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I think you are making a mistake in the denominator of the former. $P(A) = \sum_B P(A|B) P(B)$

Since rushing made me write something kinda terrible I'll pay some penance by writing out a more full solution!

The Bayes Approach to the Monty Hall Problem

In general I think you are making the Monty Hall problem a little bit more confusing when you omit the player's choice. The player chooses an arbitrary door so without loss of generality let's call it one. I believe this is what you did, but let's just be a bit more explicit. Next we write out the probability:

$P(H=1|D=3) = \frac{P(H=1) P(D=3|H=1)}{P(D=3)} $

We can calculate the numerator as $P(H=1) = \frac{1}{3}$ since each door is equally likely. We also calculate $P(D=3|H=1) = \frac{1}{2}$ since if $H=1$ Monty can pick either door without showing the grand prize. Moving on:

$P(H=1|D=3) = \frac{\frac{1}{3} \frac{1}{2}}{P(D=3|H=1)*P(H=1)+P(D=3|H=2)P(H=2)+P(D=3|H=3)P(H=3)}$

Next we have a set of terms to consider: $P(H=i)$ is $\frac{1}{3}$ for the same reason as above. Finally the conditionals: $P(D=3|H=1)=\frac{1}{2}$ as above. $P(D=3|H=2) = 1$ this is because Monty can't show door 1 since the player picked it, Monty can't show door 2 because it has the grand prize, and thus Monty must show door 3. Finally there is $P(D=3|H=3) = 0$ this is because Monty can't show the prize. Thus:

$P(H=1|D=3)= \frac{\frac{1}{3} \frac{1}{2}}{(\frac{1}{2}+ 1+ 0)(1/3)} = \frac{\frac{1}{3}\frac{1}{2}}{\frac{1}{3} \frac{3}{2}}=1/3$

Lonely Monty Hall Problem

This is not to be interpreted as the standard Monty Hall problem. Say that we revisit the problem without a player. There are again three doors, one of which has a grand prize behind it. Monty will choose and open a door that doesn't have the grand prize behind it and we are tasked with evaluating the probability that the grand prize is behind door 1.

The original write up of the question doesn't explicitly include the player so this is a fair interpretation.

Calculation via Bayes Theorem:

Let's start by just writing out the expression:

$P(H=1|D=3) = \frac{P(H=1) P(D=3|H=1)}{P(D=3)}$

Working from here $P(H=1) = \frac{1}{3}$ and $P(D=3|H=1)=\frac{1}{2}$ as above. We would then expand the denominator:

$P(D=3) = \sum_i P(D=3|H=i) P(H=i)$

Now what's different is that the player has not made a choice on door so we need to calculate the various conditional probabilities. $P(D=3|H=1) = \frac{1}{2}$ since Monty has two choices open door 2 or door 3. Similarly, $P(D=3|H=2) = \frac{1}{2}$, finally $P(D=3|H=3) = 0$ since Monty can not reveal the prize.

Overall this gives:

$P(H=1|D=3) = \frac{\frac{1}{3} \frac{1}{2}}{\frac{1}{3} \frac{1}{2}+\frac{1}{3} \frac{1}{2}+\frac{1}{3} 0} = \frac{1}{2}$