x)=\frac{P(\theta)P(x\ |\ \theta)}{\sum_\theta P(\theta )P(x\ |\ \theta)}\]
Using this theorem, we can calculate \(P(C_3\ |\ D_1,G_2)\).
\[P(C_3\ |\ D_1,G_2)=\frac{P(D_1 ,G_2\ |\ C_3)P(C_3)}{\sum_{i\in I} P(D_1 ,G_2\ |\ C_i)P(C_i)}\]
And because \(\forall i \in I\) \(P(C_i)=\frac13\), this reduces to \[P(C_3\ |\ D_1,G_2)=\frac{P(D_1 ,G_2\ |\ C_3)}{\sum_{i\in I} P(D_1 ,G_2\ |\ C_i)}\]
Meanwhile, \[P(D_1,G_2\ |\ C_3)=P(G_2\ |\ D_1,C_3)P(D_1\ |\ C_3)\].
So \[P(C_3\ |\ D_1,G_2)=\frac{P(G_2\ |\ D_1,C_3)P(D_1\ |\ C_3)}{\sum_{i\in I} P(G_2\ |\ D_1,C_i)P(D_1\ |\ C_i)}\]
First, the fact that car is in some door does not affects your first choice of door, because you don’t know where the car is. Therefore, the first choice given \(C_i\) is random.
\[\forall i\in I\ \ \ P(D_1\ |\ C_i )=\frac13\]
Yes, this is a bit complicated but significant process.
Second, the probability of existing a goat in the door where is a super car is zero. \[P(G_2\ |\ D_1,C_2)=0\]
Third, show host opened the door #2 because there is a car in the door #3. \[P(G_2\ |\ D_1,C_3)=1\]
Fourth, if you chose the door with the car, the show host can pick one of the two doors with goats and open it. \[P(G_2\ |\ D_1,C_1)=\frac12\]
So final equation is ; \[P(C_3\ |\ D_1,G_2 )=\frac{P(G_2\ |\ D_1,C_3)}{P(G_2\ |\ D_1,C_1)+P(G_2\ |\ D_1,C_3)}=\frac{1}{\frac12+1}=\frac23\]
People tend to stick to their decisions because show hosts seem to change their decisions.
But if you saw the door with a goat, you’d better change your decision!