Summarising all existing answers

(And adding a few of my points)

Explanation :

if name == "Kevin" or "Jon" or "Inbar":

is logically equivalent to:

if (name == "Kevin") or ("Jon") or ("Inbar"):

Which, for user Bob, is equivalent to:

if (False) or ("Jon") or ("Inbar"):

NOTE : Python evaluates the logical value of any non-zero integer as True. Therefore, all Non-empty lists, sets, strings, etc. are evaluable and return True

The or operator chooses the first argument with a positive truth value.

Therefore, "Jon" has a positive truth value and the if block executes, since it is now equivalent to

if (False) or (True) or (True):

That is what causes "Access granted" to be printed regardless of the name input.

Solutions :

Solution 1 : Use multiple == operators to explicitly check against each value

if name == "Kevin" or name == "Jon" or name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Solution 2 : Compose a collection of valid values (a set, a list or a tuple for example), and use the in operator to test for membership (faster, preferred method)

if name in {"Kevin", "Jon", "Inbar"}:
    print("Access granted.")
else:
    print("Access denied.")

OR

if name in ["Kevin", "Jon", "Inbar"]:
    print("Access granted.")
else:
    print("Access denied.")

Solution 3 : Use the basic (and not very efficient) if-elif-else structure

if name == "Kevin":
    print("Access granted.")
elif name == "Jon":
    print("Access granted.")
elif name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

There are 3 condition checks in if name == "Kevin" or "Jon" or "Inbar":

• name == "Kevin"
• "Jon"
• "Inbar"

and this if statement is equivalent to

if name == "Kevin":
    print("Access granted.")
elif "Jon":
    print("Access granted.")
elif "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Since elif "Jon" will always be true so access to any user is granted

Solution

You can use any one method below

Fast

if name in ["Kevin", "Jon", "Inbar"]:
    print("Access granted.")
else:
    print("Access denied.")

Slow

if name == "Kevin" or name == "Jon" or name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Slow + Unnecessary code

if name == "Kevin":
    print("Access granted.")
elif name == "Jon":
    print("Access granted.")
elif name == "Inbar":
    print("Access granted.")
else:
    print("Access denied.")

Not-empty lists, sets, strings, etc. are evaluable and, therefore, return True.

Therefore, when you say:

a = "Raul"
if a == "Kevin" or "John" or "Inbar":
    pass

You are actually saying:

if "Raul" == "Kevin" or "John" != "" or "Inbar" != "":
    pass

Since at least one of "John" and "Inbar" is not an empty string, the whole expression always returns True!

The solution:

a = "Raul"
if a == "Kevin" or a == "John" or a == "Inbar":
    pass

or:

a = "Raul"
if a in {"Kevin", "John", "Inbar"}:
    pass

Using match/case in Python 3.10 and above

Python 3.10 adds a new syntax to the language. It's officially described as "structural pattern matching", but most people call it according to the syntax: "match/case".

• Technical specification
• Motivation and rationale (i.e., why it was added, and what inspired the design)
• Official tutorial

We can use this special syntax for an example like in the question, by making one "case" that matches all the accepted usernames, and using the "wildcard" case _ in place of the else. Thus:

name = input("Hello. Please enter your name: ")
match name:
    case "Kevin" | "Jon" | "Inbar":
        print("Access granted.")
    case _:
        print("Access denied.")

Note that cases are "combined" using |, not or. This is a special syntax: Python does not try to compute "Kevin" | "Jon" | "Inbar" first (| doesn't work with strings), but instead interprets the entire line differently because it starts with case.

Simple engineering problem, let's simply it a bit further.

In [1]: a,b,c,d=1,2,3,4
In [2]: a==b
Out[2]: False

But, inherited from the language C, Python evaluates the logical value of a non zero integer as True.

In [11]: if 3:
    ...:     print ("yey")
    ...:
yey

Now, Python builds on that logic and let you use logic literals such as or on integers, and so

In [9]: False or 3
Out[9]: 3

Finally

In [4]: a==b or c or d
Out[4]: 3

The proper way to write it would be:

In [13]: if a in (b,c,d):
    ...:     print('Access granted')

For safety I'd also suggest you don't hard code passwords.

Besides some other rather rarer useful cases for the walrus operator already mentioned. This also tend to be a useful case as well.

def calc_value():
    return 43

if (v := calc_value()) == 43 and v > 42:
    print('happy short, efficient and readable code')

This works because each part of the if-statement is read separately. So (v := calc_value()) is executed and a value is assigned to v and if the first fails, you still have v in the namespace for different conditions or calculations.

If your goal is to check against a known list or key value pairs its very nice to try and use the "in" comparison operator. That way you can check your selected value against the collection with out much repetition in your code. I found this especially useful when fiddling with a password generator such as:

word = input()
password = ''
swap = {assume a dictionary of characters to be swapped out here}

for i in word:
    if i in swap:
        password += swap[i]
    else:
        password += i
print(password + 'q*s')

The "if i in swap:" here did the job of checking against a number of conditions all at once.