# What's the Expected Value of number of runs required for a hellfire ring?

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%`?

• Expected Value does not work like that. One should find the probability of a hellfire ring creation process turning successful, and go from there. Indeed, 1/p gives you in that case the number of runs to -almost- guarantee one ring, if you define, say, < 99% as "success". Commented Jun 18, 2013 at 18:45
• the `(1-10%)^10` that you are using is chance NOT to get a key/organ in 10 runs. The 10% on MP1 means that there is 90% chance of NOT getting the key/organ. So in order to find the chance of getting a key/organ in N runs on X MP, you should use the formula that you showed: `1-(1-X*10%)^N` with an exception when X = 0. So `1-(1-10%)^10 ~= 65%` because there is a 65% chance of getting a key/organ in 10 runs. Commented Jun 18, 2013 at 19:11
• btw, if you have 3 key sets, you can easily find people willing to carry you on MP10 for a set of organs, so you don't really have to try for organ parts on mp1.
– l I
Commented Jun 18, 2013 at 19:35
• The formula for expected value in the general case is Integral(0,n,P(x))=z, where P(x) is the probability function (.1), z is the target (1 key), and n is the expected value. Given that P(x) is a constant in this case, n=z/P() Commented Jun 19, 2013 at 13:47

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.

• I'm confused about where the 65% come from... apparently the limit as p approaches zero, the chance of having a success in the first 1/p trails approaches to 1-(1/e), or about 63% - my 65% is obviously approaching that, but that value seems so random. How do I make sense of the 65% and the EV of 10 runs? Commented Jun 18, 2013 at 19:18
• @user1032613: I hinted at that in my last paragraph, but a more in-depth answer is off-topic for this site. Please ask that on Math.SE and link to the question from here, and I will answer it there. Commented Jun 18, 2013 at 19:26
• Thanks very much for following site topic rules :) Have an upvote! (I posted here: math.stackexchange.com/questions/423923/…) Commented Jun 18, 2013 at 20:29

fbueckert's table still isn't quite right. Here's a fixed version:

``````+----+-------+----------+---------+--------+
| MP | 1 key | 1 portal | 1 organ | 1 ring |
+----+-------+----------+---------+--------+
|  0 |    20 |       60 |    1200 |   3600 |
|  1 |    10 |       30 |     300 |    900 |
|  2 |     5 |       15 |      75 |    225 |
|  3 |     4 |       10 |      34 |    100 |
|  4 |     3 |        8 |      19 |     57 |
|  5 |     2 |        6 |      12 |     36 |
|  6 |     2 |        5 |       9 |     25 |
|  7 |     2 |        5 |       7 |     19 |
|  8 |     2 |        4 |       5 |     15 |
|  9 |     2 |        4 |       4 |     12 |
| 10 |     1 |        3 |       3 |      9 |
+----+-------+----------+---------+--------+
``````

It looks like MP3 and MP4 never got the previous fix, while MP6 and MP7 were rounded down instead of up (except in the first column). Here's my work if anyone wants to check it:

``````+----+-------+----------+---------+--------+
| MP | 1 key | 1 portal | 1 organ | 1 ring |
+----+-------+----------+---------+--------+
|  0 | 20    | 60       | 1200    | 3600   |
|  1 | 10    | 30       | 300     | 900    |
|  2 | 5     | 15       | 75      | 225    |
|  3 | 10/3  | 10       | 100/3   | 100    |
|  4 | 5/2   | 15/2     | 75/4    | 225/4  |
|  5 | 2     | 6        | 12      | 36     |
|  6 | 5/3   | 5        | 25/3    | 25     |
|  7 | 10/7  | 30/7     | 300/49  | 900/49 |
|  8 | 5/4   | 15/4     | 75/16   | 225/16 |
|  9 | 10/9  | 10/3     | 100/27  | 100/9  |
| 10 | 1     | 3        | 3       | 9      |
+----+-------+----------+---------+--------+
``````

Sorry about answering instead of commenting. Apparently I need reputation to comment but not to answer. The MP6/7 differences are minor, but the MP3/4 differences are significant.

• Alright, have an upvote then! Commented Jul 3, 2013 at 2:34

As everyone pointed out, there is no way to guarantee a specific key drop. It gets pretty close, but it never quite gets there.

For expected value, it's just adding the chances. 10% chance of a drop means the expected value is 10. 20% means you the expected value is 5, etc.

``````Monster Power        1 key   1 portal   1 organ    1 ring
---------------------------------------------------------
No Monster Power (5%):  20         60     1,200     3,600
MP 1                    10         30       300       900
MP 2                     5         15        75       225
MP 3                     4         12        48       144
MP 4                     3          9        27        81
MP 5                     2          6        12        36
MP 6                     2          5         8        24
MP 7                     2          4         6        18
MP 8                     2          4         5        15
MP 9                     2          4         4        12
MP 10                    1          3         3         9
``````

This makes the assumption that you do all your runs at the exact same MP. I rounded up to the nearest whole value.

• what threshold are you using for "guarantee"? Commented Jun 18, 2013 at 18:31
• -1 there is no number of runs that will "guarantee" a drop of anything. Since the chances of dropping a key in one run is 1/10 at MP1, the expected number of runs for one key is 10. Where are you getting "94" from!? Commented Jun 18, 2013 at 18:43
• @fbueckert Okay, I think you got it correct assuming the EV is 1/p. Although I don't think the "EV stays constant" from MP5-MP10 is true. Rounding errors? Commented Jun 18, 2013 at 19:09
• @fbueckert although it's true that you can't do a half run, doing 36 runs on MP5 and on MP10 definite will give different chances of getting keys. Nothing is guaranteed to drop no matter how much you play. And no thanks, I don't want you to add in decimals just to make me happy. StackEx is for discussions; I'm not paying you money to boss you around. Thanks for all the help tho =) Commented Jun 18, 2013 at 22:28
• It is not "pure math" vs. impure math, it is simply correct vs. incorrect. You cannot do 1.6 runs any more than the average family can have 1.6 children; but you cannot round that up/down in the middle of a calculation, or you will find that the expected number of total children in 10 families is 10 or 20, rather than the correct answer of 16. In the same way, by rounding before multiplying you find that the expected number of runs to get a ring is the same for MP5 and MP9, which is obviously incorrect. Commented Jun 19, 2013 at 15:59