Why Do Fortuna Tier 5 Bounties have the Best Aya Drop Chance ?

[Lutesque]

So Ive only done a few bounties on both Fortuna and Deimos.... And while Deimos give me... Well... Nothing....

I got more than 6 Aya on Fortuna ...

 

So my question is.... Why ?

 Not only that.... DE said that Aya was suppose to have an the same Chance to Drop Across Every Bounty Tier on all the Open World's (Went looking for the thread... Couldn't find it)...  So then why is Fortuna Tier 5's Sticking out more than the other 14 Tiers ? 🤔

 

And you people wonder why I don't trust RNG....

And before anyone calls me out for not doing all the Bounties.... It's not just me ...

I get that RNG is RNG so there's going to be lobsided Results favouring one Bounty over the others.... But you know what... $@## that $@#@ ... Why is this Happening ? 🤔

 

Edit:

Found it...

Turns out I was wrong.... Or rather the way this was worded left me very Confused.... Still Confused infact.... 🤔....

Where's the Official Bounty Drop Table ?

[AegidiusF]

Fortuna Bounty tier 5 has the highest drop chances. I almost only do this bounty for Aya and I could get a Mag Prime set, Broar Prime set and Dakra Prime set (I already had Nova and her stuff before Prime Resurgence) some hours after the Event was launched.

It gives you 25% drop chance at phases 2 and 3, about 21% at phase 4 and 50% at final phase, and you can still get one more due to bonus rotaion at the end.

The drop tables are here www.warframe.com/droptables 

I've calculated the overall drop chance and at Fortuna Bounty tier 5 you have a 77.90% of getting at least one Aya (not even counting the bonus rotation). But very often I get 3 at just one bounty and sometimes 4. Be aware that we still have 22,10% chance of getting no Aya in the bounty.

You can easily get enough Aya to buy enough relics to get a complete set of every stuff available through relics in just 1 or 2 days and then just farm some Aya for the next rotation.

[stormy505]

It replaces the relics in those bounties before, to my knowledge fortuna always had the highest chance at getting relics before this event. Not 100% on that and if someone wants to fact check me please do.

[MqToasty]

29 minutes ago, Lutesque said:

So Ive only done a few bounties on both Fortuna and Deimos.... And while Deimos give me... Well... Nothing....

I got more than 6 Aya on Fortuna ...

So my question is.... Why ?

Honestly?  Your sample size is too small.  If you look at the drop rates on the official table:

https://n8k6e2y6.ssl.hwcdn.net/repos/hnfvc0o3jnfvc873njb03enrf56.html#cetusRewards

... you'll see that the Expected Value for all 3 Open World Level 40-60 Bounties are essentially the same.  The distribution varies slightly across Stages, meaning the Bonus is least important in Cetus and most important in Orb Vallis, but otherwise the numbers are close enough to basically be negligible.

For each area, you can expect roughly this many Aya per run in the long run if you run it many, many times:

Cetus:
0 + 0.3304 + 0.3304 + 0.2568 + 0.3878 (+ 0.3878) = 1.3054 (or 1.6932 with Bonus)

Orb Vallis:
0 + 0.25 + 0.25 + 0.2143 + 0.5 (+ 0.5) = 1.2143 (or 1.7143 with Bonus)

Cambion Drift:
0 + 0.2857 + 0.2857 + 0.2222 + 0.4348 (+ 0.4348) = 1.2284 (or 1.6632 with Bonus)

Simply put, the drop rate in Cambion Drift is slightly lower than the Cetus and Orb Vallis, but not so much that you would have been able to detect with just a few runs.  That's just RNG at work.  IMO, just pick whichever one you are familiar with and can more consistently get the Bonus for and/or run fast.  Even better if you also need another drop in that pool or need/want to curry favor with that syndicate.

[Lutesque]

1 minute ago, MqToasty said:

Honestly?  Your sample size is too small.  If you look at the drop rates on the official table:

If you read the whole Topic before responding....

49 minutes ago, Lutesque said:

And before anyone calls me out for not doing all the Bounties.... It's not just me ...

 

23 minutes ago, stormy505 said:

It replaces the relics in those bounties before, to my knowledge fortuna always had the highest chance at getting relics before this event. Not 100% on that and if someone wants to fact check me please do.

In my experience Fortuna and Cetus were always 50/50 for me.... Deimos is new to me so I can't count that one.

29 minutes ago, (NSW)AegisFifi said:

Fortuna Bounty tier 5 has the highest drop chances. I almost only do this bounty for Aya and I could get a Mag Prime set, Broar Prime set and Dakra Prime set (I already had Nova and her stuff before Prime Resurgence) some hours after the Event was launched.

It gives you 25% drop chance at phases 2 and 3, about 21% at phase 4 and 50% at final phase, and you can still get one more due to bonus rotaion at the end.

The drop tables are here www.warframe.com/droptables 

I've calculated the overall drop chance and at Fortuna Bounty tier 5 you have a 77.90% of getting at least one Aya (not even counting the bonus rotation). But Vert often I get 3 at just one bounty and sometimes 4. Be aware that we still have 22,10% chance of getting no Aya in the bounty.

You can easily get enough Aya to buy enough relics to get a complete set of every stuff available through relics in just 1 or 2 days and then just farm some Aya for the next rotation.

Thank You.... Busy Farming Aya right now but will give a quick once over when I'm done ♥️ !!!!

[MqToasty]

15 minutes ago, Lutesque said:

If you read the whole Topic before responding....

And if you understood anything I wrote, maybe you would be less confused and less distrusting of RNG?

[(NSW)BalticBarbarian]

2 hours ago, MqToasty said:

For each area, you can expect roughly this many Aya per run in the long run if you run it many, many times:

Cetus:
0 + 0.3304 + 0.3304 + 0.2568 + 0.3878 (+ 0.3878) = 1.3054 (or 1.6932 with Bonus)

Orb Vallis:
0 + 0.25 + 0.25 + 0.2143 + 0.5 (+ 0.5) = 1.2143 (or 1.7143 with Bonus)

Cambion Drift:
0 + 0.2857 + 0.2857 + 0.2222 + 0.4348 (+ 0.4348) = 1.2284 (or 1.6632 with Bonus)

While the numbers in the results are not too far off, this really isn't how probabilities work! There are plenty of damaging misconceptions about school-level statistics going around these forums already. Please don't spread any more!

[MqToasty]

7 minutes ago, (NSW)BalticBarbarian said:

While the numbers in the results are not too far off, this really isn't how probabilities work! There are plenty of damaging misconceptions about school-level statistics going around these forums already. Please don't spread any more!

Oh really now?  Then, pray tell, how do probabilities work?  Note the wording in my post: I'm giving the Expected Value for each bounty, in the long run.

[(NSW)BalticBarbarian]

4 minutes ago, MqToasty said:

Oh really now?  Then, pray tell, how do probabilities work?  Note the wording in my post: I'm giving the Expected Value for each bounty, in the long run.

The expected value is the sum of first moments - so you need to sum n*P(getting exactly n relics) for n=0...5 (assuming with bonus). Just adding up the individual probabilities may work in special cases, but that's not the general formula

[Lutesque]

6 hours ago, MqToasty said:

And if you understood anything I wrote, maybe you would be less confused and less distrusting of RNG?

What you wrote is pointless if you didn't read....

4 hours ago, MqToasty said:

Oh really now?  Then, pray tell, how do probabilities work?  Note the wording in my post: I'm giving the Expected Value for each bounty, in the long run.

They don't.... Everything is Lie and we're all going to die 😱 ?!!

 

[MqToasty]

6 hours ago, (NSW)BalticBarbarian said:

The expected value is the sum of first moments - so you need to sum n*P(getting exactly n relics) for n=0...5 (assuming with bonus). Just adding up the individual probabilities may work in special cases, but that's not the general formula

Oh boy.  Alright, I'll humor you once.  I'll even write it in long form so it's easier to follow along.  Let's take the Cetus Bounty, No Bonus case:

Odds of getting exactly 1 Aya:
((0.3304 * (1 - 0.3304) * (1 - 0.2568) * (1 - 0.3878)) * 2) + ((1 - 0.3304) * (1 - 0.3304) * 0.2568 * (1 - 0.3878)) + ((1 - 0.3304) * (1 - 0.3304) * (1 - 0.2568) * 0.3878)
= 0.4010318984300544

Odds of getting exactly 2 Aya:
(0.3304 * 0.3304 * (1 - 0.2568) * (1 - 0.3878)) + (0.3304 * (1 - 0.3304) * 0.2568 * (1 - 0.3878)) + (0.3304 * (1 - 0.3304) * (1 - 0.2568) * 0.3878) + ((1 - 0.3304) * 0.3304 * 0.2568 * (1 - 0.3878)) + ((1 - 0.3304) * 0.3304 * (1 - 0.2568) * 0.3878) + ((1 - 0.3304) * (1 - 0.3304) * 0.2568 * 0.3878)
= 0.2914078927069184

Odds of getting exactly 3 Aya:
(0.3304 * 0.3304 * 0.2568 * (1 - 0.3878)) + (0.3304 * 0.3304 * (1 - 0.2568) * 0.3878) + ((0.3304 * (1 - 0.3304) * 0.2568 * 0.3878) * 2)
= 0.0926889912940544

Odds of getting all 4 Aya:
0.3304 * 0.3304 * 0.2568 * 0.3878
= 0.0108713355684864

Expected Number of Aya per Run:
0.4010318984300544 + 0.2914078927069184 * 2 + 0.0926889912940544 * 3 + 0.0108713355684864 * 4
= 1.3054

Hmm, funny how that number is exactly the same as the 1.3054 I calculated above with the simpler method, and not "not too far off" as you claimed?  If you do not understand how this can be, then may I suggest you give your highschool math teacher a call and ask them?

Look, I do not mind being corrected if I truly made a mistake.  But if you are going to correct me, you better make sure you provide ample evidence so we can all learn something, and not attempt to sneak by with a handwave.  I really do think other forum users who may not be as adept at math deserve better than this kind of misinformation.

[MqToasty]

2 hours ago, Lutesque said:

What you wrote is pointless if you didn't read....

They don't.... Everything is Lie and we're all going to die 😱 ?!!

 

How would I have known about you being confused, not trusting RNG and wanting to see the official drop table if I did not read your post?

[(NSW)BalticBarbarian]

26 minutes ago, MqToasty said:

Hmm, funny how that number is exactly the same as the 1.3054 I calculated above with the simpler method, and not "not too far off" as you claimed?  If you do not understand how this can be, then may I suggest you give your highschool math teacher a call and ask them?

Look, I do not mind being corrected if I truly made a mistake.  But if you are going to correct me, you better make sure you provide ample evidence so we can all learn something, and not attempt to sneak by with a handwave.  I really do think other forum users who may not be as adept at math deserve better than this kind of misinformation.

Firstly, what part of the accurate definition of the Expectation (the first moment of the variable) are you considering a "handwave"? Your high school teacher might not have taught you that way, but that's what the expected value actually is. And if you are suggesting that it isn't - then you are the one who is spreading misinformation.

Secondly, your "proof by example" of using a finite-length decimal expansion to show coinciding values in one special case is a bit silly. You know better than that!

Now, had you referred to the linearity of expectation function for independent random variables (which, I admin, I forgot about. my bad.) - then your rebuke would have been reasonable.

[MqToasty]

7 minutes ago, (NSW)BalticBarbarian said:

Firstly, what part of the accurate definition of the Expectation (the first moment of the variable) are you considering a "handwave"? Your high school teacher might not have taught you that way, but that's what the expected value actually is. And if you are suggesting that it isn't - then you are the one who is spreading misinformation.

Secondly, your "proof by example" of using a finite-length decimal expansion to show coinciding values in one special case is a bit silly. You know better than that!

Now, had you referred to the linearity of expectation function for independent random variables (which, I admin, I forgot about. my bad.) - then your rebuke would have been reasonable.

You know that is not what I am talking about.  I am talking about your initial reply, claiming that my original calculations are "not too far off, this really isn't how probabilities work!".  I have already gone out of my way in showing you that you are wrong with that assertion.  So instead of tossing around jargon and posturing, how about you put up and show us your calculations?  Since you insist on pretending to know better, how about you tell us what the Expected Aya per Run is for the Bounties?

[(NSW)BalticBarbarian]

1 hour ago, MqToasty said:

o instead of tossing around jargon and posturing, how about you put up and show us your calculations?

My calculation is exactly what you've provided as the "long form" (haven't checked the exact numbers, happy to assume they are ok). I did originally say "sum n*P(getting exactly n relics)" - which is exactly what you did in your "long form" version. Do you want me to copy it over in a slightly different format?

1 hour ago, MqToasty said:

I have already gone out of my way in showing you that you are wrong with that assertion. 

You gave a single numerical example. A single example is not showing anyone anything - except for that specific example

1 hour ago, MqToasty said:

So instead of tossing around jargon and posturing

2 hours ago, MqToasty said:

then may I suggest you give your highschool math teacher a call and ask them?

"Jargon" - probably guilty. But "posturing" is all you!

 

Putting all that aside, I have already said

1 hour ago, (NSW)BalticBarbarian said:

Now, had you referred to the linearity of expectation function for independent random variables (which, I admin, I forgot about. my bad.) - then your rebuke would have been reasonable.

admitting that I've missed a way to simplify the calculation and actually providing a proof for the simplification. An actual proof, not a single example where two numbers match.

What exactly do you expect, a hand-written apology letter? Get off your high horse!

[MqToasty]

15 minutes ago, (NSW)BalticBarbarian said:

What exactly do you expect, a hand-written apology letter? Get off your high horse!

Actually, all I want is for you to admit that you are wrong, if you believe it, so other forum members who are less adept at math would not be misled by your first response.  If you do not believe it, then the onus is on you to prove that I am wrong, as I have done more than my share of calculations.

18 minutes ago, (NSW)BalticBarbarian said:

admitting that I've missed a way to simplify the calculation and actually providing a proof for the simplification. An actual proof, not a single example where two numbers match.

Saying the original method I provided is a way to "simplify the calculation" tells me that you do not understand basic probability and expected values at all.  So why pretend and posture just to be proven wrong?  Or do you still believe that it is a "coincidence" that the two numbers matched, and that it is "not proof"?  Well, I provided 6 values in my original calculation.  All you need to do is calculate the rest of the 5 your way and see if they differ.  You only need one example to prove me wrong.  Can you do it?

[(NSW)BalticBarbarian]

5 minutes ago, MqToasty said:

Saying the original method I provided is a way to "simplify the calculation" tells me that you do not understand basic probability and expected values at all. 

The "long form" method works from the primary definition of the Expected Value of a random variable - that is why it is referred to as "long form". See, for example, https://mathworld.wolfram.com/ExpectationValue.html or https://en.wikipedia.org/wiki/Expected_value So this is the primary (albeit longer) method.

The shorter method you used initially does not use the primary definition - it uses linearity (which is a consequence of the primary definition that needs to be proven). Therefore, it is a way to simplify the calculation.

If you think that you original method is the primary definition of the expected value - then you're the one who does not understand basic probability.

 

14 minutes ago, MqToasty said:

Actually, all I want is for you to admit that you are wrong

I did it twice already, including the line immediately above the line you've quoted. How many more times would you like me to do it? You're the one who's stirring up drama here!

 

 

[(PSN)robotwars7]

all i know is that the chances are too low to bother aggressively farming. I did get my first Aya on Deimos tonight though, but I don't expect to get them that often. I only want 10 to buy a cosmetic, but for the sake of those who are farming Aya to buy the Vaulted Relics, I do think one of these improvements should be made:

- Aya drop chance increased, a simple enough fix

or...

- T5 and Steel Path bounties should have a chance to give 2 or even 3 Aya, which still won't mean "giving out" primes, since the relics will still be RNG as to what you get when they open, but this way, players who do harder missions are rewarded with more chances for the relic ,and in turn the item, that they want. no guarantees, but a fairer system.

[MqToasty]

8 hours ago, (NSW)BalticBarbarian said:

The shorter method you used initially does not use the primary definition - it uses linearity (which is a consequence of the primary definition that needs to be proven). Therefore, it is a way to simplify the calculation.

If you think that you original method is the primary definition of the expected value - then you're the one who does not understand basic probability.

That alone shows me that you not only do not understand probability, but are not that good at math either.  It is the exact same formula, just applied slightly differently.  It's the difference between a * (b + c) and ab + ac.  Now if you asked nicely to begin with I would have been glad to explain it all in detail to you until you understood it, but with your recalcitrant attitude?  No thanks.  Not your professor, not your parent -- don't owe you any education.

8 hours ago, (NSW)BalticBarbarian said:

I did it twice already, including the line immediately above the line you've quoted. How many more times would you like me to do it? You're the one who's stirring up drama here!

Really?  Could have fooled me.  All I saw was you repeating "you coincidentally found something that worked, but it's no proof" and "you still don't understand probability"!  But since you are admitting you were wrong now, it's fine.  You can continue living in your fantasy world where you were only wrong by a small technicality.  But I will not hesitate to slap you down again if I find you misleading other forum-members in the future.

[MqToasty]

5 hours ago, (PSN)robotwars7 said:

all i know is that the chances are too low to bother aggressively farming. I did get my first Aya on Deimos tonight though, but I don't expect to get them that often. I only want 10 to buy a cosmetic, but for the sake of those who are farming Aya to buy the Vaulted Relics, I do think one of these improvements should be made:

- Aya drop chance increased, a simple enough fix

or...

- T5 and Steel Path bounties should have a chance to give 2 or even 3 Aya, which still won't mean "giving out" primes, since the relics will still be RNG as to what you get when they open, but this way, players who do harder missions are rewarded with more chances for the relic ,and in turn the item, that they want. no guarantees, but a fairer system.

I will admit that it is not an "easy farm", but after a few days of farming now, I am fairly confident I will be able to pick up most of the frames I am missing, if I can stick with it.  Already have Mag Prime building, currently saving up for future offerings.

[(NSW)BalticBarbarian]

4 hours ago, MqToasty said:

That alone shows me that you not only do not understand probability, but are not that good at math either.  It is the exact same formula, just applied slightly differently.  It's the difference between a * (b + c) and ab + ac.  Now if you asked nicely to begin with I would have been glad to explain it all in detail to you until you understood it, but with your recalcitrant attitude?  No thanks.  Not your professor, not your parent -- don't owe you any education.

Really?  Could have fooled me.  All I saw was you repeating "you coincidentally found something that worked, but it's no proof" and "you still don't understand probability"!  But since you are admitting you were wrong now, it's fine.  You can continue living in your fantasy world where you were only wrong by a small technicality.  But I will not hesitate to slap you down again if I find you misleading other forum-members in the future.

If one formula can be derived from another under a bunch of assumptions - it does not make them "the exact same formula, just applied slightly differently" (even if you put it in bold). It makes the latter a simplification of the former under certain assumptions (here - being able to split your random variable into a finite sum of mutually independent uniformly distributed ones).

This post of yours convinces me more than anything else that you actually don't know anything about probability - you were just given a formula by someone else and are using it everywhere hoping that it works. In this specific case, it did - but it really looks like you have no idea why. Otherwise you would not have been anywhere near as defensive about it! This time you got lucky - so congrats!

But you really should try and understand where the formulas you use come from - for your own good and for the good of others around you!

[MqToasty]

13 minutes ago, (NSW)BalticBarbarian said:

If one formula can be derived from another under a bunch of assumptions - it does not make them "the exact same formula, just applied slightly differently" (even if you put it in bold). It makes the latter a simplification of the former under certain assumptions (here - being able to split your random variable into a finite sum of mutually independent uniformly distributed ones).

😂 There were no "assumptions", and there were no "simplifications".  Simply that very equation you provided applied 5-6 times per bounty.  But if you don't get it, you don't get it.

16 minutes ago, (NSW)BalticBarbarian said:

This post of yours convinces me more than anything else that you actually don't know anything about probability - you were just given a formula by someone else and are using it everywhere hoping that it works. In this specific case, it did - but it really looks like you have no idea why. Otherwise you would not have been anywhere near as defensive about it! This time you got lucky - so congrats!

But you really should try and understand where the formulas you use come from - for your own good and for the good of others around you!

Very good advice that you should follow from now on!  I now understand why you were tossing around so much jargon.  Because you just grabbed the first equation you found and did not understand how it worked.  Sad, but not unexpected...

[Lutesque]

11 hours ago, (PSN)robotwars7 said:

all i know is that the chances are too low to bother aggressively farming. I did get my first Aya on Deimos tonight though, but I don't expect to get them that often. I only want 10 to buy a cosmetic, but for the sake of those who are farming Aya to buy the Vaulted Relics, I do think one of these improvements should be made:

I have a Resource Booster so my Real Incentive to do these Bounties is to get Mutagen Samples on Deimos and Thumpers On Cetus.... Fortuna gives me nothing so il just Feed The many many Rocks that fell out of The Exploiter's Ass to Smoke Finger and Cap my Standing that way...

So it's not as Demotivating when I complete the whole Bounty and there's no Aya.... Or when I fail Liberation Stages 😱....

First Rule Of Warframe... Never Farm For Just One Thing at a Time...

Oh and Since because Mechs are a thing now... I've been leveling my Kuva Hek and Mirage when I can then Cowering behind Void Rig when $@#& Gets Real....

You gotta make the most of your Grinds 😝

 

[(NSW)BalticBarbarian]

58 minutes ago, MqToasty said:

now understand why you were tossing around so much jargon.

If you actually look up what that "jargon" means, you will realise that it is actually mathematical terminology that forms a rigorous proof of the formulas. How exactly would you imagine that might happen if I didn't understand it?

1 hour ago, MqToasty said:

Simply that very equation you provided applied 5-6 times per bounty.

The equality between using that equation for the whole bounty and adding up results of using that equation for each of the stages is called "linearity" (look up what it means for a function to be "linear" in the mathematical sense). And it only works because the rolls from the different stages are independent of each other ("Independent random variables"). Just look up the conditions for the Expectation function to be linear.

Just because you don't understand the conditions you are using doesn't mean that you don't need them!

[MqToasty]

14 minutes ago, (NSW)BalticBarbarian said:

If you actually look up what that "jargon" means, you will realise that it is actually mathematical terminology that forms a rigorous proof of the formulas. How exactly would you imagine that might happen if I didn't understand it?

Copy and paste would be my guess.

15 minutes ago, (NSW)BalticBarbarian said:

And it only works because the rolls from the different stages are independent of each other ("Independent random variables").

Finally figured it out, eh?  And do you now realize how silly this statement is?

On 2021-11-19 at 7:01 PM, (NSW)BalticBarbarian said:

so you need to sum n*P(getting exactly n relics) for n=0...5 (assuming with bonus)

Regardless, since you've (supposedly) already conceded, I'm done with you.  Just stop misleading other forum members who won't know any better and we can go our separate ways.