The Wrong Answer of Monty Hall Problem

Question

I know the correct answer is 2/3 for switching. And, I understand the reasoning behind that answer.

But I still don't know why there exists another answer if I choose to do it another way

Let's say I pick the first door and Monty open second hall to show a goat. Should I go to first or third door?

If C3 is the event the car is behind third door and G2 is the event the goat is behind second door.

Then P(G2|C3) = 1 (because if the third door is a car, the second door must be goat)

P(C3) = 1/3 (one car among three doors)

P(G2) = 2/3 (two goats among three doors)

So P(C3 | G2) = P(G2 | C3) * P(C3) / P(G2) = 1/2

So the chance for first or third door is equal.

Can anyone help me point out my logical fallacy I made in my calculation?

Or, how can I make sure the way I do it is the right way to do it for solving Bayes rule question in the future?

Answer 1

If you want to solve the Monty Hall Problem with Bayes Theorem you need to be very careful about each term.

Our prior belief about the location of the car are uninformative thus if $P(C_i)$ is the probability that the car is behind door $i$ we get \begin{equation}P(C_1) = P(C_2) = P(C_3)= \frac{1}{3}\end{equation}

Choice of door is arbitrary so let's start by having the player pick door 1. Monty then evaluates which door to open. Monty's rules for opening a door are:

1. He will not open the player's chosen door.
2. He will not open a door if it has the car/grand prize behind it.
3. Of the remaining openable doors he will choose one at random. (Drawing from a uniform distribution of remaining doors)

The first two rules are fairly obvious to maintain the integrity of the scenario. The third rule is ensures that the player doesn't get some additional information from the pick.

Say Monty reveals door number 2 has a goat behind it. (again as you have noted this problem has a certain permutation symmetry to it so let's just pick one) We now want to update our beliefs. (Using $R_2$ as the event of revealing door 2 has a goat behind it): \begin{eqnarray} P(C_2|R_2) &=& 0\\ P(C_1|R_2) + P(C_3|R_2)&=&1\\ P(C_i|R_2)&=&\frac{P(R_2|C_i)}{P(R_2)} P(C_i)\\ \frac{P(C_3|R_2)}{P(C_1|R_2)}&=&\frac{P(R_2|C_3) P(C_3)}{P(R_2|C_1) P(C_1)}\\ \frac{P(C_3|R_2)}{P(C_1|R_2)}&=&\frac{1\cdot\frac{1}{3}}{\frac{1}{2} \cdot\frac{1}{3}}\\ \frac{P(C_3|R_2)}{P(C_1|R_2)}&=&2\\ \Rightarrow P(C_1|R_2)=\frac{1}{3}\\ \Rightarrow P(C_3|R_2)=\frac{2}{3} \end{eqnarray}

This removes the necessity of calculating the thorny $P(R_2)$. It is a very common method in using Bayes theorem in simulation due to the difficulty in calculating the denominator.

We may, of course, calculate $P(R_2)$ in this problem, using our prior estimation of $C_1$ and $C_3$ \begin{eqnarray} P(R_2) &=& P(R_2|C_1)P(C_1) + P(R_2|C_3) P(C_3)= \frac{1}{2} \cdot \frac{1}{3}+1 \cdot\frac{1}{3}\\ P(R_2)&=&\frac{1}{6} +\frac{1}{3} = \frac{1}{2} \end{eqnarray}

Plugging this into your expression gives the expected $\frac{2}{3}$.

As others have noted what you solved with $P(R_2)=\frac{2}{3}$ corresponds to the situation where Monty will open door number 2 regardless of whether a goat is behind it or not. It also corresponds to Monty not opening door 2, and asking whether to switch between 1 and 3. Essentially you aren't using the information that Monty has in your estimation of $P(R_2)$.