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 %