Expected value of number of tiles needed to fill up a given area I randomly thought of this question in the car: Given that you have a $100ft$ $*$ $150ft$ field and that $1.5ft$ $*$ $1.5ft$ tiles are randomly placed within the perimeter of the field with overlaps allowed, what is the expected value of the number of tiles needed to fill the area.
I thought it would be simple enough, but I tried everything and was not able to make any headway. Any thoughts?

 A: The simple answer is that if the tiles are required to be placed entirely within the field, it will take an infinite number of tiles to cover the whole field.  The corners can only be covered if the tile is placed precisely over one point, which occurs with probability zero.  
A harder question is if you allow the centers of the tiles to be anywhere within the field.  Then the corners are still four times less likely to be covered than center points by any given tile.  You can get a rough idea by covering the field with a $1.5 \times 1.5$ foot grid, which is about $67 \times 101$ locations.  Each location has a chance of about $1/(67\cdot 101)$ of being covered by a given tile.  We can now use a Poisson distribution to say the expected number of tiles covering a spot when we have placed $n$ tiles is $\frac n{6767}$ so each spot has a $e^{-\frac n{6767}}$ chance of being uncovered.  The chance that no spot is uncovered (ignoring correlations) is $(1-e^{-\frac n{6767}})^{6767}$.  If you want a chance of $p$ that no spot is uncovered, we do 
$$(1-e^{-\frac n{6767}})^{6767}=p\\
6767 \log(1-e^{-\frac n{6767}})=\log p\\
e^{-\frac n{6767}}\approx -\frac {\log p}{6767}\\
-\frac n{6767}\approx \log\left(-\frac {\log p}{6767}\right)\\
n\approx -6767\log\left(-\frac {\log p}{6767}\right)$$
For $p=0.5$ you need $n\approx 62163$ and for $p=0.9$ you need $n \approx 74912$  Basically you need to cover the field about $10-12$ times deep to cover it all.
A: Here is an approach via Monte Carlo simulation.  It seems hard to write a program that keeps track of exactly what parts of the field that are covered, so we discretize (if that's a word) the problem by covering the field with a grid of points at 0.1 ft intervals, and we consider the field covered if all the grid points are covered.  So what we end up with is an approximate solution to an approximate version of the original problem.
With all those caveats in mind, here is the algorithm.  Cover the field with a grid of points at 0.1 ft intervals, from (0,0) to (100,150).  All the points are originally "uncovered". Generate a random X from 0 to 100 and a random Y from 0 to 150.  Mark all the grid points within the 1.5 ft square centered at (X,Y) as covered.  Repeat until all the grid points are covered.  Code in Python 2.7 follows:

    # 1 unit of measurement = 0.1 foot

import math
import random

def ndraws():
    "return the number of draws required to cover 1000 X 1500 grid"
    covered = [[False for j in range(1501)] for i in range(1001)]
    ncovered = 0
    ntrials = 0
    while ncovered < 1501 * 1001:
        ntrials += 1
        x = random.uniform(0, 1000)
        y = random.uniform(0, 1500)
        ilo = max(0, math.trunc(x-6.5))
        ihi = min(1001, math.trunc(x+7.5))
        jlo = max(0, math.trunc(y-6.5))
        jhi = min(1501, math.trunc(y+7.5))
        for i in range(ilo, ihi):
            for j in range(jlo, jhi):
                if not covered[i][j]:
                    covered[i][j] = True
                    ncovered += 1
    return ntrials

# set the random number seed, for reproducibility
random.seed(1234)
# run 100 trials and report the number of draws required in each trial
for i in range(100):
    print ndraws()

In 100 repetitions, the average number of 1.5 by 1.5 squares required was 127,220.  A 95% confidence interval is 123,435 to 131,006.
A: Maybe consider square regions $1.5 \times 1.5$ whose centers make grid with spacing $0.75$ apart. Then consider the probability of picking point inside, $\frac{9}{240000}$ then the repeated probabilities to fill the rectangle. Consider now as the $.75$ sides are generalized to any size.
