32 replies to this topic

[Knightlazarus]

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:

It's alot more simple than people think.

[Kalandros]

There's an easier way to get +20 than spending for 4903290550 HD Bradiums/Carniums.
Its a method that's been in game since upgrading was coded into the game.

[Knightlazarus]

There's an easier way to get +20 than spending for 4903290550 HD Bradiums/Carniums.
Its a method that's been in game since upgrading was coded into the game.

I can do that actually, but I know better than to carry it out and actually do it. That scale of cheating is stupid, and the GM's would murder anyone involved.

[Kalandros]

Huh?
I don't get what you're saying but I simply meant to have hundreds of +10 of the item you want to +20 and go step by step - upgrade a 100 valk shields to +11, upgrade all those that made it to +12, etc

Still takes HD Bradiums/Carniums but much less - just takes many copies of the item, the same way one would to +10 from +9 before the safe upgrades.

[majorBoobage]

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:

Oh you... you're just so cute!!!

[TrashBag]

Damn I dunno which is harder, winning the lotto, or getting a +20.

[Vehona]

Definitely +20 is harder If renouille calculated chances right (I have no reason to doubt about correctness), the chance to get +20 is 1 / 5 billions. FYI, world population is ~7 billions so if everyone give a single attempt to refine to +20, only 1 guy will manage it - I'm quite sure that 2+ people to win a lotto is real

Edited by Vehona, 12 May 2012 - 02:15 PM.

[Hrothmund]

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

This is whats referred to as the "gamblers curse" and affects anyone who gambles on stuff.

Huh?
I don't get what you're saying but I simply meant to have hundreds of +10 of the item you want to +20 and go step by step - upgrade a 100 valk shields to +11, upgrade all those that made it to +12, etc

Still takes HD Bradiums/Carniums but much less - just takes many copies of the item, the same way one would to +10 from +9 before the safe upgrades.

O.x

That would take substantially MORE ores than upgrading a single item. Laz is making more sense than you at this point.

Edited by Hrothmund, 12 May 2012 - 02:31 PM.