Probability of getting a defined value by freely assembling the top 3 results of a D4+D6+D8+D10 roll I'm trying to compute the probability of getting a given value with the following rules :


*

*Roll one 4 sided dice + one 6 sided dice + one 8 sided dice + one 10 sided dice.

*Pick the 3 largest results.

*Choose a value by selecting one dice, assembling 2 dice or even summing all 3 dice. 



Example:
  
  
*
  
*I roll 2 on the 4 sided dice
  
*I roll 1 on the 6 sided dice
  
*I roll 6 on the 8 sided dice
  
*I roll 5 on the 10 sided dice 
  
*I pick the 3 largest results : 2,6,5
  
*Then, I can choose any of the following values: 2, 6, 5, 2+6=8, 2+5=7, 6+5=11, 2+6+5=13
  

What is the probability of getting a 2 or a 15 or even a 24 with this rules?
In fact, I am searching for a way to generalize this problem so that I can change the number of dice or the number of sides of each dice or even the number of top dice to keep. I created a computer program to obtain this probabilities using brute force, but I would like a more elegant and faster solution. 
 A: too long for a comment apparently:
just by examining sums of dice you can come up with a lot of restrictions:


*

*any total die sum over 18 must include the 10 sided die (4+6+8=18) 

*any total die sum over 20 must include the 8 sided die (4+6+10=20)

*any total die sum over 22 must include the 6 sided die (4+8+10=22)

*any three die sum over 10 must contain either the 8 or 10 sided dice(4+6=10)

*any three die sum over 12 must contain either the 6 or 10 sided dice(4+8=12)

*any three die sum over 14 must contain either the 6 or 8 sided dice(4+10=14)

*any three die sum over 16 must contain either the 4 of 8 sided dice(6+10=16)

*any three die sum over 18 must contain either the 4 or 6 sided dice(8+10=18)

*any three die sum over 22 must contain the 6 sided die(5 or 6 only, 4+8+10=22)


anyways I'll let you do the actual math. 
A: Comment: I don't immediately see a method that isn't tedious. 
If I were using enumerative methods such as those suggested so far,
I would almost surely leave out some cases on my first try. 
As a check on
your work, it is easy to see that the mean of the sum of three
should be 12.
Also, in a simulation of a million runs, I got the following answers
to four places; at least two or three places should be accurate.
s3
     3      4      5      6      7      8      9     10     11     12     13 
0.0038 0.0109 0.0219 0.0365 0.0520 0.0674 0.0805 0.0916 0.0974 0.0987 0.0941 
    14     15     16     17     18     19     20     21     22     23     24 
0.0852 0.0738 0.0594 0.0458 0.0325 0.0216 0.0134 0.0075 0.0039 0.0016 0.0005 


I hope someone will suggest a clever combinatorial method.
