RRRGGB

with rng as it is, i doubt anyone could accurately estimate how many chromes it would take. that said, odds are really not in favour to get RRRGGB with that armour, but i think its still quite possible compared some other combination i could think of.
personally, if i really needed those colours, i'd rather sell that armour and buy an armour/ev or armour chest. might be cheaper in the long run :)
gl

I am doing a eldrich battery build, so I need armour or eva + es
I know it ain't impressive, but a " good " armour/es cost around 40 exalt, and I wanted to use a 6L befor i got that kinda wealth

And since syrioforel's calculation, I will aim for 400 cromos befor I try RRRGGB

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Solidwolf wrote:

I am doing a eldrich battery build, so I need armour or eva + es
I know it ain't impressive, but a " good " armour/es cost around 40 exalt, and I wanted to use a 6L befor i got that kinda wealth

And since syrioforel's calculation, I will aim for 400 cromos befor I try RRRGGB

Lol.

No.

You should take the average amount of chroms required, divided it by 2, convert it to exa and chaos combination and then host a lottery that gives that as payout.

Linking, coloring, socketing, eternaling, etc things yourself is stupid in this game since a lottery never fails and is on average cheaper as well.

For things that are easily to color, socket or fuse, then sure do it yourself, but for harder things like the thing you are trying to achieve: host a lottery.

They revealed a while back that it's the "attributes required" that governs this. It's not a static percent based only on item type. The higher your att req, the more likely it is to have that color. The .10-.45-.45 is just an approximation to the actual probability.

It's a fairly good approximation on items that level and slightly off on the highest level items (vaal, etc).

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Kirielis wrote:

It's a fairly good approximation on items that level and slightly off on the highest level items (vaal, etc).

Yes, but even a slight variation... Lets try say .08-.46-.46 will have a large effect on the expected number needed. (p=.46 now)

p^3 * (1-2p)^3 * 6! / (3!2!) = 60 * p^3 * (1-2p)^3= .002990 predicting 335 orbs needed.

About twice as much for a single percentage point off.

My point is... we need the actual formula used if we're ever going to make informed choices. Right now, it's more responsible to just say "lots" and leave it at that.

Don't listen to the ppl saying do a lottery for three off colors. Thats not that hard ot get on your own

p^3 * (1-2p)^3 * 6! / (3!2!) = 60 * p^3 * (1-2p)^3

This is wrong by the way... it assumes you'll be just as happy with RRRGGG as RRRGGB or even RRRBBB.

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Shagsbeard wrote:

p^3 * (1-2p)^3 * 6! / (3!2!) = 60 * p^3 * (1-2p)^3

This is wrong by the way... it assumes you'll be just as happy with RRRGGG as RRRGGB or even RRRBBB.

It does no such thing. The formula is correct provided socket coloring proceeds by the model I have indicated: that G/B roll with the same probability p, and that sockets are independently colored.

See: 3 R's are needed, each happening with probability (1-2p). 2 G's, each have probability p, and 1B, each having probability p.

That makes (1-2p)^3*p^3. Then there are 60 = 6!/(3!2!) rearrangements of the symbols RRRGGB.

The attribute requirement thing is why I gave a formula instead of a probability at first.

EDIT:

Shags, the calculation that would identify RRRBBB with RRRGGG with RRRGBB with RRRGGB would be

(2p)^3 * (1-2p)^3 * 6!/(3!3!)

No... but go on thinking you're right if it makes you feel better.

The correct formula is

.1^3 * .45^3 * C(3,6)*C(2,3)*C(1,1). So you're off by a factor of 3.