Probability of painting the whole wall with bars of 1x10 pixel length

Question

Today I faced an interesting question. Let me tell the full story, my colleauge was in an muesum and saw a specific art. The painter was writing words on a wall until the wall was nearly filled up to 100% with red paint. It was not live, but you saw the red wall and a cinematic how the painter was working. So he asked me, how many words you have to write until the wall is filled to 100%. Since this can't be solved in a correct matter, we abstracted it.

We have a wall of 50 x 10 Pixels in size. One word consists of 10x1 Pixels and you can write them from top to bottom or left to right or even diagonal. As shown below.

You can start your word or line from any point of the map, so there are 500 starting points. It's also allowed to cross the border and finish the word on the opposite.

So for the first word you have 500 pixels to start and you will for sure not touch any other pixel, since there are none. so you will take 1/50 of the space. The second word has may touch some or even all of your pixels (in case its the identical line). so 10/500 is already black. To me its like drawing a balls out of an urne... if you pick 10 balls out of 500 where 490 are white and 10 are black you still have a good chance of picking only white ones :) 490/500 equals 98% for the first pick. which leaves us 10/499 spots left, since I wont hit my first pixel again ...

There must be some better formula behind all this, can you guys/girls help me out? While I try to do it on my own here :)?

Answer 1

Not really an answer, but there's no way to fit it in a hint. I wrote a python script to simulate the experiment:

from random import randint,seed

def color(direction, x, y):
    if direction == 0:
        return { (a%width,y) for a in range(x,x+10) }
    if direction == 1:
        return { (x,a%height) for a in range(y,y+10) }
    if direction == 2:
        return {((x+a)%width, (y+a)%height) for a in range(10)}
    if direction == 3:
        return {((x-a)%width, (y-a)%height) for a in range(10)}

width = 50
height = 10
trials = 1000
average = 0

seed()

for _ in range(trials):
    count = 0
    pixels = set()
    while len(pixels)<width*height:
        direction = randint(0, 3)
        x = randint(0,49)
        y = randint(0,9)
        pixels |= color(direction,x,y)
        count += 1
    average += count
print(average/trials)

I've run it several times, and it's always produced answers in the range $(330,340).$ This is considerably smaller than $4\cdot50H_{50}\approx900,$ which I knew was an overestimate, but it's also a lot smaller than I can come to grips with. The expected number of words needed to cover a particular pixel is $50$. So at the end, if we have a couple of uncolored pixels left, that can't be covered by the same word, it takes 100 words, on average, to color both of them. I would expect it to take much longer than $340$ words to color all the pixels.