How is the per-breed perfect chance calculated?

Guaranteed-inherited IVs are already perfect, so only the remaining required IVs must roll max. If R IVs still need to roll perfect and each has chance p, the per-breed success chance is p raised to the power R.

Why is the expected number of attempts one over the chance?

Each breed is an independent trial with success probability s. The number of trials until the first success follows a geometric distribution, whose mean is exactly 1/s. So a 1 percent per-breed chance averages 100 breeds.

What is the chance of at least one success in N attempts?

With per-breed chance s, the chance of failing every one of N independent breeds is (1 − s) to the power N. Subtracting that from 1 gives the at-least-once probability within N attempts.

What value of p should I use for one IV?

If a random IV is uniform over V possible values and you want the single maximum, p is 1/V. For a stat with 32 possible values, p is about 0.031. If parents pass higher floors, raise p to match your effective odds.

Does inheritance reduce how many IVs must roll?

Yes. Each guaranteed-inherited perfect IV is one fewer that must roll randomly. Increasing guaranteed inheritance by one cuts the random requirement and multiplies your per-breed odds by 1/p, dramatically reducing attempts.

Final Fantasy XIV Breeding & IV Calculator

Email me this result

Get this tool's output sent to your inbox, plus one useful tool a week. No spam, unsubscribe any time.

Breeding for a perfect roll is a probability problem: some stats are inherited for free, the rest must roll their maximum, and you repeat until one offspring gets everything right. This calculator gives the exact odds and expected breeds.

How it works

Let R be the number of IVs that still need to roll perfect after inheritance, and p the chance any one IV rolls max. The per-breed success chance is:

s          = p ^ R
expected   = 1 / s                  (mean of a geometric distribution)
P(≥1 in N) = 1 − (1 − s) ^ N

Inheritance matters enormously: every guaranteed-perfect IV removes one factor of p from s, so each extra inherited stat multiplies your per-breed odds by 1/p and divides expected attempts by the same.

Example and tips

Suppose you need 5 perfect IVs, 2 are guaranteed-inherited, and each random IV has a 1/32 chance to hit max. Then R = 3 and s = (1/32)³ ≈ 0.0000305, giving about 32,768 expected breeds — clearly impractical. Inheriting a third stat drops R to 2 and cuts expected breeds to about 1,024. The lesson: chase inheritance first, brute-force RNG last.