If you don't get this post, you don't need to; it's pure numbers, which I know a lot of people aren't too good with. To summarize right away: Rho should never,

There is a new post about Aion under this post that I just posted today. You can go read that instead of this. (Aaro Note~ : This would have been a perfect place to wh0re out some publicity for the new "new post" box on the side. Try harder, Pearz~)

Background:

Rho did 40 tries for a test. She wanted to see if the Destruction Magic expertise (shot-based) affected shot as a gun. She used the highest and lowest damages, and then compared to the modifier change of highest and lowest damage over the these 40 shots.

Argument:

Rho did not believe that you could have just taken all 40 tries, done the average damages of them, and then already get to see the difference in damage.

Rho's Claim:

The averages of the 40 shots would be flawed and not be able to show that ~15% difference in such small numbers, because if the numbers were rounded twice, the difference could NOT be seen.

My claim:

The averages of 40 shots would show the difference.

Base Assumption:

Rounded every chance you got: IE 105% (shot-type) x 10 LNG vs not rounding until end

Expertise Modifier +2 Difference

This proof requires 2 parts, with 2 assumptions and the 2 different common rounding methods.

Rounding method 1: rounded DOWN every time a decimal happens (to remove decimals).

Rounding method 2: rounded DOWN 0-0.4, 0.5+ UP

Assumption 1.

Part 1: (rounding twice)

Base 105%

((105% x 10) + 0) * (0.8 - 1)

= 10 * (0.8 - 1)

0.8 = 8

0.81 = 8

0.82 = 8

0.83 = 8

0.84 = 8

0.85 = 8

0.86 = 8

0.87 = 8

0.88 = 8

0.89 = 8

0.9 = 9

0.91 = 9

0.92 = 9

0.93 = 9

0.94 = 9

0.95 = 9

0.96 = 9

0.97 = 9

0.98 = 9

0.99 = 9

1 = 10

Average of 21 tries = ((8x10) + (9x10) +10)/21

= 8.571428571428571

With Modifier

((105% x 10) + 2) * (0.8 - 1)

= (10 + 2) * (0.8 - 1)

= 12 * (0.8 - 1)

0.8 = 9

0.81 = 9

0.82 = 9

0.83 = 9

0.84 = 10

0.85 = 10

0.86 = 10

0.87 = 10

0.88 = 10

0.89 = 10

0.9 = 10

0.91 = 10

0.92 = 11

0.93 = 11

0.94 = 11

0.95 = 11

0.96 = 11

0.97 = 11

0.98 = 11

0.99 = 11

1 = 12

Average of 21 tries = ((9x4) + (10x8) +(11x8) + 12)/21

= (32 + 80 + 88 + 12)/12

= 10.0952380952381

On 21 tries, your average case shows (10.0952380952381-8.571428571428571)/8.571428571428571) = 0.177777777784 % increase in damage

Part 2: (rounding only at the end)

Without Modifier

105*10 = 10.5

10.5 * (0.8-1)

0.8 = 8

0.81 = 8

0.82 = 8

0.83 = 8

0.84 = 8

0.85 = 8

0.86 = 9

0.87 = 9

0.88 = 9

0.89 = 9

0.9 = 9

0.91 = 9

0.92 = 9

0.93 = 9

0.94 = 9

0.95 = 9

0.96 = 10

0.97 = 10

0.98 = 10

0.99 = 10

1 = 10

Average of 21 tries = ((8x6) + (9x10) (10x5))/21

=(48 + 90 + 50)/21

= 8.952380952380952

With Modifier

(105*10) + 2 = 12.5

12.5 * (0.8-1)

0.8 = 10

0.81 = 10

0.82 = 10

0.83 = 10

0.84 = 10

0.85 = 10

0.86 = 10

0.87 = 10

0.88 = 11

0.89 = 11

0.9 = 11

0.91 = 11

0.92 = 11

0.93 = 11

0.94 = 11

0.95 = 11

0.96 = 12

0.97 = 12

0.98 = 12

0.99 = 12

1 = 12

Average of 21 Tries = ((10x8) + (11x8) + (12x5))/21

= (80 + 88 + 60)/21

= 10.85714285714286

On 21 tries, your average case shows (10.85714285714286-8.952380952380952)/8.952380952380952) = 0.2127659574468086 % increase in damage

So, with rounding method 1, you see a 17% change in damage if you round twice, whereas rounding once gives you 21% change. To have reached 10% change in damage or less while rounding twice would require:

(10.0952380952381 - A)/A = 0.1

A = 9.177489177489182

Assuming that you got odds 2nd set of numbers, for ((105% x 10) + 0) * (0.8 - 1) = 9.177489177489182 Average

For 9.177489177489182 to be averaged out, you must roll 192 minimum within the 21 tries. Rolling PURE 9s on 21 tries = 189, and would still require you to have rolled a 10 3 times on 1/20 odds = 0.0125% chance of this happening, not including the odds of pure 9s on everything else.

If we assume the 8.952380952380952 hit average,

(A - 8.952380952380952)/8.952380952380952 = .1

A = 9.847619047619047

For ((105% x 10) + 2) * (0.8 - 1) to reach 9.847619047619047 we need 206.8 MAX. One of the best cases for this would be with highest odds of this occurring would be: (9x11) + (10x5) + (11 x 4) + 12 = 205. This requires 9, which has 4/21 odds to show up 11 times. 4/21 = 0.19% occurring 11 times in 21 tries is 2.75 times odds. So, you take 0.19/19 = 10% chance (take only 75% of change of next /19). Roughly 2.5% chance of this occurring.

So, end consensus: best case you're looking at is 2.5% chance that using averages method will net you something wrong with only a 21-size sample. With your 40-size sample, the odds of this occurring is even LOWER.

-----------------------------------------------------------------------------------

Assumption 2.

Part 1: (rounding twice)

Base 105%

((105% x 10) + 0) * (0.8 - 1)

= 10 * (0.8 - 1)

0.8 = 8

0.81 = 8

0.82 = 8

0.83 = 8

0.84 = 8

0.85 = 9

0.86 = 9

0.87 = 9

0.88 = 9

0.89 = 9

0.9 = 9

0.91 = 9

0.92 = 9

0.93 = 9

0.94 = 9

0.95 = 10

0.96 = 10

0.97 = 10

0.98 = 10

0.99 = 10

1 = 10

Average of 21 tries = ((8x5) + (9x10) +(10x6)/21

(40+90+60)/21

= 9.047619047619048

With Modifier

((105% x 10) + 2) * (0.8 - 1)

= (10 + 2) * (0.8 - 1)

= 12 * (0.8 - 1)

0.8 = 10

0.81 = 10

0.82 = 10

0.83 = 10

0.84 = 10

0.85 = 10

0.86 = 10

0.87 = 10

0.88 = 11

0.89 = 11

0.9 = 11

0.91 = 11

0.92 = 11

0.93 = 11

0.94 = 11

0.95 = 11

0.96 = 12

0.97 = 12

0.98 = 12

0.99 = 12

1 = 12

Average of 21 tries = ((10x8) + (11x8) + (12x5))/21

= (80 + 88 + 60)/21

= 10.85714285714286

On 21 tries, your average case shows (10.85714285714286-9.047619047619048)/9.047619047619048) = 0.1999999999999999 % increase in damage

Part 2: (rounding only at the end)

Without Modifier

105*10 = 10.5

10.5 * (0.8-1)

0.8 = 8

0.81 = 9

0.82 = 9

0.83 = 9

0.84 = 9

0.85 = 9

0.86 = 9

0.87 = 9

0.88 = 9

0.89 = 9

0.9 = 9

0.91 = 10

0.92 = 10

0.93 = 10

0.94 = 10

0.95 = 10

0.96 = 10

0.97 = 10

0.98 = 10

0.99 = 10

1 = 11

Average of 21 tries = ((8+ (9x10) (10x9) + 11)/21

=(8 + 90 + 99 + 11)/21

= 9.904761904761905

With Modifier

(105*10) + 2 = 12.5

12.5 * (0.8-1)

0.8 = 10

0.81 = 10

0.82 = 10

0.83 = 10

0.84 = 11

0.85 = 11

0.86 = 11

0.87 = 11

0.88 = 11

0.89 = 11

0.9 = 11

0.91 = 11

0.92 = 12

0.93 = 12

0.94 = 12

0.95 = 12

0.96 = 12

0.97 = 12

0.98 = 12

0.99 = 12

1 = 13

Average of 21 Tries = ((10x4) + (11x8) + (12x8) + 13)/21

= (40 + 88 + 96 + 13)/21

= 11.28571428571429

On 21 tries, your average case shows (11.28571428571429-9.904761904761905)/9.904761904761905) = 0.1394230769230769 % increase in damage

I'm not gonna calculate the odds of that 0.1999999999999 going down to 0.10 or less, 'cause if 0.17 was already only gonna drop down to 10% with 2.5% chance, .19 would be even lower odds.

-----------------------------------------------------------------------------------

So, end of story: the odds of those 40 shots not showing over 10% difference in damage would be extremely low. Though it's not a perfect win 'cause I also said in clan chat that as you round every step the oscillation gets larger (shown in 2nd assumption for rounding method). It wasn't true for Method 1 'cause no one but programs would use that way of rounding...

~Pearz

*ever*challenge me, number-wise.There is a new post about Aion under this post that I just posted today. You can go read that instead of this. (Aaro Note~ : This would have been a perfect place to wh0re out some publicity for the new "new post" box on the side. Try harder, Pearz~)

Background:

Rho did 40 tries for a test. She wanted to see if the Destruction Magic expertise (shot-based) affected shot as a gun. She used the highest and lowest damages, and then compared to the modifier change of highest and lowest damage over the these 40 shots.

Argument:

Rho did not believe that you could have just taken all 40 tries, done the average damages of them, and then already get to see the difference in damage.

Rho's Claim:

The averages of the 40 shots would be flawed and not be able to show that ~15% difference in such small numbers, because if the numbers were rounded twice, the difference could NOT be seen.

My claim:

The averages of 40 shots would show the difference.

Base Assumption:

Rounded every chance you got: IE 105% (shot-type) x 10 LNG vs not rounding until end

Expertise Modifier +2 Difference

This proof requires 2 parts, with 2 assumptions and the 2 different common rounding methods.

Rounding method 1: rounded DOWN every time a decimal happens (to remove decimals).

Rounding method 2: rounded DOWN 0-0.4, 0.5+ UP

Assumption 1.

Part 1: (rounding twice)

Base 105%

((105% x 10) + 0) * (0.8 - 1)

= 10 * (0.8 - 1)

0.8 = 8

0.81 = 8

0.82 = 8

0.83 = 8

0.84 = 8

0.85 = 8

0.86 = 8

0.87 = 8

0.88 = 8

0.89 = 8

0.9 = 9

0.91 = 9

0.92 = 9

0.93 = 9

0.94 = 9

0.95 = 9

0.96 = 9

0.97 = 9

0.98 = 9

0.99 = 9

1 = 10

Average of 21 tries = ((8x10) + (9x10) +10)/21

= 8.571428571428571

With Modifier

((105% x 10) + 2) * (0.8 - 1)

= (10 + 2) * (0.8 - 1)

= 12 * (0.8 - 1)

0.8 = 9

0.81 = 9

0.82 = 9

0.83 = 9

0.84 = 10

0.85 = 10

0.86 = 10

0.87 = 10

0.88 = 10

0.89 = 10

0.9 = 10

0.91 = 10

0.92 = 11

0.93 = 11

0.94 = 11

0.95 = 11

0.96 = 11

0.97 = 11

0.98 = 11

0.99 = 11

1 = 12

Average of 21 tries = ((9x4) + (10x8) +(11x8) + 12)/21

= (32 + 80 + 88 + 12)/12

= 10.0952380952381

On 21 tries, your average case shows (10.0952380952381-8.571428571428571)/8.571428571428571) = 0.177777777784 % increase in damage

Part 2: (rounding only at the end)

Without Modifier

105*10 = 10.5

10.5 * (0.8-1)

0.8 = 8

0.81 = 8

0.82 = 8

0.83 = 8

0.84 = 8

0.85 = 8

0.86 = 9

0.87 = 9

0.88 = 9

0.89 = 9

0.9 = 9

0.91 = 9

0.92 = 9

0.93 = 9

0.94 = 9

0.95 = 9

0.96 = 10

0.97 = 10

0.98 = 10

0.99 = 10

1 = 10

Average of 21 tries = ((8x6) + (9x10) (10x5))/21

=(48 + 90 + 50)/21

= 8.952380952380952

With Modifier

(105*10) + 2 = 12.5

12.5 * (0.8-1)

0.8 = 10

0.81 = 10

0.82 = 10

0.83 = 10

0.84 = 10

0.85 = 10

0.86 = 10

0.87 = 10

0.88 = 11

0.89 = 11

0.9 = 11

0.91 = 11

0.92 = 11

0.93 = 11

0.94 = 11

0.95 = 11

0.96 = 12

0.97 = 12

0.98 = 12

0.99 = 12

1 = 12

Average of 21 Tries = ((10x8) + (11x8) + (12x5))/21

= (80 + 88 + 60)/21

= 10.85714285714286

On 21 tries, your average case shows (10.85714285714286-8.952380952380952)/8.952380952380952) = 0.2127659574468086 % increase in damage

So, with rounding method 1, you see a 17% change in damage if you round twice, whereas rounding once gives you 21% change. To have reached 10% change in damage or less while rounding twice would require:

(10.0952380952381 - A)/A = 0.1

A = 9.177489177489182

Assuming that you got odds 2nd set of numbers, for ((105% x 10) + 0) * (0.8 - 1) = 9.177489177489182 Average

For 9.177489177489182 to be averaged out, you must roll 192 minimum within the 21 tries. Rolling PURE 9s on 21 tries = 189, and would still require you to have rolled a 10 3 times on 1/20 odds = 0.0125% chance of this happening, not including the odds of pure 9s on everything else.

If we assume the 8.952380952380952 hit average,

(A - 8.952380952380952)/8.952380952380952 = .1

A = 9.847619047619047

For ((105% x 10) + 2) * (0.8 - 1) to reach 9.847619047619047 we need 206.8 MAX. One of the best cases for this would be with highest odds of this occurring would be: (9x11) + (10x5) + (11 x 4) + 12 = 205. This requires 9, which has 4/21 odds to show up 11 times. 4/21 = 0.19% occurring 11 times in 21 tries is 2.75 times odds. So, you take 0.19/19 = 10% chance (take only 75% of change of next /19). Roughly 2.5% chance of this occurring.

So, end consensus: best case you're looking at is 2.5% chance that using averages method will net you something wrong with only a 21-size sample. With your 40-size sample, the odds of this occurring is even LOWER.

-----------------------------------------------------------------------------------

Assumption 2.

Part 1: (rounding twice)

Base 105%

((105% x 10) + 0) * (0.8 - 1)

= 10 * (0.8 - 1)

0.8 = 8

0.81 = 8

0.82 = 8

0.83 = 8

0.84 = 8

0.85 = 9

0.86 = 9

0.87 = 9

0.88 = 9

0.89 = 9

0.9 = 9

0.91 = 9

0.92 = 9

0.93 = 9

0.94 = 9

0.95 = 10

0.96 = 10

0.97 = 10

0.98 = 10

0.99 = 10

1 = 10

Average of 21 tries = ((8x5) + (9x10) +(10x6)/21

(40+90+60)/21

= 9.047619047619048

With Modifier

((105% x 10) + 2) * (0.8 - 1)

= (10 + 2) * (0.8 - 1)

= 12 * (0.8 - 1)

0.8 = 10

0.81 = 10

0.82 = 10

0.83 = 10

0.84 = 10

0.85 = 10

0.86 = 10

0.87 = 10

0.88 = 11

0.89 = 11

0.9 = 11

0.91 = 11

0.92 = 11

0.93 = 11

0.94 = 11

0.95 = 11

0.96 = 12

0.97 = 12

0.98 = 12

0.99 = 12

1 = 12

Average of 21 tries = ((10x8) + (11x8) + (12x5))/21

= (80 + 88 + 60)/21

= 10.85714285714286

On 21 tries, your average case shows (10.85714285714286-9.047619047619048)/9.047619047619048) = 0.1999999999999999 % increase in damage

Part 2: (rounding only at the end)

Without Modifier

105*10 = 10.5

10.5 * (0.8-1)

0.8 = 8

0.81 = 9

0.82 = 9

0.83 = 9

0.84 = 9

0.85 = 9

0.86 = 9

0.87 = 9

0.88 = 9

0.89 = 9

0.9 = 9

0.91 = 10

0.92 = 10

0.93 = 10

0.94 = 10

0.95 = 10

0.96 = 10

0.97 = 10

0.98 = 10

0.99 = 10

1 = 11

Average of 21 tries = ((8+ (9x10) (10x9) + 11)/21

=(8 + 90 + 99 + 11)/21

= 9.904761904761905

With Modifier

(105*10) + 2 = 12.5

12.5 * (0.8-1)

0.8 = 10

0.81 = 10

0.82 = 10

0.83 = 10

0.84 = 11

0.85 = 11

0.86 = 11

0.87 = 11

0.88 = 11

0.89 = 11

0.9 = 11

0.91 = 11

0.92 = 12

0.93 = 12

0.94 = 12

0.95 = 12

0.96 = 12

0.97 = 12

0.98 = 12

0.99 = 12

1 = 13

Average of 21 Tries = ((10x4) + (11x8) + (12x8) + 13)/21

= (40 + 88 + 96 + 13)/21

= 11.28571428571429

On 21 tries, your average case shows (11.28571428571429-9.904761904761905)/9.904761904761905) = 0.1394230769230769 % increase in damage

I'm not gonna calculate the odds of that 0.1999999999999 going down to 0.10 or less, 'cause if 0.17 was already only gonna drop down to 10% with 2.5% chance, .19 would be even lower odds.

-----------------------------------------------------------------------------------

So, end of story: the odds of those 40 shots not showing over 10% difference in damage would be extremely low. Though it's not a perfect win 'cause I also said in clan chat that as you round every step the oscillation gets larger (shown in 2nd assumption for rounding method). It wasn't true for Method 1 'cause no one but programs would use that way of rounding...

~Pearz