32 replies to this topic

[Raharja]

So i tried to +12 my +11 Ygg Crown today. After about 550 good tries using HD Craniums, it ended up at +10 (yes +10, hard luck, no?). My question is, Oda/Heim, are you sure today's refinement rate wasn't bugged? i've been +12-ing many items already and it took me no more than 200 tries on every single try of it. There is of course, this fallacy that incremental tries will seemingly skew my chance of success, while it will always stay at 10%. But then, even with 10% chance and 95% confidence, it should only take about


N = Ceiling( log(0.05) / log(0.9) ) = 29

tries. Even if i'd infer my prior statistics, this try is like off at the magnitude of 3.

happens, but was i really screwed -- i mean, really?

(it took me so many HD Craniums, emptied my account's outstanding KP and topped up for more, i don't even care to count how many exactly I've solemnly burnt. What sure is, of the 66.4-ish M zeny I spared for this try -- by the time I'm writing this, it reads 16,333,674 11,333,674.)

Edited by Raharja, 11 May 2012 - 02:20 PM.

[Knightlazarus]

So i tried to +12 my +11 Ygg Crown today. After about 550 good tries using HD Craniums, it ended up at +10 (yes +10, hard luck, no?). My question is, Oda/Heim, are you sure today's refinement rate wasn't bugged? i've been +12-ing many items already and it took me no more than 200 tries on every single try of it. There is of course, this fallacy that incremental tries will seemingly skew my chance of success, while it will always stay at 10%. But then, even with 10% chance and 95% confidence, it should only take about


N = Ceiling( log(0.05) / log (0.9) ) = 29

tries. Even if i'd infer my prior statistics, this try is like off at the magnitude of 3.

happens, but was i really screwed -- i mean, really?

(it took me so many HD Craniums, emptied my account's outstanding KP and topped up for more, i don't even care to count how many exactly I've solemnly burnt. What sure is, of the 66.4-ish M zeny I spared for this try -- by the time I'm writing this, it reads 16,333,674.)

Bad luck is bad.

[DrAzzy]

Wow, that's harsh.

The problem with your math, though, is that it would only be true if your ygg crown stayed at +11 instead of falling to +10.

That's 29 tries from +11 to +12, which means 28 failures, each of which would take an average of 10 HD's to get the +11 back.

Based on that number assuming average luck on the +11ing, you'd be left with 309 HD's - presumably you also had bad luck with the +11ing as well.

You're probably in well under the 5th percentile in terms of upgrade luck - but someone's gotta be in the bottom few percent, and it looks like today, you're that person.

Maybe some other day, you'll be in in the top 1% by getting a +10->12 in 2 HD's.


Also.... HD Craniums? I think you're looking for HD Carnium.

Edited by DrAzzy, 11 May 2012 - 11:08 AM.

[majorBoobage]

Well that sucks, but I always refer to this:
http://wiki.ragmar.c...rading_Above_10

Basically in the simulation of going from +11 to +12:
Average = 109.96
Most = 647

So while you are getting unlikely, it's not the worst it could be.

[DrAzzy]

Uhhhhhh

Do note that most/least there are not absolute most/least - as least is 1 per upgrade, and most is infinite. They specify that it's within three standard deviations, but there's still a small (under 1%) chance that it could take more than that....

[majorBoobage]

Uhhhhhh

Do note that most/least there are not absolute most/least - as least is 1 per upgrade, and most is infinite. They specify that it's within three standard deviations, but there's still a small (under 1%) chance that it could take more than that....

This is a simulation of 500,000 attempts at upgrading the weapon as described. So while it is possible that it is more than that, it's completely statistically insignificant. These results are absolutely valid as a corollary to in-game upgrading. The one exception to this is if the rates that most believe to be true (10% chance from level to level) are not actually correct, since it's never been announced officially.

[Raharja]

Wow, that's harsh.

The problem with your math, though, is that it would only be true if your ygg crown stayed at +11 instead of falling to +10.

That's 29 tries from +11 to +12, which means 28 failures, each of which would take an average of 10 HD's to get the +11 back.

Based on that number assuming average luck on the +11ing, you'd be left with 309 HD's - presumably you also had bad luck with the +11ing as well.

You're probably in well under the 5th percentile in terms of upgrade luck - but someone's gotta be in the bottom few percent, and it looks like today, you're that person.

Maybe some other day, you'll be in in the top 1% by getting a +10->12 in 2 HD's.


Also.... HD Craniums? I think you're looking for HD Carnium.

Uhhhhhh

Do note that most/least there are not absolute most/least - as least is 1 per upgrade, and most is infinite. They specify that it's within three standard deviations, but there's still a small (under 1%) chance that it could take more than that....

Yes, sorry for the typo.

Using Markov Model (not the Hidden one):

[+12]

up(0.1) / down(0.9)

[+11]

up(0.1)

[+10]

the "standard" amount of tries should be (reverse probabilistic model) :

N = "0 loop + 1 loop + 2 loops ..."
N = 10 ( 0.1 * 1 + 9/10 * 10 ( 0.1 * 1 + 9/10 * 10 ( 0.1 *1 + 9/10 * 10 ( ... ))))

which yields a number of about ~110 if i'm not mistaken

Edited by Raharja, 11 May 2012 - 11:34 PM.

[Kalandros]

took me about 800 for a single success at +13.

[DrAzzy]

110 is expected from the trivial calculation as well (ie, 10 attempts avg to get +10 ->11, 10 attempts to get +11->12, so 10 HD for the +11->12, and 100 more to get to +11 10 times)

Edited by DrAzzy, 11 May 2012 - 12:08 PM.

[Hrothmund]

Also.... HD Craniums? I think you're looking for HD Carnium.

HD Craniums for when he was hitting his head on his desk to avoid breakage

[miliardo]

I got lucky the other day on upgrading 2 things to 12 with only 6 hds

[Chigikogou]

HD Craniums are painful, imagine having that piece of crystal hammered in your head..

but seriously dude, I know what you feel, the reason why my Proxy Garb remained +10 (>w<)

[MotherOneDog]

HD Craniums are painful, imagine having that piece of crystal hammered in your head..

but seriously dude, I know what you feel, the reason why my Proxy Garb remained +10 (>w<)

Ohhhhhh yeaaaah. I like how you call it Proxy Garb. Sounds way more baus than Nid Garb <3

+1 to you.

[renouille]

This is a simulation of 500,000 attempts at upgrading the weapon as described. So while it is possible that it is more than that, it's completely statistically insignificant. These results are absolutely valid as a corollary to in-game upgrading. The one exception to this is if the rates that most believe to be true (10% chance from level to level) are not actually correct, since it's never been announced officially.

I'd say it's more of an indication of problems with your simulation, judging by how wildly wrong some of the numbers are.

[majorBoobage]

I'd say it's more of an indication of problems with your simulation, judging by how wildly wrong some of the numbers are.

Oh you...

[Akshaysanwal]

Your Math sucks big time XD
What Azzy is the right thing.

[Raharja]

Your Math sucks big time XD
What Azzy is the right thing.

judging from this I doubt you even understand the math

[renouille]

Oh you...

Unfortunately I'm not kidding. The +11->+12 upgrade can be modeled as a Markov chain with these states and transitions.



Calculating where you'll be after 647 transitions is straightforward:

octave:1> A = [ .9 .1 0; .9 0 .1; 0 0 1 ]
A =

   0.90000   0.10000   0.00000
   0.90000   0.00000   0.10000
   0.00000   0.00000   1.00000

octave:2> A^647
ans =

   0.00237   0.00024   0.99739
   0.00215   0.00022   0.99763
   0.00000   0.00000   1.00000

Reading from the second row:
+10 0.215%
+11 0.022%
+12 99.763%

So that's a 0.237% chance of failure after 647 attempts, which is an order of magnitude more likely than a card drop. You're telling us that in 500000 trials, not once did you exceed 647 attempts. The probability of this occurring is 5.6x10^-516, or 0.0000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000056%
I think it's more likely that there's something wrong with your simulation.

This is assuming that your 647 is indeed for +11 to +12. It might actually be for +10 to +12 (since your page says least=2), in which case a maximum of 647 is even more unlikely.

(You also have least=40 for +13. Getting three successes in a row at 10% is 0.1%. You weren't able to hit a 1/1000 once in 500000 tries? Really?)


In contrast, here's a simulation that's easy to understand:

$ cat sim1.lua
local print = print
local random = math.random

local function upgrade()
	local level = 11
	local count = 0
	while level < 12 do
		count = count + 1
		if random(10) == 1 then
			level = level + 1
		elseif level ~= 10 then
			level = level - 1
		end
	end
	return count
end

math.randomseed(os.time())
local fail = 0
for i = 1, 500000 do
	if upgrade() > 647 then fail = fail + 1 end
end
print(fail)

$ luajit sim1.lua
1220

1220/500000 = 0.244%



Then there's the matter of average number of ores used. The actual expected values, given by a simple recurrence relation, are:
+10 -> +11 => 10 (10 from +10)
+11 -> +12 => 100 (110 from +10)
+12 -> +13 => 910 (1020 from +10)
+13 -> +14 => 8200 (9220 from +10)
+14 -> +15 => 73810 (83030 from +10)
+15 -> +16 => 664300 (747330 from +10)
+16 -> +17 => 5978710 (6726040 from +10)
+17 -> +18 => 53808400 (60534440 from +10)
+18 -> +19 => 484275610 (544810050 from +10)
+19 -> +20 => 4358480500 (4903290550 from +10)

You can extend the simulation:

$ cat sim2.lua
local print = print
local random = math.random

local function upgrade(level, target)
	local count = 0
	while level < target do
		count = count + 1
		if random(10) == 1 then
			level = level + 1
		elseif level ~= 10 then
			level = level - 1
		end
	end
	return count
end

math.randomseed(os.time())
for j = 10, 13 do
	local sum = 0
	local k = j + 1
	for i = 1, 500000 do
		sum = sum + upgrade(j, k)
	end
	print(string.format("+%d -> +%d => %f", j, k, sum / 500000))
end

for k = 12, 14 do
	local sum = 0
	for i = 1, 500000 do
		sum = sum + upgrade(10, k)
	end
	print(string.format("+10 -> +%d => %f", k, sum / 500000))
end

$ luajit sim2.lua
+10 -> +11 => 10.004268
+11 -> +12 => 100.186342
+12 -> +13 => 913.884986
+13 -> +14 => 8207.019994
+10 -> +12 => 109.927122
+10 -> +13 => 1020.979678
+10 -> +14 => 9226.278846

Pretty close to the actual values, right?

But your page says:

+13 => 963.74
+14 => 16384

Edited by renouille, 12 May 2012 - 01:15 AM.

[Maximusss]

Pwned by the math person 8D

[Bascojiin]

Unfortunately I'm not kidding. The +11->+12 upgrade can be modeled as a Markov chain with these states and transitions.



Calculating where you'll be after 647 transitions is straightforward:

octave:1> A = [ .9 .1 0; .9 0 .1; 0 0 1 ]
A =

   0.90000   0.10000   0.00000
   0.90000   0.00000   0.10000
   0.00000   0.00000   1.00000

octave:2> A^647
ans =

   0.00237   0.00024   0.99739
   0.00215   0.00022   0.99763
   0.00000   0.00000   1.00000

Reading from the second row:
+10 0.215%
+11 0.022%
+12 99.763%

So that's a 0.237% chance of failure after 647 attempts, which is an order of magnitude more likely than a card drop. You're telling us that in 500000 trials, not once did you exceed 647 attempts. The probability of this occurring is 5.6x10^-516, or 0.0000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000056%
I think it's more likely that there's something wrong with your simulation.

This is assuming that your 647 is indeed for +11 to +12. It might actually be for +10 to +12 (since your page says least=2), in which case a maximum of 647 is even more unlikely.

(You also have least=40 for +13. Getting three successes in a row at 10% is 0.1%. You weren't able to hit a 1/1000 once in 500000 tries? Really?)


In contrast, here's a simulation that's easy to understand:

$ cat sim1.lua
local print = print
local random = math.random

local function upgrade()
	local level = 11
	local count = 0
	while level < 12 do
		count = count + 1
		if random(10) == 1 then
			level = level + 1
		elseif level ~= 10 then
			level = level - 1
		end
	end
	return count
end

math.randomseed(os.time())
local fail = 0
for i = 1, 500000 do
	if upgrade() > 647 then fail = fail + 1 end
end
print(fail)

$ luajit sim1.lua
1220

1220/500000 = 0.244%



Then there's the matter of average number of ores used. The actual expected values, given by a simple recurrence relation, are:
+10 -> +11 => 10 (10 from +10)
+11 -> +12 => 100 (110 from +10)
+12 -> +13 => 910 (1020 from +10)
+13 -> +14 => 8200 (9220 from +10)
+14 -> +15 => 73810 (83030 from +10)
+15 -> +16 => 664300 (747330 from +10)
+16 -> +17 => 5978710 (6726040 from +10)
+17 -> +18 => 53808400 (60534440 from +10)
+18 -> +19 => 484275610 (544810050 from +10)
+19 -> +20 => 4358480500 (4903290550 from +10)

You can extend the simulation:

$ cat sim2.lua
local print = print
local random = math.random

local function upgrade(level, target)
	local count = 0
	while level < target do
		count = count + 1
		if random(10) == 1 then
			level = level + 1
		elseif level ~= 10 then
			level = level - 1
		end
	end
	return count
end

math.randomseed(os.time())
for j = 10, 13 do
	local sum = 0
	local k = j + 1
	for i = 1, 500000 do
		sum = sum + upgrade(j, k)
	end
	print(string.format("+%d -> +%d => %f", j, k, sum / 500000))
end

for k = 12, 14 do
	local sum = 0
	for i = 1, 500000 do
		sum = sum + upgrade(10, k)
	end
	print(string.format("+10 -> +%d => %f", k, sum / 500000))
end

$ luajit sim2.lua
+10 -> +11 => 10.004268
+11 -> +12 => 100.186342
+12 -> +13 => 913.884986
+13 -> +14 => 8207.019994
+10 -> +12 => 109.927122
+10 -> +13 => 1020.979678
+10 -> +14 => 9226.278846

Pretty close to the actual values, right?

But your page says:

Can you please write my Math Classes?

[Maknificent]

I actually was wondering the same thing. Ever since they implemented that "safety" feature it's been harder for me to +12 anything. This is coming from a person with a lot of +14 gears

[Bascojiin]

Out of curiosity:

How many tries would you need and how many $ would it cost you to upgrade something from +0 -> +20 AND +10 -> +20

[Niji]

Unfortunately I'm not kidding. The +11->+12 upgrade can be modeled as a Markov chain with these states and transitions.



Calculating where you'll be after 647 transitions is straightforward:

octave:1> A = [ .9 .1 0; .9 0 .1; 0 0 1 ]
A =

   0.90000   0.10000   0.00000
   0.90000   0.00000   0.10000
   0.00000   0.00000   1.00000

octave:2> A^647
ans =

   0.00237   0.00024   0.99739
   0.00215   0.00022   0.99763
   0.00000   0.00000   1.00000

Reading from the second row:
+10 0.215%
+11 0.022%
+12 99.763%

So that's a 0.237% chance of failure after 647 attempts, which is an order of magnitude more likely than a card drop. You're telling us that in 500000 trials, not once did you exceed 647 attempts. The probability of this occurring is 5.6x10^-516, or 0.0000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000056%
I think it's more likely that there's something wrong with your simulation.

This is assuming that your 647 is indeed for +11 to +12. It might actually be for +10 to +12 (since your page says least=2), in which case a maximum of 647 is even more unlikely.

(You also have least=40 for +13. Getting three successes in a row at 10% is 0.1%. You weren't able to hit a 1/1000 once in 500000 tries? Really?)


In contrast, here's a simulation that's easy to understand:

$ cat sim1.lua
local print = print
local random = math.random

local function upgrade()
	local level = 11
	local count = 0
	while level < 12 do
		count = count + 1
		if random(10) == 1 then
			level = level + 1
		elseif level ~= 10 then
			level = level - 1
		end
	end
	return count
end

math.randomseed(os.time())
local fail = 0
for i = 1, 500000 do
	if upgrade() > 647 then fail = fail + 1 end
end
print(fail)

$ luajit sim1.lua
1220

1220/500000 = 0.244%



Then there's the matter of average number of ores used. The actual expected values, given by a simple recurrence relation, are:
+10 -> +11 => 10 (10 from +10)
+11 -> +12 => 100 (110 from +10)
+12 -> +13 => 910 (1020 from +10)
+13 -> +14 => 8200 (9220 from +10)
+14 -> +15 => 73810 (83030 from +10)
+15 -> +16 => 664300 (747330 from +10)
+16 -> +17 => 5978710 (6726040 from +10)
+17 -> +18 => 53808400 (60534440 from +10)
+18 -> +19 => 484275610 (544810050 from +10)
+19 -> +20 => 4358480500 (4903290550 from +10)

You can extend the simulation:

$ cat sim2.lua
local print = print
local random = math.random

local function upgrade(level, target)
	local count = 0
	while level < target do
		count = count + 1
		if random(10) == 1 then
			level = level + 1
		elseif level ~= 10 then
			level = level - 1
		end
	end
	return count
end

math.randomseed(os.time())
for j = 10, 13 do
	local sum = 0
	local k = j + 1
	for i = 1, 500000 do
		sum = sum + upgrade(j, k)
	end
	print(string.format("+%d -> +%d => %f", j, k, sum / 500000))
end

for k = 12, 14 do
	local sum = 0
	for i = 1, 500000 do
		sum = sum + upgrade(10, k)
	end
	print(string.format("+10 -> +%d => %f", k, sum / 500000))
end

$ luajit sim2.lua
+10 -> +11 => 10.004268
+11 -> +12 => 100.186342
+12 -> +13 => 913.884986
+13 -> +14 => 8207.019994
+10 -> +12 => 109.927122
+10 -> +13 => 1020.979678
+10 -> +14 => 9226.278846

Pretty close to the actual values, right?

But your page says:

o my god my head just exploded.....

[Akshaysanwal]

judging from this I doubt you even understand the math

Sure. /swt

[Vehona]

Btw, how accurate are values of success chances for 10+ items? I mean, I doubt that anybody could state that known values are correct. All those chances were calculated in an experiment, right?