I can't quite tell who your response is directed at.

**You.**

the 17.3% is a best case scenario because all it requires it that at least ONE of the IVs is maxed. anything above 1 is a bonus.

**That's a worst case scenario. If it were the best case scenario, there would be possibility of improvement. You are assuming you only catch single flawless Ditto. The result will actually be lower because if you catch a double flawless Ditto, you don't need to catch 5 other single flawless, you only need to catch 4.**

the way you want to calculate it simply breaks down the 17.3% into its constituents. the logic applied above still applies, you still need to calculate how many dittos per case per IVs remaining need to be caught.

With your method you'd end up having to catch more, because now you're going for 2 IV maxed dittos, 3 IV maxed dittos, etc. you can't really suggest that with a lower chance (15.9% <-> 17,3%) you're going to need fewer ditto to be successful?

**No, you would need to catch less. I'm not saying "You have to go for this configuration of double flawless and four single flawless." I'm saying "You can go for 6 single flawless (your idea), OR you can go for a double flawless Pokemon and 4 single flawless, OR..." The "OR" operator in probability is similar to addition in algebra. Because probability can never be negative, and because you have a non-zero probability that a multi-flawless Ditto will pop up, this decreases the number of Ditto needed. You start out needing X with only single-flawless, and then you increase the probability to get even more flawless stats, meaning you'll need to catch more single-flawless later. This means you'll need some number less than X on average.**

I'd prefer to catch two dittos, each with a differing matched stat and try hatch a 2 IV maxed ditto.

**You cannot breed Ditto.**

Finally:

"To find the actual probability, you'd need to find the probability of six Dittos at 1 perfect IV each, OR five Dittos of 1 and one Ditto of 2"

no, because the 17.3% expects at least 1. anything with more than 1 can be considered a bonus.

**Yes, the 17.3% is at least 1. Therefore, any calculations that claim exactly 1 using this % will overestimate the number needed.**

"Then there's also the fact that finding a 4 flawless Ditto when you already have 5 1 flawless Dittos is the same as finding a 1 flawless Ditto."

The chances of winning the lottery is 50:50, either you do or you don't. no, wait, that isn't right...

statistics has no memory.

**I never said it did. What I said was to imagine this scenario:**

I catch 5 Dittos with flawless IVs in 5 different stats (all but Speed). Now in the last Ditto I catch, I'll finish my "set" if I catch a single-flawless Ditto (Speed), but I'll also finish the set with a 4-flawless Ditto (HP, Attack, Defense, Speed). If I were to catch that same 4-flawless Ditto as the first Ditto with a flawless IV, I'd only need to catch Ditto for Special Attack and Special Defense.

chance of finding a ditto with perfect speed, for example, is roughly 3%.

**Yes, 3.125% is the chance of finding a Ditto with 31 Speed.**

chance of finding a ditto with 4 perfect random IVs=

6!/4!(6-4)! * 1/32^4 * 31/32^2 = 0,000013425

which comes works out at 0,001343%.

which is greatly smaller than 3%!

**But that chance is included in the 3.125%. 1/32 gives you the probability that Speed will be perfect, but it doesn't also make the statement that none of the other stats are perfect. The actual chance of getting just perfect Speed, but imperfect others is (1/32) * (31/32)^5. This is about 2.666% chance. The probability of finding a Ditto with perfect Speed and 2-5 other perfect IVs is therefore 3.125-2.666, or about .459%. This probability will serve to lower the average amount of Ditto needed to be caught before getting at least one Ditto for each flawless IV in every stat.**