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:
$P(H=1|D=3) = \frac{\frac{1}{3} \frac{1}{2}}{\frac{1}{3} \frac{1}{2}+\frac{1}{3} \frac{1}{2}+\frac{1}{3} 0} = \frac{1}{2}$
Best Answer
Consider two simple variations of the problem:
For a contestant to know the probability of his door choice being correct, he has to know how many positive outcomes are available to him and divide that number by the amount of possible outcomes. Because of the two simple cases outlined above, it is very natural to think of all the possible outcomes available as the number of doors to choose from, and the amount of positive outcomes as the number of doors that conceal a car. Given this intuitive assumption, even if the host opens a door to reveal a goat after the contestant makes a guess, the probability of either door containing a car remains 1/2.
In reality, probability recognizes a set of possible outcomes larger than the three doors and it recognizes a set of positive outcomes that is larger than the singular door with the car. In the correct analysis of the problem, the host provides the contestant with new information making a new question to be addressed: what is the probability that my original guess is such that the new information provided by the host is sufficient to inform me of the correct door? In answering this question, the set of positive outcomes and the set of possible outcomes are not tangible doors and cars but rather abstract arrangements of the goats and car. The three possible outcomes are the three possible arrangements of two goats and one car behind three doors. The two positive outcomes are the two possible arrangements where the first guess of the contestant is false. In each of these two arrangements, the information given by the host (one of the two remaining doors is empty) is sufficient for the contestant to determine the door that conceals the car.
In summation:
We have a tendency to look for a simple mapping between physical manifestations of our choices (the doors and the cars) and the number of possible outcomes and desired outcomes in a question of probability. This works fine in cases where no new information is provided to the contestant. However, if the contestant is provided with more information (ie one of the doors you didn't choose is certainly not a car), this mapping breaks down and the correct question to be asked is found to be more abstract.