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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%.

Now, back to the answer
There are 6 ways to find exactly one rare candy after a battle. You can have your first Pokemon find one and the rest not find one. You can have your second Pokemon find one and the rest not find one, etc. Because each pickup is independent of the others, the chance of each way happening is 0.8%*(1-0.8%)*(1-0.8%)*(1-0.8%)*(1-0.8%)*(1-0.8%), which is about 1/130. Because there are 6 ways to find exactly one rare candy, the chance of finding exactly one rare candy is 6 times that expression, or about 1/22.
The chance of finding at least one rare candy after a battle is 1-p(finding no rare candies). Finding no rare candies means all 6 of your Pokemon didn't find a rare candy. Therefore, the chance of finding at least one rare candy is 1-(1-0.8%)*(1-0.8%)*(1-0.8%)*(1-0.8%)*(1-0.8%)*(1-0.8%), or about 1/21.
The average number of rare candies you'll find after one battle is the sum of the average number of rare candies that each of your Pokemon finds after one battle. You have 6 Pokemon, and each of them finds, on average, 0.8% of a rare candy after a battle. Therefore, the average number of rare candies you'll find is 6 times that expression, or about 1/21.
Note that the second and third answers aren't exactly the same even though they're very close to each other.

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I just realized that I misinterpreted 8% as 1/8- I guess I was used to 10% being 1/10. I'm in Geometry and still made that mistake... Well, anyway, thanks for answering (and the lengthy talk on probability -lol). I never thought about it that way! :)

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