model sketch
I'll try to summarize the model we're using so far and here some additional reading that might help:
Spatial randomness
Poisson distribution
Assumptions for mining:
1) There exists a base unit area to which one and only one claim can be assigned.
2) All base units do have the same probability p to get a claim assigned and are independent of each other.
Hence a mining field (we can call it spot) is made up of n = n1*n2 base units.
With a finder of range r we are searching r^2*pi such base units contemporarily.
The probability to get at least one find would correspond then to a binomial distribution with number r^2*pi attempts on p.
If we do assume that the base unit is a 1mx1m square, then we would search about 9503 such units with a finder of range 55m.
To model this with a binomial distribution is quite laborious. If p is very small and n is very large, then the binomial distribution converges to a Poisson distribution with parameter l = n*p, the density of the points in the mining field.
Moreover, the probability to get a certain number of finds in a a subarea of the mining field will be proportional to the probability of a find in the mining field. Hence, we don't need to know n and p but only l and do assume that for the area that we are searching some l' proportional to l exists. Let's use the following notation:
Po(l,k) = l^k*exp(-l)/k!
to denote a Poisson distribution with parameter lambda = l. Po(l,k) is then the probability to get exactly k finds in the search area with density l.
Sry, but all this was necessary to define and describe spatial randomness. Now let's start with our observations.
Solution
The only thing we do know is the probability of at least one find in the search area covered by the finder. Hence, we only know
h = hit rate = P(at least one find in search area) = 1-Po(l',0).
From this we can calculate the density l' for the search area (please note that this is not the whole mining field).
l' = -ln(1-h)
When rebombing (image shows optimal situation with find on perimeter)
[br]Click to enlarge[/br]
part of the old area is searched again (dark) plus a new area (light). Hence, the probability to get at least one find in the second bomb is a combination of the hit rate of those two areas.
Let's call the search area A, the part of the old area O and hence the new area N becomes A-O. This can be also written as
O = A*w,
N = A*(1-w), with a w >= 0 and <=1.
A rebomb will occur number hit rate times, i.e. h times. Hence overall hit rate will be
h' = h(1+h2)/(1+h), where
h = hit rate
h2 = hit rate of second bomb.
Hence the only thing that we need to find is h2.
Immortal uses an approximation here in assuming that only the hit rate of the new area is relevant not accounting for additional finds in the part of the old area O.
Since we know, that we deal with Poisson distributions, we can model this, assuming that area O is not too small. Otherwise the approximation with the Poisson distribution will not work and we might have to switch back to a binomial one.
h2 = 1- P(XO + XN = 0),
where XO and XN are respective random variables for the number of finds in O and N.
As XO and XN are independent we can write this as
h2 = 1-P(XO=0)*P(XN=0)
P(XN=0) = exp(-l'*(1-w))
P(XO=0) is a conditional probability as we already had a find in O. Hence
P(XO=0) = l'*w*exp(-l'*w)/(1-exp(-l'*w))
h2 is then
h2 = 1 - l'*w*exp(-l')/(1-exp(-l'*w))
Examples:
h= 27% with 55m finder
l' = 0.31471
w = 20% (optimal case when double bombing)
then
h2 = 24.7%
overall = 26.5%
w = 30% (mean case)
then
h2 = 23.5%
overall = 26.3%
w = 0% (no rebombing)
then
h2 = 27% (limiting case as lim x/(1-exp(-x)) = 1 for x ->0 )
overall = 27%
w = 100% (double bombing the same first drop)
then
h2 = 14.9% (limiting case)
overall = 24.4%
Comments:
Although it looks easy, it was rather difficult to find the right approach, at least for me. I was trapped by several errors. One was that the search area for the 2nd bomb contains part of the old are with one find and part of a new one. As total area is still the same I’ve thought that this would be the same situation as for the first drop but with one claim missing. This is indeed not the case as proven above, but why? The new area adds information that was not available before and hence we do have a new situation in the 2nd drop. To combine old with new information to get a common probability was the main issue to solve.
