Diablo – What’s the Expected Value of number of runs required for a hellfire ring

diablo-3

On Monster Power 1, the drop rate for each key 10%. If I do 10 runs, the chance of having one key dropped in these 10 runs is 1-(1-10%)^10, or about 65%. After getting 3 keys, I can try to get a loot from a uber with a chance of 10%. I will need 3 loot from uber bosses to craft one ring. How do I calculate the Expected Value of total number of runs I will need?

Edit: from wikipedia, I just learned the Mean of a Geometric Distribution is 1/p. So the Expected Value of getting a key drop on Monster Power 1 should be 10 runs. But then, if 10 is the mean, why would 1-(1-10%)^10 ~= 65%?

Best Answer

Your formula is correct: if the chances of getting a ring in one run are 1/10, then the expected number of runs is 10. And if that is the case, then the expected number of runs to get 3 rings is 3*10=30 runs.

Once you have all three keys, you craft them into a portal, which you use to get a chance at one of three organs. Assuming these organs also have a 10% drop rate, that means you'll need an average of 10 portals ( = 30 keys = 300 key-runs) per organ. Since you need 3 organs, your overall expected number of runs is 900 for all the keys you need, plus another 30 to get the organs, for a total of 930 (at MP1).

In general, if k is the probability of a key dropping and o is the probability of an organ dropping, your expected number of runs (including both keys and organs) will be 9/po + 3/o.


Note that the error in your logic is that you appear to be assuming the expected value is the same as the point at which you have 50% chance of succeeding - it's not. Take coin-flips for example; you have a 50% chance of getting heads after only one flip, but the expected number of flips to see a heads is 2.