Lets say I am trying to paint a wall by randomly (uniform) shooting at it with a paintball gun. On average, how many paintballs will it take so that 99% of it has been covered?
The first paintball is easy, on average it paints N1%=(area of paint mark left)/(area of wall) of the wall.
The second paintball gets a little tricky because there is a chance it will hit the area of the first paintball or overlap with it. The overlap seems to be very tricky to solve (specially since circles) but let's just simplify it for now and say that overlap takes away half a paintball's worth of area on average. so something like N2% = N1% - (1/2 (area of paintball)/(area of wall))/(area of wall). In other words, on average the second paintball will paint as much of the wall as the first paintball, minus a little bit due to the chances of overlapping.
The third paintball is when it starts to get very tricky as you have to take into account the scenario where the second paintball overlapped the first one and the scenario where the second paintball hit a different part of the wall.
After that, the amount of overlap scenarios that you need to take into account seems to rise exponentially as each paintball could or could not have hit one of the previous ones. And you have to take into account every scenario.
Is there an easier way to solve this problem analytically? or is it as complicated as it sounds? (I am willing to tolerate a lot of assumptions like the half paintball overlap, ignoring edge effects, my wife not murdering me, etc...)
If the entire room is to be painted (so that we can ignore edges) then the probability that a particular point on the wall/floor/ceiling is not covered by paint after $n$ paint balls have been fired is given by: $$ P = \left( 1 - \frac{A_{paint}}{A_{wall}}\right)^n $$ where the $A$'s refer to the area of the paint from one paintball and the area of all the interior of the room. If you choose $n$ so that $P=0.01$, the expected fraction of wall area to be covered by paint would be 0.99.