13.05.2016 - 17:59
Plain and simple, is there an equation for calculating the average atk/def value of a unit/stack, which is able to applied to changes in values from strategies?
---- #UniBoycott
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14.05.2016 - 13:40
I know that the numbers displayed are not correct, but I wanted an equation that I could use on a spreadsheet I wished to create, but thanks for your response!
---- #UniBoycott
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14.05.2016 - 21:30
when the game start you can see the change in the unit tab, usually in the top right. By using that you can determine the attack/defence of a unit. then you can calculate the attack/defence of a unit then multiply it by the number of that certain unit.
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15.05.2016 - 14:43
Actually a better measure of the real average attack a unit will do is: So the average attack of a standard tank is (8+1)/2 = 4.5
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njab Account verwijderd |
15.05.2016 - 15:51 njab Account verwijderd
I ask for actual formula. Something dirty is behind that method.
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15.05.2016 - 16:08
How can a formula be dirty or otherwise?
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njab Account verwijderd |
16.05.2016 - 03:46 njab Account verwijderd
It's dirty if you just display formula's result (e.g. multiply by 2 and add 1) but not explain it.
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njab Account verwijderd |
16.05.2016 - 12:15 njab Account verwijderd
That's an approximation, not actual maths.
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17.05.2016 - 11:47
Ok. Let a unit have max attack A. Then when attacking it will roll a randomly selected value (let this be α), so that 1 ≤ α ≤ A. Because any value between 1 and A has an equal probability of occuring it can be taken that the the average attack (in this case both the mean and median) will be exactly the midpoint of these two end values. Therefore you can use the formula for calculating an arithmetic mean to calculate the average attack: add the numbers you want to take the mean of and then divide by how many there are. Therefore the average attack is (1+A)/2 I hope that constituted sufficient explanation.
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njab Account verwijderd |
17.05.2016 - 12:14 njab Account verwijderd
The original formula is A(A+1)/2 * 1/A (sum of all the roll numbers divided by how many of them are here) and it equals to what you said. What I meant with my post is that you just posted conducted formula and not the raw one.
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17.05.2016 - 12:51
Indeed, series sums is a more precise way of doing it, but I would never try to explain it all like that, because it would just confuse people. I would prefer to make sure everything's explained too, but people don't appreciate it much.
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