Monty Hall Optimization in Python 3.5
If you were using python 2.x, there would be no need to do doors = [i for i in range(1, num_doors+1)], because range just returns a list. So just do:

doors = range(1, num_doors+1)


(similar for host_doors).

Since you are using python 3.x, range will return a range object, but you can just call list on it:

doors = list(range(1, num_doors+1))




Minor point: there is no need for the intermediate variable probability, just return successes / trials.



Python has an official style-guide, PEP8, which recommends not having unnecessary blank lines (such as between your if and else block).

Also, you should add one space in front between # and the actual comment.

It is also customary to use _ as a loop variable that is unused.



You can make your code slightly more efficient by only drawing the user's switched choice when actually switching:

def monty_hall (num_doors, trials, switch):
    successes = 0
    for _ in range(trials):
        # determine behind which door the prize is
        doors = list(range(1, num_doors+1))
        prize_door = np.random.choice(doors)

        # door the guest chooses. (Fixed to 1)
        guest_choice = 1
        doors.remove(guest_choice)

        # determine which door the host opens before guest is allowed to switch
        # this will be faster for larger lists than copying the `doors` list
        host_choice = prize_door
        while host_choice == prize_door:
            host_choice = np.random.choice(doors)
        doors.remove(host_choice)

        # determine which door the guest chooses if switching
        if switch:
            guest_choice = np.random.choice(doors)

        # check win condition
        if guest_choice == prize_door:
            successes += 1

    return successes / trials


It might actually be faster (for larger number of doors it will be), not to copy the doors for the host choice, but just take a random door from doors and take another one if it is the prize_door:

host_choice = prize_door
while host_choice == prize_door:
    host_choice = np.random.choice(doors)
doors.remove(host_choice)


For small lists this will be better:

host_doors = doors[:]
host_doors.remove(prize_door)
host_choice = np.random.choice(host_doors)
doors.remove(host_choice)




Overall the most costly part of this code is probably the removes, which can be quite costly for large lists if removing near the beginning (like for guest_choice). It would be better to store the doors as a set (for which deleting is O(1)). But that makes choosing one door harder (you have to do np.random.choice(tuple(doors)) everytime you draw). You should time both approaches and see what works best for your data.