How to estimate your score for an HSK practice exam? I'm preparing for my first HSK exam in October - probably level 1 - so I completed a practice exam. I know the number of correct answers, but I don't how to convert these into a score. 

From what I've read, it's not a simple matter of dividing the number of correct answers by the total number of questions and then multiplying by the maximum score. Apparently the score is calculated by comparing with other students.

Where can I learn more about this? Is there at least some sort of heuristic, like: if you answer X percent correctly, you'll probably pass?
Brief Answer
To pass any level of the new HSK, you have to beat 74.75% of the total.
That is to say, for this case

I had 11/20 correct answers in the listening part and 19/20 in the reading part.

if 74.75% of the total had done worse than that (i.e. as the weights of listening and reading are 1:1, such many people had correctly answered less than 30 questions), you will pass the exam.
For &quot;Is there at least some sort of heuristic, like: if you answer X percent correctly, you'll probably pass?&quot;, the answer is maybe. There may be some historical data that you can refer to. They serve as important references, but you cannot 100% trust them, if the HSK exam you take is specially more difficult or easier than former ones.
Why 74.75%?
From the official website of the HSK, we know
+------------------+--------------------+---------+
| HSK Score(Total) | HSK Score(Section) | Percent |
+------------------+--------------------+---------+
|              400 |                100 |      99 |
|              277 |                 69 |      90 |
|              250 |                 63 |      80 |
|              231 |                 58 |      70 |
|              215 |                 54 |      60 |
|              200 |                 50 |      50 |
|              185 |                 46 |      40 |
|              169 |                 42 |      30 |
|              150 |                 37 |      20 |
|              123 |                 31 |      10 |
+------------------+--------------------+---------+

And from its description:

A section score of HSK (Elementary-Intermediate) is a scale score with 50 as the score mean and 15 as the standard deviation, while the total score is a scale score with 200 as the score mean and 60 the standard deviation ... The right column indicates the percentage occupied by those whose scores are lower than the corresponding scores in the standard HSK reference group.

It is easy to verify via the table that a section score meets the normal distribution (with with 50 as the score mean and 15 as the standard deviation), so is the total score (with 200 as the score mean and 60 the standard deviation).
Besides, this page introduces a calculation method. In brief, it is

HSK=50+15×Z

Where Z can be considered as a Gaussian variate with 0 as the mean and 1 as the standard deviation, and then the HSK for a section is a N(50, 15^2) Gaussian variate. By observation, we can find that, there're some regular patterns for the setting of the total score. If an exam has 4 section, the setting for total score is a N(200, 60^2) variate. Consider these passing scores

HSK level 1-2: listening (100), reading (100). Total (200). Passing: total score &gt;= 120.
HSK level 3-6: listening (100), reading (100), writing (100). Total (300). Passing: total score &gt;= 180.

We can conclude that at level 1-2, the total score is a N(100,30^2) variate, and at level 3-6, N(150, 45^2). Thus we can see that in the standard Gaussian cumulative distribution function:

Φ[ (120-100)/30 ] = 0.7475
Φ[ (180-150)/45 ] = 0.7475

PS: it seems in the old HSK, the weights of listening, grammar, reading, and comprehensive sections are with this ratio 1:0.6:1:0.8. However if we try the weighting formula for the old HSK total score

HSK = [ (Z1+0.6×Z2+Z3+0.8×Z4) / 3.4 ] ×60 +200

where Z1,Z2,Z3,Z4 i.i.d. ~ N(50,15^2), the final HSK will be a N(200, 30.566^2) variate, which is a contradiction to another definition N(200, 60^2). Anyway, as the math model hasn't been rigorously stated, my answer considers N(200, 60^2) as the correct one because its CDF value examples are well listed.