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User Info: Auron402

Auron402
10 years ago#1
I am in the process of updating my FAQ on the Ancient Cave, and I would like to include the specific probabilities of finding Iris items. I KNOW that I have seen this information somewhere, but I can't seem to find it right now. I recall that this is a decreasing function -- that every time you find an Iris item, the probability of finding the next one decreases. What I am seeking are the specific numerical probabilities of finding each of the nine Iris items that appear in red treasure chests.

If you have this information, please reply. I will see that you get due credit for your contribution to my FAQ as well as my sincere appreciation.

User Info: StapledShut

StapledShut
10 years ago#2
Well, of course it's going to decrease with there being fewer of them. Or do you mean that it decreases by more than normal?
I'll put it in a scenario to clarify. For simplicity's sake (Note to the Nit-Pickers: this means it's a hypothetical situation, one that I know is not directly indicative of reality), let's say there's 91 items that can appear in a red chest, plus 9 Iris Treasures, all with equal probabilities. This would put the probability of one being in a given chest at 9%. After one is found, it would reduce to 8/99 or 8.081%. Do you mean it just decreases like that, or that the probability is programmed to step down in addition to the natural decrease?

User Info: ARTEMlS

ARTEMlS
10 years ago#3
I don't know the exact probabilities unfortunately, but considering the experience I made the probability goes down a lot the more Iris items you got.

I also made the experience that the probability also goes down a lot, if you have beaten the Master.

But this is only my observation. I can't say for sure, that it's right.
Final Fantasy 3 - FC: 4510 8027 2171 (Europe)

User Info: Auron402

Auron402
10 years ago#4
The second scenario. The reference that I have seen (but cannot find right now) claimed that there is a specific assigned probability of a given red chest containing an Iris item. Once an Iris item is found, the new probability of a given red chest containing an Iris item is significantly lower than the probability of finding the first one -- MORE than you would experience by merely removing one possible outcome from the sample space.

Again, if anyone out there has the actual numbers that I seek, I would appreciate a response. Thank you.

User Info: StapledShut

StapledShut
10 years ago#5
Okay. Sorry, I don't have that information, but I was itching to know that bit anyway. I have this obsession with quantifying everything I can.

User Info: Gunty_

Gunty_
10 years ago#6
Sorry for the late reply, but the mechanism works like this:

One of the first things that is done when generating a new AC floor is the first checkpoint for Iris Treasures. The first checkpoint determines if the next floor will potentially contain an Iris Treasure. Here, the game generates a random number from 0 to 255 and checks if it's lower than 5. If it's lower than 5, (5/256 = 1.95%) that game proceeds to the next checkpoint.

At the next checkpoint, the game determines what type of Iris Treasure is possibly generated. To do this, the game again generates a random number from 0 to 255 and divides it in 9 domains, for every Iris Treasure one. From the domain of the second random number, the Iris Treasure type is determined.
Next, the game checks if the Iris Treasure to be generated was already found by you; if you found it before than you will end up with nothing. This obviously happens more often when you've found more Iris Treasures.

The last thing that might prevent the Iris Treasure from appearing, is the number of treasure chests in the to be generated floor. If there is no room for treasure chests, the Iris Treasure simply cannot appear. This isn't determined by a set chance, but can be derived from statistical data. If I recall correctly, about 2% of all rooms contain no chests, although I'm afraid I've lost exact data about it.

For <u>n</u> Iris Treasures found, these are odds of passing each checkpoint:

Pass 1: Chance to pass: 5 / 256 = 1.95%
Pass 2: Average chance to pass: (9-n) / 9 = 100% - n*11.1%
Pass 3: Statistical chance to pass: ~98%

Yielding the total chance:

P(n) = 49 / 23850 * (9-n)

P(0) = 1.85 %
P(1) = 1.64 %
P(2) = 1.44 %
P(3) = 1.23 %
P(4) = 1.03 %
P(5) = 0.82 %
P(6) = 0.62 %
P(7) = 0.41 %
P(8) = 0.21 %

When ignoring pass 3 (since I'm not too sure about this value) this transforms to:

P(n) = (9-n) / 477

P(0) = 1.89 %
P(1) = 1.68%
P(2) = 1.47 %
P(3) = 1.26 %
P(4) = 1.05 %
P(5) = 0.84%
P(6) = 0.63 %
P(7) = 0.42 %
P(8) = 0.21 %

User Info: Gunty_

Gunty_
10 years ago#7
For the record, the number n used in the formula above excludes the Ancient Jelly. So the probability of finding Iris Treasures does not depend on whether you beat the Ancient Cave or not.

User Info: StapledShut

StapledShut
10 years ago#8
So it does decrease incrementally. It just seems like it decreases by more because of that first checkpoint, as you call it, isolating the probability from all other treasures. Interesting.

User Info: ARTEMlS

ARTEMlS
10 years ago#9
Thank you for clearing that up!

The rumor that the probability decreases came from the official German Nintendo hotline. They told me even that it's absolute impossible to get Iris Items after beating the Master. When I had beaten him the first time I only missed one item. Unfortunately I wasn't very lucky and after several weeks of not finding it I started to believe that there might be something true within it and started a completely new game.

Now I feel really stupid, because I wasted so much time to start again collecting the complete Iris Items and blue chest stuff from the beginning. I just wished I'd have had a little bit more luck.
Final Fantasy 3 - FC: 4510 8027 2171 (Europe)

User Info: Auron402

Auron402
10 years ago#10
Gunty -- First of all, let me thank you for providing the information that I sought. You gave me exactly what I needed -- and quite a bit more. I find it remarkable that you dug up information this specific.

My background is in mathematics, and I'm trying to make sense of the formulas that you provided. Is it possible that when you wrote your probability formulas that you interchanged the last two digits in 256? For example, your denominator in your first formula was 23850. This number equals 265*90, not 256*90. It seems that this denominator should be 256*90=23040. Likewise on your second formula, 5/(265*9)=1/477, and it seems that this should be 5/(256*9)=5/2304. I am assuming that your number 256 (=2^8) is correct, since it is a power of 2.

I realize that this changes to probabilities only slightly but I'm trying to make sense of what you wrote. If you used 265 instead of 256, it all makes sense.
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