262144 rank 1 enchants wow the cost is insane.

givearandomname · June 2013

never thought about it but killing time atm.

so made chart of how many of each tier enchants you need to get 1 rank 10.

rank 10 = 1
rank 9 = 4
rank 8 = 16
rank 7 = 64
rank 6 = 256
rank 5 = 1024
rank 4 = 4096
rank 3 = 16384
rank 2 = 65536
rank 1 = 262144

if you look at thos numbers u would need 262144 rank 1 eanchants to reach the rank 10 mark thats with no failure to fuse. lets say you don't want to fail any combines.

you would need 87354 colesent wards at $10.00 each from store or $3.34 from auction house<useing current prices of ad to zen echange on dragon server> that would cost you $873,380

now given most people don't use wards on the lower ones broke cost down per tier.

rank 10 = 1

zen store

zen to ad price on ah dragon server
rank 9 = 4 = 1 wards

$10.00

$ 3.34
rank 8 = 16 = 4 wards

$40.00

$13.36
rank 7 = 64 = 16 wards

$160.00

$ 53.44
rank 6 = 256 = 64 wards

$640.00

$ 213.76
rank 5 = 1024 = 256 wards

$2560.00

$ 855.04
rank 4 = 4096 = 1024 wards

$10,240.00

$ 3420.16
rank 3 = 16384 = 4069 wards

$40,690.00

$13,590.46
rank 2 = 65536 = 16384 wards

$163,840.00

$54,722.56
rank 1 = 262144= 65536 wards

$655,360.00

$218,890.24

can double check my math pretty sure its all right and if so the cost of enchants is way more then most people will ever be able to afford this actualy puts pay to win to shame.

dkcandy · June 2013

Then ask yourself "How did those players with Perfect/Level 10 enchants get their?" not by legit means!

Best way to get rank 10 is to buy rank 7 and work from there but it will cost a mint.

tarmalen · June 2013

So when I get a rank 5 drop I am RICH?!?!?!

kiralyn · June 2013

tarmalen wrote: »

So when I get a rank 5 drop I am RICH?!?!?!

Yeah, good thing that stuff much higher than rank 1 drops. That chart means little, since it doesn't represent the situation at all.

desjardinii · June 2013

Biggest flaw in you math is that who really uses wards below rank 7. Or well maybe 6. Anyhow...step away from the calculator...put it down slowly...

sunnified · June 2013

LOL?
#1. at lvl 60 rank 3 and rank 4 drops.. so buying these will be relatively cheaper.
#2. no1 is so dumb to use coal wards for fusing rank 1-rank 7.. u use green wards.. that will prbably cut ur costs by a large chunk..

before u do things like this think logically. no doubt u made my day.

myschaella · June 2013

Welcome to PW and this is the most obvious of the cash grabs but you can easily find the rest. In the end it will be PW that ruins this game unfortunately.

traciatim1 · June 2013

The one fatal flaw like others have pointed out is that you only use a green ward where the cost of the ward is less than the cost of a new rune, and you'd only use a blue ward when the cost and chance of failure of using green wards or new runs doesn't make sense . . . Yet still, it makes you kind of wonder how so many people have top rank runes now, doesn't it?

No wipe from Beta? Worst . . . idea . . . ever.

givearandomname · June 2013

desjardinii wrote: »

Biggest flaw in you math is that who really uses wards below rank 7. Or well maybe 6. Anyhow...step away from the calculator...put it down slowly...

I am aware of it and could of just inclued thos tiers but chose to do them all as have nothing better to do while sitting in que.

even just doing rank 7 and on thats still $210.00 in wards and if you buy the enchants your looking at alot more not going to go into full details on cost of the enchants but rank 5 go for around 5000 ad each and higher ranks go for more. and you need alot of them.

so you do the math if you bought all the enchants and wards I would say you would be well over $1000.00 per enchant

tarmalen · June 2013

traciatim1 wrote: »

The one fatal flaw like others have pointed out is that you only use a green ward where the cost of the ward is less than the cost of a new rune, and you'd only use a blue ward when the cost and chance of failure of using green wards or new runs doesn't make sense . . . Yet still, it makes you kind of wonder how so many people have top rank runes now, doesn't it?

No wipe from Beta? Worst . . . idea . . . ever.

Having 100 alt accounts to "farm" for c-wards. Massive exploiting and duping that was eventually fixed (well some of them). Selling things on the AH for ridiculous prices.

The most important was the never fail on a fusing exploit that could let you rank up enchants without failure.

mconosrep · June 2013

I can't imagine there is anyone who hasn't figured out that the vast majority of the higher tier runes were not created by legitimate means.

wesgar · June 2013

Anyone that uses Coalescent wards for anything greater than 1% success rate fuses, (like what you see with the shards for armor and enhancements, and maybe high level Enchants, higher than rank 8) is wasting money or AD.

if you want a 'real' math problem, you'd need to figure out the percentage of failure for each rank, (it goes up as the ranks do,) then make an estimate based on the rate of failure, as to how many of the green wards, (which cost a dime's worth of Zen, or 3k - 4k ish of AD,) that you would actually use attempting to fuse the # of runes that you would need in order to get the next rank.

That would be much more accurate. What you have posted, doesn't reflect the reality at all.

Also, I'd start at 'rank 4' as those are the enchants that commonly drop while running level 60 content and in theory, the 4096 enchants that you would need could be farmed, perhaps by a guild or a dedicated farmer / group of farmers over a year or two, / several months.

Wish I could design the equation, but, I don't have the skills, the problem would include things like this though:

At Rank 4 it would take 4096 enchants to make a Rank 10 enchant.

you would need to successfully fuse those Rank 4's into 1024 Rank 5's.

There is an X% chance of failure each time you try to fuse a rank 4, into a rank 5.

Take that X% failure rate, of 1024 fuse attempts, that result would give you an approximate # of how many of the green wards you would use up.

You could extrapolate the cost of the wards, to see how much you'd have to spend on green wards, to get the 1024 Rank 5's that you'd need - then you'd have to do the whole thing again for each rank till you got to Rank 10.

G'luck, there's your semi coherent story problem for the day. But if you did it, it would give a much more accurate representation of how people actually fuse their runes - ain't nobody smart using Coalescent wards, to fuse rank 4's into rank 5's.

sludge01 · June 2013

I double checked you math, and I can do you one better.. I calculated the number of Teir-1 enchants required to get to Tier-10 IF you never used any wards.. If you didn't spend ANY money, you would use over 150,000,000,000 (150 BILLION) Rank 1 Enchants to get a Tier-10. Now, if that's no PROOF of pay to win, or cheaters, I truly don't know what is. Check my math if you like. I'm unsure of success rate of higher than rank 7, cause that's the highest I've ever seen.

Success Require Ratio Real Number
Rank 1 262144
Rank 2 65536 95.00% 62259.2 1.05 275,941
Rank 3 16384 90.00% 14745.6 1.11 306,601
Rank 4 4096 85.00% 3481.6 1.18 360,707
Rank 5 1024 60.00% 614.4 1.67 601,179
Rank 6 256 40.00% 102.4 2.50 1,502,947
Rank 7 64 20.00% 12.8 5.00 7,514,735
Rank 8 16 10.00% 1.6 10.00 75,147,345
Rank 9 4 5.00% 0.2 20.00 1,502,946,910
Rank 10 1 1.00% 0.01 100.00 150,294,690,976

chronomancer · June 2013

It will cost you like a ton of money to make one rank 10 enchant. Let's say you start off with a Rank 5.

4 x rank 5 = 1 rank 6 = $0 + $10 ward = Cost $10
4 x rank 6 = 1 rank 7 = $10 x 4 + $10 ward = $50
4 x rank 7 = 1 rank 8 = $50 x 4 +$10 ward = $210
4 x rank 8 = 1 rank 9 = $200 x 4 + $10 ward = $810
4 x rank 9 = 1 rank 10 = $810 x 4 + $10 ward = $3250

Holy sheets.

sludge01 · June 2013

I just redid the calucations using your math. Assuming you are smart, you would only use a covalesent ward on rank 8-10.. You would use preservation wards on rank 1-7.. Here's the numbers required to make a rank 10 Ward, using wards.. Calculate yourself how much zen/real money it would take. Hint: It's a LOT

Success Used Pres Used Cov
Rank 1

262144

Rank 2

65536

95.00%

68,985
Rank 3

16384

90.00%

18,204
Rank 4

4096

85.00%

4,819
Rank 5

1024

60.00%

1,707
Rank 6

256

40.00%

640
Rank 7

64

20.00%

320
Rank 8

16

10.00%

16
Rank 9

4

5.00%

4
Rank 10----1

1

sludge01 · June 2013

Actually, if you started at Rank-5, it would cost you on average, $1,340 dollars in wards to make a single rank 10..

observerer · June 2013

This thread is entertaining. Thanks for the info guys...

kevinf08 · June 2013

Is rank 10 the highest you can achieve with enchantments?

degraafination · June 2013

But just remember, it's not P2W.

mconosrep · June 2013

degraafination wrote: »

But just remember, it's not P2W.

True its E2W (Exploit2Win) since hardly anyone actually for the top tier enchantments....

chronomancer · June 2013

The dungeons are designed to be E2W. It's that broken.

morkelb78 · June 2013

In retro spec, if they wiped beta, the market / game would start off worse then you are now. case in point. The people who bought the hero's pack have 2 mill diamonds the people that bought the next one down have 600k diamonds, very few would waste diamonds this time around. On top of that they would have to refund Zen. This would put the people that spent alot of money on this game at such and advantage it wouldn't be funny.

seneca671 · June 2013

Sometimes I'm really glad I have better things to do than care if I have the highest ranking gear and enchants. Do the end-game instances a couple of times, then double back to Foundry and goofing around. Turning the game into an analysis/grind to get the best stuff... hell, I could get a menial job if I wanted to grind like that, and I'd get paid to do it!

healhamsta · June 2013

You don't need rank 10 enchants to win. /thread

chronomancer · June 2013

Seriously I don't think anyone has a rank 10 yet... it's so not worth the cost.

hkiewa · June 2013

chronomancer wrote: »

Seriously I don't think anyone has a rank 10 yet... it's so not worth the cost.

You would be wrong

mconosrep · June 2013

chronomancer wrote: »

Seriously I don't think anyone has a rank 10 yet... it's so not worth the cost.

Wrong on both counts since it cost most of the people who got them (almost) nothing.........

haelra · June 2013

Well, the rate of consumption of a given tier to produce the next tier is r=(4*p+(1-p))/p where p is the probability of success. Basically, for a given attempt, you will have a p chance of consuming four lower tiers for the success outcome, and 1-p chance of consuming one lower tier in the failure outcome. Only successes produce the higher tier, so the number of those produced per attempt is p.

Tier 1 -> 2 : p=95% r=4.05
Tier 2 -> 3 : p=90% r=4.11
Tier 3 -> 4 : p=85% r=4.18
Tier 4 -> 5 : p=60% r=4.67
Tier 5 -> 6 : p=40% r=5.50
Tier 6 -> 7 : p=20% r=8.00
Tier 7 -> 8 : p=10% r=13.00
Tier 8 -> 9 : p=5% r=23.00
Tier 9 -> 10 : p=1% r=103.00

Multiply them together and you get 440,021,236.05 rank 1's to make a rank 10, on average, without using wards.

ganjaman1 · June 2013

haelra wrote: »

Well, the rate of consumption of a given tier to produce the next tier is r=(4*p+(1-p))/p where p is the probability of success. Basically, for a given attempt, you will have a p chance of consuming four lower tiers for the success outcome, and 1-p chance of consuming one lower tier in the failure outcome. Only successes produce the higher tier, so the number of those produced per attempt is p.

Tier 1 -> 2 : p=95% r=4.05
Tier 2 -> 3 : p=90% r=4.11
Tier 3 -> 4 : p=85% r=4.18
Tier 4 -> 5 : p=60% r=4.67
Tier 5 -> 6 : p=40% r=5.50
Tier 6 -> 7 : p=20% r=8.00
Tier 7 -> 8 : p=10% r=13.00
Tier 8 -> 9 : p=5% r=23.00
Tier 9 -> 10 : p=1% r=103.00

Multiply them together and you get 440,021,236.05 rank 1's to make a rank 10, on average, without using wards.

Now imagine how much bank and inventory space you'll need by this logic to store them before fusing

kalizaar · June 2013

haelra wrote: »

Well, the rate of consumption of a given tier to produce the next tier is r=(4*p+(1-p))/p where p is the probability of success. Basically, for a given attempt, you will have a p chance of consuming four lower tiers for the success outcome, and 1-p chance of consuming one lower tier in the failure outcome. Only successes produce the higher tier, so the number of those produced per attempt is p.

Tier 1 -> 2 : p=95% r=4.05
Tier 2 -> 3 : p=90% r=4.11
Tier 3 -> 4 : p=85% r=4.18
Tier 4 -> 5 : p=60% r=4.67
Tier 5 -> 6 : p=40% r=5.50
Tier 6 -> 7 : p=20% r=8.00
Tier 7 -> 8 : p=10% r=13.00
Tier 8 -> 9 : p=5% r=23.00
Tier 9 -> 10 : p=1% r=103.00

Multiply them together and you get 440,021,236.05 rank 1's to make a rank 10, on average, without using wards.

Expect men in black to show up at your door tonight to silence you from showing how scary real math can be when applied to games.

dkcandy · June 2013

haelra wrote: »

Well, the rate of consumption of a given tier to produce the next tier is r=(4*p+(1-p))/p where p is the probability of success. Basically, for a given attempt, you will have a p chance of consuming four lower tiers for the success outcome, and 1-p chance of consuming one lower tier in the failure outcome. Only successes produce the higher tier, so the number of those produced per attempt is p.

Tier 1 -> 2 : p=95% r=4.05
Tier 2 -> 3 : p=90% r=4.11
Tier 3 -> 4 : p=85% r=4.18
Tier 4 -> 5 : p=60% r=4.67
Tier 5 -> 6 : p=40% r=5.50
Tier 6 -> 7 : p=20% r=8.00
Tier 7 -> 8 : p=10% r=13.00
Tier 8 -> 9 : p=5% r=23.00
Tier 9 -> 10 : p=1% r=103.00

Multiply them together and you get 440,021,236.05 rank 1's to make a rank 10, on average, without using wards.

Or 1 Exploit and accounts not banned for using said exploit.

262144 rank 1 enchants wow the cost is insane.

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