scoobmx

10-25-2012, 01:07 PM

In another thread I had made a suggestion that I think was largely ignored (maybe it wasn't really a suggestion as much as just analysis), but here goes again.

Hypothesis: It would be beneficial in quests where a low drop-rate item that is sought after potentially drops in many different chests (such as seals in certain quest series as well as named loot in Caught in the Web) for said item to be condensed down to dropping only in 1 chest with a drop-rate that equals the sum of the drop-rates of all the chests it could have dropped in.

Proof: Pulling loot in DDO follows a Bernoulli process, where for each chest that drops a given item, you have probability p of obtaining when you open it, as it calls a random number generator on the loot table. For n chests each having probability p of dropping this item, the chance of pulling exactly one of this item is:

C(n,1)*p*(1-p)^(n-1)

Where C(n,k) = n!/(k!*(n-k)!) and this follows the binomial distribution. The mean, representing the average number of said item dropping in one run through, is n*p. The variance is n*p*(1-p). Comparing 2 cases where we keep the mean n*p constant, but change the number of chests n, we can show that the more chests there are, the higher variance loot drops are:

1 chest with rate p: mean is p and variance is p*(1-p)

2 chests with rate p/2 for each: mean is again p but variance is p*(1-p/2)

Continuing: since n*p is constant, the only part of variance that changes is the (1-p) factor, which continues to increase.

So the higher number of chests there are, keeping the mean drop rate constant, the more the game punishes unlucky people and rewards lucky people.

For reference, in a realistic case, you wouldn't mind pulling multiples either, so the probability of pulling AT LEAST one of said item would be:

1 - n*Integral(t^(n-1),t,0,1-p)

Hypothesis: It would be beneficial in quests where a low drop-rate item that is sought after potentially drops in many different chests (such as seals in certain quest series as well as named loot in Caught in the Web) for said item to be condensed down to dropping only in 1 chest with a drop-rate that equals the sum of the drop-rates of all the chests it could have dropped in.

Proof: Pulling loot in DDO follows a Bernoulli process, where for each chest that drops a given item, you have probability p of obtaining when you open it, as it calls a random number generator on the loot table. For n chests each having probability p of dropping this item, the chance of pulling exactly one of this item is:

C(n,1)*p*(1-p)^(n-1)

Where C(n,k) = n!/(k!*(n-k)!) and this follows the binomial distribution. The mean, representing the average number of said item dropping in one run through, is n*p. The variance is n*p*(1-p). Comparing 2 cases where we keep the mean n*p constant, but change the number of chests n, we can show that the more chests there are, the higher variance loot drops are:

1 chest with rate p: mean is p and variance is p*(1-p)

2 chests with rate p/2 for each: mean is again p but variance is p*(1-p/2)

Continuing: since n*p is constant, the only part of variance that changes is the (1-p) factor, which continues to increase.

So the higher number of chests there are, keeping the mean drop rate constant, the more the game punishes unlucky people and rewards lucky people.

For reference, in a realistic case, you wouldn't mind pulling multiples either, so the probability of pulling AT LEAST one of said item would be:

1 - n*Integral(t^(n-1),t,0,1-p)