Drop Rates of some items is rediculous!

[Yasminou]

Oh, okay. LOL
Let's math this out. (And feel free to correct my math, others out there who are better at this than I am.) In Void Ark, there are a possible 21 items to drop on each boss, according to the loot tables and two chests per boss.

Probability = (# of desirable outcomes)/(# of possible outcomes)
So the probability of a drop is 2/21 (two desirable outcomes out of a possible 21 drops). That means that you have a 4.76% chance to see a specific item of loot drop in a chest.

Odds = (# of desirable outcomes)/(# of undesirable outcomes) So if you are looking for a specific piece of loot, then you have 1 desirable outcome out of a possible 20 outcomes per chest. Odds are seeing the 1 piece you want become 5%

However, the odds of prior drops do not influence the odds of future drops, as someone else linked to know as Gambler's Fallacy. The more you run VA and see other drops happening, doesn't mean that the 1 desirable drop you want is more likely to happen. For example, if you flip a coin you have a 50-50 chance of getting heads and a 50-50 chance of getting tails. If you get 10 heads in a row, that still means that on your next flip the odds are still 50-50. If you see the same loot in a row, or even dropping at the same time, then that doesn't mean that there is anything wrong with RNG. It just means that you're unlucky so far.

I am not sure to understand your computation. I agree that each boss has 21 possible drops, but the number of desirable outcomes is 1 if you aim to get a particular piece (even though there are two independent drops) or 2 if you aim to get a piece among 2 (for instance healer's and caster's bodies). To see that your formula is incorrect, imagine that each chest had 22 drops instead of 2, you would have a strictly greater than 1 probability.

Each piece can drop from two bosses (A and B) and each boss's chest has two drops (which can be the same), let's say A1/A2 and B1/B2. The probability of not seeing a particular piece is that the outcomes of A1, A2, B1 and B2 are not this piece, hence (20/21)^4. Therefore, the probability of seeing it at least once is 1-(20/21)^4 which is roughly 0.1773.

If someone aims to get any of two pieces for a same role, then it gets more complicated depending whether this two pieces can drop from a same boss or not. Let's assume one wants the healer's body or the healer's legs. Legs drop from the 2d boss (outcomes B1, B2) and the last boss (outcomes D1, D2). Bodies drop from the 3rd boss (outcomes C1, C2) and the last boss (outcomes D1, D2). If we expect not to see any of these pieces then the probability is (20/21)^4 (20/21 for B1, B2, C1, C2) times (19/21)^2 (19/21 for D1, D2). Therefore the probability of seeing at least one of these piece is 1-(19/21)^2*(20/21)^4 which is roughly 0.3265.
Finally, if both pieces do not drop from the same bosses, let's say healer's head (1st and 2d boss so outcomes A1, A2, B1, B2) and healer's chest (outcomes C1, C2, D1, D2), then the probability of seeing at least one of these pieces drop is 1-(20/21)^8=0.3232.

Cheers.