Return same position for similar values in a list

I have sorted a list in descending order, I am trying to find a way in which similar numbers re-appearing in the list will return equal position and the next unique number to return the next position. Example in this samplelist = [50, 40, 40, 30, 30, 20, 10] 50 should return 1, 40 returns 2, next 40 returns 2, 30 returns 4, next 30 returns 4 then 20 returns 6, 10 returns 7. Note that this will be a dynamic list. This is how far I have gone but it isn’t quite right.

samplelist = [10, 20, 30, 30, 40, 40, 50]
samplelist.sort(reverse=True)
print(samplelist)
appearances = {}
for index, value in enumerate(samplelist):
    appearances[value] = samplelist.count(value)
for index, value in enumerate(appearances.items()):
    print(f'Position for {value[0]} = {index + value[1]}')

Answer

This can also be done using just the list.index utility in python, which gives the index of the first occurrence of an element.

So, your program must do this :

appearances[value] = samplelist.index(value)

instead of this :

appearances[value] = samplelist.count(value) # It just returns the count of the value in samplelist