A Basic Caluation on Titanium Ore Prospecting [Maths]
Warning, Incorrect Content:
Thanks to Luj to point out my calculation error,
more detail in the comments…
Sorry Folks 
Warning, Geek post :
We had a little talk on the Titanium Ore Prospecting on JMTC’s IRC channel…
I had come up with the calculation below
In wowhead, an epic gem drop rate on Titanium Ore Prospecting is 4-5%
Some conclude the chance of an epic gem drop is 4.5*6 (6 of epic gem) = 27%
Although amount of the equation agree, the calculation is wrong. You can’t apply this logic to other calculation. I’ll post the correct one here.
Expectation of a Ore Prospecting:
let P = a certain Epic Gem Drop
1-P = a certain Epic Gem didn’t drop
therefore
A prospect without a Epic Gem = (1-p)^6 = 0.758
A prospect without at least one Epic Gem = 0.2413 (3% left?)
Expectation of a prospect = 1 x (Chance of 1 gem Drop) + 2 x (Chance of 2 gems Drop) …. + 6 x (Chance of 6 gems Drop)
= 
Correct me if I am wrong 
(No Ratings Yet)
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November 12th, 2009 at 6:18 pm
“Although amount of the equation agree, the calculation is wrong.”
Can you come up with an example where the two calculations arrive at different answers? If not, I don’t think you can fairly call the first calculation wrong.
Also, is it possible to get all six gems from one prospect? If not, I don’t believe either calculation is correct.
November 16th, 2009 at 11:59 am
Can you come up with an example where the two calculations arrive at different answers?
Sure, say a operation had 50% drop rate on item A and Item B, given A and B is independent.
In the first calculation, it’ll drop either A or B is 100%, since A + B = 50%+50% = 100%
On the second calculation (Mine), that would be, 25% dropping both A and B, 25% only drop A, 25% only drop B, and 25% drop nothing.
The reason that I believe the drop rate is independent: We had report on 2 epic gem dropping on one prospect. Note that, it’s a rare case, since the chance would be (6C2) x (0.045)^2 x (0.955)^2, that would be around 2.216%.
And you second question.
the chance of dropping 6 epic gem is 9.9 * 10^6. This pose not much impact on the calculation, but the previous 2.216% is significant enough.
The reason that the two calculation more the less yield the same amount is that. the 4.5% is far from the 50%, the larger the chance, the more significant the error will be.
The purpose of this post is to remind the calculation error, so that it won’t be used on other aspect, e.g. grinding drops on mobs.
November 19th, 2009 at 3:18 am
Hello again and thank you for the response.
I do not believe that example works. We are interested in expected value, not the probability of different drop combinations (both, only A, only B, nothing). In the example given, we still wind up with the expected number of gems being the same under both calculations.
I can state my point from the earlier comment differently: if two functions have the same value across all points in the domain, they are the same function (despite surface differences in how they are written). I believe this is case with the two calculation methods in this post.
November 19th, 2009 at 2:26 pm
I guess, I may made a mistake on the calculation on the expectation.
)
So it is, if the drop rate are indepent, the expectation can be just added up.
So you are correct for that.. (Time to revision on problabilty
Thanks for pointing that out…
*Revisioning on expected value chapter*