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Rashmika Clavain
Gallente
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Posted - 2010.04.08 12:02:00 -
[1]
Hi, I am tryingto factor in the average csot of datacore losses for an invented job relevant to my skills. However math percentages were never my strong point!
So, for example, I have a chance of sucess on a cruiser hull of 31.5%
A cruiser hull uses 8 datacores of type x at price y. So, in a 10 run attempt, I'll need 80 type x datacores. With the 31.5% rate, I'll waste 54.8 datacores over 10 attempts:
(80-(80*0.315))=54.8
Therefore, for every job I attempt on avergage I'll lose 5.48 datacores.
There the true average datacore cost per invented is 13.48 datacores.
However that figure seems too low?
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Omg Corn
Gallente
Alcoholocaust.
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Posted - 2010.04.08 12:09:00 -
[2]
Edited by: Omg Corn on 08/04/2010 12:16:17
Just divide the number of datacores by the probability (probability %/100)
31.5% = .315
8 datacores / .315 = 25.4 datacores per success.
Edit because I wanted to be first and didn't post everything :D :
you can guesstimate this to be correct. 31.5% is a little less than 1 in 3 successes, which means for every 24 (8*3) datacores you use you will get one success.
The way you worked it out, you get the number of datacores wasted for 10 runs - 54.8. Since 3.15 of those runs will succeed, you will waste 54.8/3.15 = 17.4 datacores per success, plus the 8 that you didn't waste, which equals to 25.4
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Akita T
Caldari
Caldari Navy Volunteer Task Force
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Posted - 2010.04.08 12:16:00 -
[3]
 Originally by: Rashmika Clavain
So, in a 10 run attempt, I'll need 80 type x datacores. With the 31.5% rate, I'll waste 54.8 datacores over 10 attempts
Sort of correct.
Originally by: Rashmika Clavain
Therefore, for every job I attempt on avergage I'll lose 5.48 datacores.
Still "sort of" correct, but already a bit wrong.
That's a pointless measure, you should only care about datacores used per SUCCESS, not datacores used per ATTEMPT.
So it's 80 datacores used for 3.15 successes, or ~25.4 datacores per success on average.
Originally by: Rashmika Clavain
There the true average datacore cost per invented is 13.48 datacores.
And this is where you went completely wrong.
Like pointed out above, it's 80/3.15 = ~25.4, not 8*(2-.315)=13.48
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Rashmika Clavain
Gallente
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Posted - 2010.04.08 12:17:00 -
[4]
Edited by: Rashmika Clavain on 08/04/2010 12:22:36
Ahh okies. As I said I thought my final figure of 13 ish was way too low (God knows what it represents... if anything).
Thanks.
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Emporer Norton
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Posted - 2010.04.08 14:50:00 -
[5]
I usually overestimate on my costs a bit so for cruiser figure 1 in 4 success to be safe so (8*A)+(8*B)*4=totatal cost to invent or can do (8*A)+(8*B)*10/total runs invented for cost/run for that 10 jobs
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