discuss:
The exact formula is as follows:
Where
* HPmax is the number of hit points the Pokémon has at full health,
* HPcurrent is the number of hit points the Pokémon has at the moment,
* rate is the catch rate of the Pokémon,
* bonusball is the multiplier for the Poké Ball used, and
* bonusstatus is the multiplier for any status ailment the Pokémon has (2 for sleep and freeze, 1.5 for paralyze, poison and burn).
Given this formula, the maximum value for a (if the Pokémon could have 0 HP) would be catch rate × bonusball × bonusstatus. The minimum value for a (for a Pokémon with full health) would be ⅓ × catch rate.
If a is greater than or equal to 255, then the Pokémon is caught. If not, then calculate b as follows:
Then generate 4 random numbers between 0 and 65535, inclusive. If the random numbers are less than or equal to b, then the Pokémon is caught; otherwise the ball shakes n times, where n is the number of random numbers that are less than b. Note that b ≥ 65535 if a ≥ 255.
Therefore, the probability p of catching a Pokémon, given the values a and b calculated above, is:
The second expression for p may be expanded as follows:
Since (216 - 1)4 ≈ 264, we can approximate p with the following expression:
The percentage error in this approximation approaches 0 as a approaches 255, and does not exceed 0.02%.
For a constant probability p, the probability P that a player can capture the Pokémon with no more than r tries is:
Note that this is the cumulative probability function for a geometric distribution. The expected value of r is 1/p, that is to say, on average, a Pokémon that can be caught with probability p will be caught with 1/p tries.
The inverse problem, the number of tries, r, needed to have a probability P of capturing a Pokémon is:
Reply
