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Mining Entropia Universe mining tactics, tips and equipment discussion.

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Old 11-02-2010, 07:36   #1
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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...-TT-returns-90
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/...ng-payout.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) = 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)

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 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.



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; 01-25-2011 at 06:15.
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Old 11-02-2010, 07:39   #2
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And back to bombs

Total Bombs: 62930

0 amp, 304 bombs
TT spent this run: 318 (finder & bombs only, untaxed area)
TT returns: 343
% this run: 108%

Total TT Spent: 91133
Total TT Returns: 78245
Total TT % return: 85.86%

Projects Highest single loot (including first 60k TT spent) = 477pedder

Rgds

Ace
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Old 11-02-2010, 08:01   #3
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welcome back . How many bombs have you dropped with the new finder till now ?
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Old 11-02-2010, 08:03   #4
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Quote:
Originally Posted by falkao View Post
welcome back . How many bombs have you dropped with the new finder till now ?
Ty

Since i started no crossover when bombing, i have dropped 2906 bombs, not enough yet for you to do the number crunching

Rgds

Ace
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Old 11-02-2010, 09:43   #5
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Quote:
Originally Posted by ace flyster View Post
Ty

Since i started no crossover when bombing, i have dropped 2906 bombs, not enough yet for you to do the number crunching

Rgds

Ace
indeed, if you're using the 54m finder but are not rebombing anymore, then we will need at least 47k drops as the expected difference is small (26.2% to 25.4%)

can you please correct the following typos:

P(XO=0) = l'*w*exp(-l*'(1-w))/(1-exp(l'*(1-w))

h2 is then

h2 = 1 - l'*w*exp(-l')/(1-exp(l'*(1-w))

to

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))
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Old 11-02-2010, 13:02   #6
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Is this the first "project" back on EF?

Interesting read.
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Old 12-17-2010, 20:10   #7
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I havent posted my last few runs, life been hectic

But here we go

0 amp, 304 bombs
TT spent this run: 318 (finder & bombs only, untaxed area)
TT returns: 248
% this run: 78%

Total TT Spent: 91451
Total TT Returns: 78493
Total TT % return: 85.83%

Projects Highest single loot (including first 60k TT spent) = 477pedder

Rgds

Ace
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Old 12-17-2010, 20:11   #8
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0 amp, 304 bombs
TT spent this run: 318 (finder & bombs only, untaxed area)
TT returns: 248
% this run: 78%

Total TT Spent: 91769
Total TT Returns: 78741
Total TT % return: 85.80%

Projects Highest single loot (including first 60k TT spent) = 477pedder

Rgds

Ace
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Old 12-17-2010, 20:12   #9
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0 amp, 304 bombs
TT spent this run: 318 (finder & bombs only, untaxed area)
TT returns: 357
% this run: 112%

Total TT Spent: 92088
Total TT Returns: 79098
Total TT % return: 85.89%

Projects Highest single loot (including first 60k TT spent) = 477pedder

Rgds

Ace
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Old 12-17-2010, 20:13   #10
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0 amp, 304 bombs
TT spent this run: 318 (finder & bombs only, untaxed area)
TT returns: 435
% this run: 137%

Total TT Spent: 92406
Total TT Returns: 79533
Total TT % return: 86.07%

Projects Highest single loot (including first 60k TT spent) = 477pedder

Rgds

Ace
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