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Project, TT% returns, 90%?


Project, TT% returns, 90%?
I have moved this back to EF as i much prefer it, i will not update all posts , only the last one, but you can find them here:
http://www.planetcalypsoforum.com/fo...TTreturns90 Big thankyou to Falkao and Immortal for their Rebombing statistical anaylsis, i will try to gather it all together and post here for easy access
There is a lot of discussion at the moment at how long does it take to get a decent % return. So this log is to keep track of that
Pre VU10 i was use to minimum 90% returns, since i have seen a different shift in tt returns in mining, i thought i would share my findings
My last 58k spent in mining has given back 83.69% tt returns, this is a lot lower than I expected.
This blog will start from now, at 60k spent and lets see how long it takes to get to the previously seen 90% returns if at all possible now in mining?
Now i am aware short term could give a high number so this blog will run for at least 10k peds spent. I no longer can mine as much as use to due to rl commitments so this will be slow i am afraid
Starting values
TT spent = 60433
TT returns = 50574
TT % return = 83.67%
Refining costs not included nor any hunting mobs that get in my way, which are not too many, maybe 10 peds per run
Will be done with mainly 101 amps or no amps and with vrx2k or if i hit a biggy then vrx3k as i am close to maxing it, two levels to go
All posts are welcome
Rgds
Ace
Running Total
Total TT Spent: 97496
Total TT Returns: 84219
Total TT % return: 86.38%
Hit rate sample when rebombing claims:
Total bombs = 19152
No. of finds = 4805
Hit rate = 25.08%
Rate of SOOTO Finds
No. of Bombs = 36688
No. of SOOTO's = 7
Rate = 5241 bombs per find
Competition has started
Win 1000 EFD
Guess how many bombs untill the next SOOTO find
Finally i got an SOOTO find,
Bomb number 7409, gratz to Wolverine for guessing 8500 and being closest, 1000 EFD on your way, will post it in your post where you guessed
Steffel 69
Sibolovin 363
Esme 568
Das 666
Harm 938
BBpunisher 1001
Lyrisiana 1234
Exsolvo 1500
UnaAlconbury 2356
Kosmos 2500
Khaso 2674
JC 2790
Immortal 3000
Leona 3333
Mandred 3500
HinDragoz 4250
overburn 4873
captee 5000
Wolverine Queen 8500
Sofia 12600
CMS 13013
Useful info: Thankfyou Falkao for this
Useful Links:
http://www.entropiaforum.com/forums/...ngpayout.html
http://www.entropiaforum.com/forums/...analysis.html
http://www.entropiaforum.com/forums/...ml#post2512718
Originally Posted by falkao
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) = 1Po(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(1h)
When rebombing (image shows optimal situation with find on perimeter)
Click to enlarge
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 AO. This can be also written as
O = A*w,
N = A*(1w), 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 = 1P(XO=0)*P(XN=0)
P(XN=0) = exp(l'*(1w))
P(XO=0) is a conditional probability as we already had a find in O. Hence
P(XO=0) = l'*w*exp(l'*w)/(1exp(l'*w))
h2 is then
h2 = 1  l'*w*exp(l')/(1exp(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/(1exp(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.
Other bits of random info:
Finder decay effects claim sizes, this was shown in a competition myself and Xen were in, more details in this thread
Testing Ores Refresh Rate in Mining
http://www.planetcalypsoforum.com/fo...s+refresh+rate
Last edited by ace flyster; 01252011 at 06:15.
