Why does round(1/2) returns 0 but round(7/2) returns 4 in Python?

Hello I was trying to determine if a element of a given list was part of the second half of the list. For example if it had 7 elements

my_list = ['a','b','c','d','e','f','g']

my function would only return True for elements ‘e’, ‘f’ and ‘g’.

The solution was pretty simple:

def fashionably_late(arrivals, name):
    
    return (arrivals.index(name) + 1 > round(len(arrivals)/2))

My solution had a bug though because of how rounding works, I just want to understand why

However this behavior from python is perplexing, why is 7/2 (i.e 3.5 ) rounded to 4 the same way as 9/2 (i.e. 4.5) to 4.

for i in range(1,11):
    print(i , round(i/2))
​

    1 0
    2 1
    3 2
    4 2
    5 2
    6 3
    7 4
    8 4
    9 4
    10 5

#The way I expect it to work is: 

for i in range(1,11):
    print(i , i//2 + i %2)
​
1 1
2 1
3 2
4 2
5 3
6 3
7 4
8 4
9 5
10 5

As a side note I am aware of the math.ceil() function.

Answer

From the docs

if two multiples are equally close, rounding is done toward the even choice (so, for example, both round(0.5) and round(-0.5) are 0, and round(1.5) is 2)

This convention is often called statistician’s rounding, as it produces more accurate results over large aggregate data (always rounding up on 0.5 increases the average of the data by 0.5, but rounding either way using a consistent rule keeps the mean more consistent, on aggregate)