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# What would the chance of getting a Rare Candy with Pickup?

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I am planning to farm some Rare Candy and thought Pickup would be a great way of doing so. According to Bulbapedia, at level 51+ in S/M, there's an 8% chance of picking one up after applying the 10% chance of getting anything at all. If I'm planning on using 6 Pokémon with Pickup, what would the chance of one holding a Rare Candy after a battle? Would it be 1/8 x 1/10 = 1/80 x 6 Pokes = 3/40 battles resulting in the item?

EDIT: I just realized that I made a mistake when changing 8% into a fraction- it's 8/100, not 1/8. Anyway, the question's been answered (thanks, sumwun!) Feel free to answer if you've got a different way of doing it.

reshown

First, the math lesson
a% means a/100. For example, 8% means 8/100, or about 1/13.
p(a) means the probability that an outcome a happens. For example, p(pickup activating) is 10%.
If a and b are different possible outcomes of the same experiment, then p(a or b) = p(a)+p(b). For example, p(pickup activating or not activating) = p(pickup activating)+p(pickup not activating) = 10%+90% = 1.
The probability that an outcome a doesn't happen is 1-p(a). For example, p(pickup not activating) = 1-p(pickup activating) = 1-10% = 90%.
If a and b are possible outcomes of different experiments, then p(b) = p(a)*p(b given a happened). For example, p(finding rare candy) = p(pickup activating)*p(finding rare candy given pickup activating) = 10%*8% = 0.8%.
Two experiments are independent if their outcomes don't affect one another. For example, if you had two Pokemon with pickup, then the chances that the first finds an item and the chance that the second finds an item are independent.
If a and b are possible outcomes of two different independent experiments, then p(a and b) = p(a)*p(b). For example, p(both pickups activating) = p(first Pokemon's pickup activating)*p(second Pokemon's pickup activating) = 10%*10% = 1%.