Unfortunately in my case I have no idea of the positions of the lost characters, but I keep trying, because it has been very useful for me to learn new skills and resources.
In which case, I think the search space is something like (58chars*47positions)*(58chars*47pos)*(58chars*47pos)*(58chars*47pos)... (58*47)4 = 5.5220891*1013 possible combinations 
---snip---
the first 2 chars always start with '5H', '5J', '5K', 51-2 = 49 positions and 49-4 = 45 characters known
if exact position of 4 lost characters are known = 58^4 = 11,316,496 iterations
if exact position is unknown = 58^4 * 4-combination
= (58^4) * ( 49! / 4! (49-4)! ) = (58^4) * (46*47*2*49)
= (58^4) * 211876 = 2,397,693,906,496 iterations = 2.39 * 1012
do I get this right?
I am pretty sure you made a small mistake there. You do ignore the order of the 4 missing words with your statement: ( 49! / 4! (49-4)! )
You do iterate through each position with your words, but you have to iterate within the words too.
This should be the correct calculation:
There are 58 choose 4 possible combinations to pick the 4 correct chars from the charspace (without considering the order).
Now consider the first two characters as already known. Only looking at the private key without the first two chars here:
To put the first unknown char into the correct place, you can choose between 1) before of each already existing char (45 possibilities) and 2) behind the last one (45+1).
For the second one you choose either before each char in your privkey (46 possibilities) or behind the last one (46+1)...
Overall there are 46 * 47 * 48 * 49 possible combinations to place the 4 chars into a private key with 45 known characters (ignoring the first two) into the right order. This assumes the already known characters are in a correct order already.
The total search space (for the private key without the first 2 chars) therefore is 46 * 47 * 48 * 49 * (58 choose 4) = 2.1574231*1012.
Multiplied with 4 (combinations the priv key can start with) = total amount of combinations = 8.6296925 * 1012
Considering the birthay paradox, there is a 50% chance to find the correct private key after 1/2 of this space.
Feel free to correct me if i have made a mistake!