Calculating a particular probability

Factual Questions
1

So I’m farming an enemy for loot in this game. The enemy normally drops crap, but there’s a chance it will drop rare items. Chances are 0.001 for rare armor, 0.001 for a rare weapon, and 0.007 for a rare rune. I would calculate this using a binomial trial formula. For 100 runs I get about a 55% chance that I get at least 1 item.

Great.

Two questions:

It so happens that this enemy actually drops 4 items each time it dies. Normally I would just say that I have four times as many trials, but I can equip some runes that create a 15% chance of dropping 8 items instead of 4. How on earth do I calculate now? At first I thought of making the number of trials some kind of expected value but that won’t work because of nCk uses only integers AFAIK.

It is possible that the game restricts itself to one rare item per drop, max. How does this affect the calculation? I can’t simply reduce “trial” to “kill” because it would be a new trial every time an individual piece of loot is created, but on a particular drop, if I do get a piece of loot then the rest of the loot may not matter and shouldn’t be counted.

2

Comparing

100% chance of dropping 4 items: 100% * 4 items = expected value of 4 items
85% chance of dropping 4 items and 15% chance of dropping 8 items: 85%*4 + 15%*8 = expected value of 4.6 items

so the runes give you an “extra” .6 items per kill.

Also, assuming success or failure of the three types of drops are calculated separately, more than one rare dropping simultaneously is very rare (compare .001 chance with .001 * .001 = .00001 chance), so the effect of the cap seems negligible to me.

3

The expected value thing doesn’t quite work here, as you’ve noted. Rather, suppose you kill N enemies, each of which has an 85% probability of dropping 4 items and a 15% chance of dropping 8 items. Then the cumulative probability of getting no items at all when killing N enemies will be

(probability of exactly 4M items being dropped)*(probability that 4M items contain no rare items)

summed over all M from N to 2N. (Note that the number of items dropped is always divisible by 4.) The probability of exactly 4M items being dropped after killing N enemies is (if I’ve done my math right)

(0.15)[sup]M - N[/sup](0.85)[sup]2N - M[/sup]( N choose (M - N) )

and the probability of 4M items containing no rare items (assuming all “drops” are independent) is (0.991)[sup]4M[/sup]. (The number 0.991 is the probability of a crap item being dropped, given the probabilities of the good stuff.) So you need to calculate the number

(0.991)[sup]4M[/sup](0.15)[sup]M - N[/sup](0.85)[sup]2N - M[/sup]*( N choose (M - N) )

summed over M from N to 2N. This is equivalent to

(0.991)[sup]4N[/sup] (0.85 - (0.991)[sup]4[/sup]*0.15)[sup]N[/sup]

without any summation necessary. (Expand out that second factor as a polynomial to see this.) You can then subtract this number from 1 to find the probability of getting at least one rare item. This gives a 64.6% chance of getting a rare item after killing 25 enemies, an 87.4% chance after 50 enemies, and a 98.4% chance after 100 enemies.

Whether or not this answer changes under a “no multiple rare drops” rule depends on how said rule is implemented. As long as the probability of 4 given items containing no rare items is still (0.991)[sup]4[/sup], then the above calculation still goes through. It’s easy to envision a software rule that would give a result like that — something like “if there are multiple rare items, replace all but one with crap”. The only wrinkle I can think of is when the enemy drops 8 items; are these regarded as a single set of 8 by the software, or as two sets of 4? If the latter, then the above result goes through; if the former, then the probabilities of getting rare items will be diminished somewhat (though, as iwakura43 points out, the effect would be very small.)

4

Very clear, thank you.