This might be late, but I feel like I could contribute.
If you don’t want to reroll the question is simple. The question is: “Which result was rolled by which unit?”. There are several ways of doing this I will first describe the simple and then the longer (Which will give the same result).
Simple: Since most of the result where misses, they are not interesting, the only question is where to assing the 3. There is 1 in 4 units that could have been the unit rolling a 3, terefore it is 1 in 4 chance that unit was the artillery.
Longer:
In basic uniform probability, we first start with the assumption that every result is equally likely. Every string of lottery numbers have the same odds. all we have to do then is to count the number of outcomes we like, and compare that to the number of possible outcomes when assigning the results. I will call your units I1, I2, I3 and A. There will be 24 ways of assinging the dies, and 6 of them will assign the 3 to the art. The ways of assigning them are:
A: 3 I1: 6 I2:8 I3:9 = 1 Hit
A: 3 I1: 6 I2:9 I3:8 = 1 Hit
A: 3 I1: 8 I2:6 I3:9 = 1 Hit
A: 3 I1: 8 I2:9 I3:6 = 1 Hit
A: 3 I1: 9 I2:6 I3:8 = 1 Hit
A: 3 I1: 9 I2:8 I3:6 = 1 Hit
A: 6 I1: 3 I2:8 I3:9 = 0 Hit
A: 6 I1: 3 I2:9 I3:8 = 0 Hit
A: 6 I1: 8 I2:3 I3:9 = 0 Hit
A: 6 I1: 8 I2:9 I3:3 = 0 Hit
A: 6 I1: 9 I2:3 I3:8 = 0 Hit
A: 6 I1: 9 I2:8 I3:3 = 0 Hit
A: 8 I1: 3 I2:6 I3:9 = 0 Hit
A: 8 I1: 3 I2:9 I3:6 = 0 Hit
A: 8 I1: 6 I2:3 I3:9 = 0 Hit
A: 8 I1: 6 I2:9 I3:3 = 0 Hit
A: 8 I1: 9 I2:3 I3:6 = 0 Hit
A: 8 I1: 9 I2:6 I3:3 = 0 Hit
A: 9 I1: 3 I2:6 I3:8 = 0 Hit
A: 9 I1: 3 I2:8 I3:6 = 0 Hit
A: 9 I1: 6 I2:8 I3:9 = 0 Hit
A: 9 I1: 6 I2:9 I3:8 = 0 Hit
A: 9 I1: 8 I2:6 I3:3 = 0 Hit
A: 9 I1: 8 I2:3 I3:6 = 0 Hit
In total. That is 6 possible hits, of 24 possible scenarios. Which is 1 in 4.
