Shard drop rate is toooooo low :(

"Winterwolves;c-2393960" wrote:

You are not listening so there is not much point.

You appear to have deleted or changed the post that was replying to, but it basically had no correct math in it at all.

Not deleted...it disappeared suddenly...

I rewrote it above...

So, point out where is the uncoreect math...

You surely agree with the math of the individual attempt.

(0/1) = 67%
(1/1) = 33%

Now see the 5 collective attempts as 5 attempts per day....and do it 1000 days = 5000 attempts
>>> 67% x 5000 attempts = 3350 failed attempts.

Since i do 5 attempts per day so these 3350 failed attempts will be archieved in 3350 attempts ÷ 5 attempts/day = 670 days.

You see?

Spending 2800 crystals on character shards is spending crystals like crazy. That is an absolute last resort thing to do, and only the big spenders should use crystals like this.

Some advice:

I saw you said you do 1 gal war battle per day.
Complete gal war each day. The whole track. Piles of that currency and 25(?) crystals per day. Once you have done it a bunch you get to just sim it.

Use crystals on energy only, and pick 2 characters to refresh the node on. Preferably accelerated characters. Just refresh them once per day to cut the farm time in half. (Not because of a change in drop rate, because doing 10 sims per day is twice as many as 5).

When you get a pile of crystals don't blow it on gear and shards. Use it to refresh more energy types each day (only for up to 3x per day, until the refresh cost goes up). You will get more gear and shards per crystals this way.

"GegeGerard24;c-2393965" wrote:

"Winterwolves;c-2393960" wrote:

You are not listening so there is not much point.

You appear to have deleted or changed the post that was replying to, but it basically had no correct math in it at all.

Not deleted...it disappeared suddenly...

I rewrote it above...

So, point out where is the uncoreect math...

You surely agree with the math of the individual attempt.

(0/1) = 67%
(1/1) = 33%

Now see the 5 collective attempts as 5 attempts per day....and do it 1000 days = 5000 attempts
>>> 67% x 5000 attempts = 3350 failed attempts.

Since i do 5 attempts per day so these 3350 failed attempts will be archieved in 3350 attempts ÷ 5 attempts/day = 670 days.

You see?

Why do you keep on with the collective attempts? Each attempt is an individual 33% chance.

You can't lump 5 attempts together and say 0/5 is the 67% fail. You have about a 14% chance of a 0 out of 5.

The 1/5 has 1 drop and 4 not drops.
The 2/5 has 2 drops and 3 not drops.
Etc

Why would you do 1000 days if you are doing 5 per day? Again with the collective thing messing up your math.

You need to do about 1000 attempts to get to the 330 shards. Not 1000 days. A group of 5 attempts is not the same as 1 attempt.

A player who does not refresh the node would need 200 days in order to do 1000 attempts. A player who refreshes the node once per day will only need 100 days.

"nfidel2k;c-2393963" wrote:
The mistake was transferring the 33% drop rate to the success of the full set.

I don’t know if you were going with the chance to roll at least one success out of five, which would be 33%; but the inverse of that would be the chance to roll at least one failure out of five, which would be 67%. So the chance of rolling 1, 2, 3, 4, or 5 out of 5 is 33% technically, but the 67% inverse is the chance to roll 0, 1, 2, 3, or 4 out of 5. Not just roll a 0 out of 5.

The chance to roll 0 out of 5 is the chance on a single roll (67%) carried to the power of the number of rolls. So 0.67 x 0.67 x 0.67 x 0.67 x 0.67, or (0.67)^5, or approximately 13.5%.

No...and it is out of 6, not out of 5.

There are 6 samples (0/5), (1/5), (2/5), (3/5), (4/5) and (5/5)

But we know that the probabily of scoring is 33% <<< everyone seems to agree with this.

What is the probability of scoring...the probability THAT we score.

In individual score...we can say.
1st attempt = (1/1) <<< scores
2nd attempt = (0/1)
3rd attempt = (0/1)
4th attempt = (1/1) <<< scores
5th attempt = (0/1)

Or we can say (2/5) if we want to say it as 5 collective attempts.

Now, we are all agree that in a large data, we get 33% of success rate.
Lets say, 1000 data.

In individual attempt = 1000 attempts have to be made.
In 5 collective attempts = 200 attempts have to be made.

But since we can do 5 attempts per day (without refresh)...these 1000 attempts are actually those 200 5-collective-attempts...are they not? You see?

So...how do you now say...what is applied to those 1000 attemps cant be apllied to those 200 5-collective-attempts?

It is just a new way of saying it....just a term.
The datas are the same, no matter how you group them, in group of 1 or in group of 5.

1000 individual attempts = 200 5-collectives attempts.

Thus what can be aplied to the first can be applied to the later too.

All seem to agree to 67% x 1000 individual attempts = 670 fail attemps.

So...67% x 200 5-collectives-attempts = 134 fail attempts

So i will have (0/5) 134 times...
And i will have the amount of (1/5) + #(2/5) + #(3/5) + #(4/5) + # (5/5) = 66 times.

Just do normal 1000 individual sampling that you already know...

Like this....
(1/1)
(0/1)
(0/1)
(0/1)
(0/1)
(0/1)
(0/1)
(0/1)
(0/1)
(0/1)
(0/1)
(1/1)
(1/1)
(0/1)
(0/1)
•
•
•
(1/1) <<< until the 1000th data.

But now, group it into 5.

(1/1) group 1 = (1/5)
(0/1)
(0/1)
(0/1)
(0/1)

(0/1) group 2 = (0/5)
(0/1)
(0/1)
(0/1)
(0/1)

(0/1) group 3 = (2/5)
(1/1)
(1/1)
(0/1)
(0/1)
•
•
•
(1/1) <<< until the 1000th data.

I am sure...that you will have close to 134 of group 2 (failed group) and 66 of success group which contain (1/5) and/or (2/5) and/or (3/5) and/or (4/5) and/or (5/5).

BTW, i am out.
I have presented the math as good as i could. And i hope you see it.

I dont want to make enemies by argueing to much while the truth is only in GOD's hand and the programmers'es hands.

Thanks for the good discussion.

Your math is still wrong.

People want to get the shards faster...

...then they end up with a 7 star character sitting on the bench without gear.

That's not how you play f2p.

He Dunning-Krugered all of us.

And I think the math degree is impressive Darjelo. It’s definitely not a “Certificate of Completion” thing, it takes work. Very cool.

I just realized that God is one of CG's programmers responsible for drop rate.

"GegeGerard24;c-2393964" wrote:

"crzydroid;c-2393959" wrote:

So if you're getting 55 a day from GAC, that means you are in Carbonite 5. If you get even to Carbonite 1, that is 20 more crystals a day right there. If you are brand new that is one thing. But don't claim to speak to all f2p. Plenty of f2p are in Kyber and finish at the top of their fleet payout. They earn 600-700+ crystals a day. They can easily afford refreshes.

F2P in my shoes, in my condition...
Dont talk about kyber with 600-700 crystal a day x_x

Yes...at 25 crystal per refresh and so much things (sold in crystals) to buy in the shipment...i wonder if an F2P player (again, in my shoes) would waste his crystals in unsure/uncertain attempt in nodes.

Sure, in your shoes, at the start of the game, refreshing character nodes would seem daunting. But a lot of people would be expected to advance and grow and many have. So my comments were more a response to what seemed like a blanket statement about "f2p." People mean a lot of different things when they say "f2p" on this site. Mostly, it seems they mean "me." There are a lot of f2p in this game at different stages of progress. Back when arena was more important and the game was more meta-driven, the only way for an f2p to get in on the new meta during the first run of the event was to refresh nodes multiple times. So it is very possible.

And not to beat a dead horse about the math(s), but the 33% drop rate, as others have pointed out, refers to the chance of dropping a shard per each attempt. This also amounts to the proportion of successful drops out of any number, N. For an individual sample, x, the chance of a shard is 0.33. So the expected value of shards over N attempts is N*0.33.

Yes, over time, and over individuals, 67% of the total drops are expected to be failures, though this can vary from sample to sample. But that just means that out of 1000 attempts, the expected value of successes is 330. So if you do five a day, that's an average across individuals and character farms of 200 days.

The chance of getting 0 out of 5 is NOT 67%. The porportion of expected failures out of 5 is 67%. 5*0.67 = 3.35 expected failures. Which means 1.65 expected successes, so any random draw from among individuals and across days, we should expect one or two successes out of five. But some of the observations will be different! As pointed out, the chance of exactly 0/5 is 0.67^5 = 13.5%.

You are right that there are six possible outcomes, but rather than the chance of 0 holding at 67% across attempts, the chance of any single outcome (e.g. 0/5) decreases as the number of outcomes increases. And by single here, I mean that there is only one way you can get 0/5, and that is for all attempts to be 0. There are five ways to get 1/5, however. You can have a success on either attempt 1, attempt 2, etc. So the relative odds of 1/5 go up (i.e., is not just 0.33/5).

People have listed the exact probability tables for you, and you were even pointed towards the binomial distribution, which for your convenience is:

(N!/(x!*(N-x)!)*p^x*q^(N-x)

where N is the total number of trials, x is the number of success, and p and q are the odds of success and failure, respectively (where q = 1-p).