Solution: The total number of ways to choose 3 distinct letters is $inom263$. The favorable cases involve selecting 3 letters that correspond to exactly one placeholder symbol. Assuming each symbol is defined by a unique combination of 3 letters (with no overlap in symbol definitions), there are 10 favorable combinations. Thus, the probability is $rac10{inom263} = rac102600 = rac1260$. $oxed{\dfrac1260}$ - Dyverse