MadScientist said: Never do simulations for something when you can calculate the probability directly.This problem is a model of "drawing from an urn without replacement". The number of items N=80 and the number of successes K=1 (the legendary token). The applicable probability distribution is the hypergeometric distribution. The mean number of successes after n tries is given by n*K/N, which is n/80 in our case.Strategy 1: Draw 10 tokens on 8 different vaults. The expected number of legendary tokens is 1/8 * 8 = 1.Strategy 2: Draw 80 tokens from one vault. The expected number of legendary tokens is 80/80 = 1.As you can see, the expected number of LTs is exactly the same for the two strategies.
IIAlonditeII said: MadScientist said: Never do simulations for something when you can calculate the probability directly.This problem is a model of "drawing from an urn without replacement". The number of items N=80 and the number of successes K=1 (the legendary token). The applicable probability distribution is the hypergeometric distribution. The mean number of successes after n tries is given by n*K/N, which is n/80 in our case.Strategy 1: Draw 10 tokens on 8 different vaults. The expected number of legendary tokens is 1/8 * 8 = 1.Strategy 2: Draw 80 tokens from one vault. The expected number of legendary tokens is 80/80 = 1.As you can see, the expected number of LTs is exactly the same for the two strategies. Except, the flaw with Strategy 1 is that you can draw... 0 in all 8 vaults. Strategy 2 cannot fail, and has the added benefit of leaving you with a well-stocked horde for the next vault in any event where you draw the token earlier than expected. IE, if you expected to draw it by 10 or so, you'd already have 70 tokens saved up for the next time.