- Jan 3, 2019
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[reposted from July 12, 2019]
Hi all. I have seen countess rune drop chances for past versions of the game posted here, but never have seen an update for 1.13b through 1.14d. This is probably because the runes are so bountiful from countess and LK that people either don't care as much about the numbers, or just head to LK. But I thought it was still info that I wanted to know, so I calc'd her updated chances, which is an interesting math problem.
The same data is reflected in my general runefinding calculator HERE. But I wanted to generate some tables and post the countess's data separately. Additionally, @drmalawi brought to our attention some of the shortcomings in the methodology previously used for "time to find or cube a rune". I wanted to expound on this a little further in this sheet as well. More discussion and explanation on this will be in the "Cubing Math" section below.
And also, big thank you to ubeogesh (reddit/discord user) who helped me a ton in calculating these, and indeed had come to same direct drop odds before me.
The Sheet
Link is HERE
The Tables
These tables are also available in the spreadsheet linked above, and perhaps are easier to read in the sheet.
Average runs for direct drop of rune
Runs to find or cube at least 1 of rune with 63.2% confidence level
Note: Cubing is considered for the runes 2 below the one in question.
Long term runs per cubed rune (extremely high number of runs done)
Note: Cubing is considered for the runes 2 below the one in question.
Background Info
The countess is particularly interesting when calculating rune drop chances. In fact, other than ubeogesh's own spreadsheet, I have not seen correct odds for 1.13b+ anywhere. The german drop calculator is reasonably close, but still pretty far off. The reason it is so difficult is because the countess drops from two TC's: Countess Item, and Countess Rune. She attempts to drop an item 5 times, then attempts to drop a rune UP TO 3 times. Any monster in D2 is limited to 6 drops. So if countess successfully drops 5 items, then she will only attempt to drop 1 rune, but if she drops 3 or less items (due to nodrop) then she will attempt to drop 3 runes. This is what makes countess drop more runes on p1 than on p7 (almost double the amount).
The nature of these conditional drop attempts is what makes the math so challenging, and the drop calculators so incorrect. One further wrinkle is when countess drops from her ITEM TC, that item CAN be a rune. This TC can drop up to Lo, whereas the rune TC can only drop up to Ist. The chances of a good rune coming from her item TC are very small, and should not be considered for running purposes (meaning don't run countess hoping for her to drop that Lo). However it does make a difference in her overall odds, and this is included in the sheet and in the tables.
Cubing Math
TLDR: If you want to see how many runs you might expect to do to find or cube a rune from countess, use the "Runs to find or cube at least 1 of rune with 63.2% confidence level" numbers.
If you want to do like 2000+ runs and want to know how many Ists you can hope to cube at the end of the runs, use the "Long term runs per cubed rune" numbers.
Longer Version:
As I mentioned in the first paragraph, the way we had calculated time to cube or find a rune in the past was a bit flawed. @drmalawi explains all this, as well as a nice method for correct calculation in his youtube video HERE (seems he has now set this to private).
However the answer obtained from this old method of calculating time to find or cube DOES accurately calculate one thing. It correctly shows the average time to find or cube a rune over infinitely many runs. I will try and explain why this is an important distinction. Let's say you want to cube an Ist from the countess. You drop a Mal rune. Then a few runs later you drop an Um rune. Then several runs later you drop an Ist directly. Well the Mal and Um did not help you at all on your way to that single Ist. However if you were doing a very large amount of runs to cube several Ists, then that Mal and that Um would indeed get used to cube an Ist eventually. So basically if you do enough runs, you "waste" less runes, because you get to cube them, and your average runs per cubed Ist goes down.
Ok so to sort this out a bit, I provided the numbers for time to find or cube at least 1 rune with 63.2% confidence level. Confidence level (CL) just means "this percent chance you will succeed". I used 63.2% because this is often the CL we associate with "average" in D2. Most all single drops in D2 follow a binomial probability distribution, and in this distribution, "average" is the same as 63.2% CL. However, "time to find or cube at least one of a rune" does not have a binomial probability distribution, so I cannot say that 63.2% CL is average. The true average is tough to calculate and, in my opinion, is not helpful anyways for this case. 63.2% CL provides the same chance to succeed as "average" in other cases, so I thought it was a nice number to use here.
Finally, what if I want to know the time to find or cube 2 Ists? Or 3 Ists? Or 6 for that pitzerker phase blade? Well you COULD use malawi's complementary probability method for more runes, but the number of combinations you have to calc gets very big. At some high number of Ists the "long term average runs" becomes very accurate, but I don't know how many Ists that requires. I also don't plan to try and figure it out
. I would say to myself "well we know it's between 176 and 136 for expectation." And I would be fine with that. Maybe if you want 6 Ists for the phase blade, shoot for something like 150x6 = 900 runs. After all, it is only an average or 63.2% CL. It very well may take many more runs, or many less. But I do like seeing the numbers and getting a practical idea of what to expect. If someone else digs further into that math, I would of course enjoy seeing it, but don't deem it of enough value to warrant a lot of effort.
Anyways that's it! Hope the sheet and tables are helpful and/or interesting. And of course any discussion or questions are welcome!
Hi all. I have seen countess rune drop chances for past versions of the game posted here, but never have seen an update for 1.13b through 1.14d. This is probably because the runes are so bountiful from countess and LK that people either don't care as much about the numbers, or just head to LK. But I thought it was still info that I wanted to know, so I calc'd her updated chances, which is an interesting math problem.
The same data is reflected in my general runefinding calculator HERE. But I wanted to generate some tables and post the countess's data separately. Additionally, @drmalawi brought to our attention some of the shortcomings in the methodology previously used for "time to find or cube a rune". I wanted to expound on this a little further in this sheet as well. More discussion and explanation on this will be in the "Cubing Math" section below.
And also, big thank you to ubeogesh (reddit/discord user) who helped me a ton in calculating these, and indeed had come to same direct drop odds before me.
The Sheet
Link is HERE
The Tables
These tables are also available in the spreadsheet linked above, and perhaps are easier to read in the sheet.
Average runs for direct drop of rune
Rich (BB code):
RUNE Players 1/2 Players 3/4 Players 5/6 Players 7/8
-------- ------------ ------------ ------------ ------------
Lum 34 46 56 61
Ko 51 69 84 92
Fal 66 89 109 119
Lem 99 133 163 179
Pul 131 175 215 235
Um 196 262 322 353
Mal 192 257 316 346
Ist 288 386 473 519
Gul 29745 18070 16082 15445
Vex 44617 27104 24123 23168
Ohm 42353 25729 22899 21992
Lo 63530 38593 34348 32988Runs to find or cube at least 1 of rune with 63.2% confidence level
Note: Cubing is considered for the runes 2 below the one in question.
Rich (BB code):
RUNE Players 1/2 Players 3/4 Players 5/6 Players 7/8
-------- ------------ ------------ ------------ ------------
Fal 56 75 92 101
Lem 81 108 133 145
Pul 110 147 180 197
Um 126 169 207 227
Mal 135 181 222 243
Ist 176 236 289 317
Gul 382 507 618 674Long term runs per cubed rune (extremely high number of runs done)
Note: Cubing is considered for the runes 2 below the one in question.
Rich (BB code):
RUNE Players 1/2 Players 3/4 Players 5/6 Players 7/8
-------- ------------ ------------ ------------ ------------
Fal 41 54 66 73
Lem 58 78 95 104
Pul 79 106 130 142
Um 95 127 155 170
Mal 104 139 170 186
Ist 136 183 224 245
Gul 326 431 524 571Background Info
The countess is particularly interesting when calculating rune drop chances. In fact, other than ubeogesh's own spreadsheet, I have not seen correct odds for 1.13b+ anywhere. The german drop calculator is reasonably close, but still pretty far off. The reason it is so difficult is because the countess drops from two TC's: Countess Item, and Countess Rune. She attempts to drop an item 5 times, then attempts to drop a rune UP TO 3 times. Any monster in D2 is limited to 6 drops. So if countess successfully drops 5 items, then she will only attempt to drop 1 rune, but if she drops 3 or less items (due to nodrop) then she will attempt to drop 3 runes. This is what makes countess drop more runes on p1 than on p7 (almost double the amount).
The nature of these conditional drop attempts is what makes the math so challenging, and the drop calculators so incorrect. One further wrinkle is when countess drops from her ITEM TC, that item CAN be a rune. This TC can drop up to Lo, whereas the rune TC can only drop up to Ist. The chances of a good rune coming from her item TC are very small, and should not be considered for running purposes (meaning don't run countess hoping for her to drop that Lo). However it does make a difference in her overall odds, and this is included in the sheet and in the tables.
Cubing Math
TLDR: If you want to see how many runs you might expect to do to find or cube a rune from countess, use the "Runs to find or cube at least 1 of rune with 63.2% confidence level" numbers.
If you want to do like 2000+ runs and want to know how many Ists you can hope to cube at the end of the runs, use the "Long term runs per cubed rune" numbers.
Longer Version:
As I mentioned in the first paragraph, the way we had calculated time to cube or find a rune in the past was a bit flawed. @drmalawi explains all this, as well as a nice method for correct calculation in his youtube video HERE (seems he has now set this to private).
However the answer obtained from this old method of calculating time to find or cube DOES accurately calculate one thing. It correctly shows the average time to find or cube a rune over infinitely many runs. I will try and explain why this is an important distinction. Let's say you want to cube an Ist from the countess. You drop a Mal rune. Then a few runs later you drop an Um rune. Then several runs later you drop an Ist directly. Well the Mal and Um did not help you at all on your way to that single Ist. However if you were doing a very large amount of runs to cube several Ists, then that Mal and that Um would indeed get used to cube an Ist eventually. So basically if you do enough runs, you "waste" less runes, because you get to cube them, and your average runs per cubed Ist goes down.
Ok so to sort this out a bit, I provided the numbers for time to find or cube at least 1 rune with 63.2% confidence level. Confidence level (CL) just means "this percent chance you will succeed". I used 63.2% because this is often the CL we associate with "average" in D2. Most all single drops in D2 follow a binomial probability distribution, and in this distribution, "average" is the same as 63.2% CL. However, "time to find or cube at least one of a rune" does not have a binomial probability distribution, so I cannot say that 63.2% CL is average. The true average is tough to calculate and, in my opinion, is not helpful anyways for this case. 63.2% CL provides the same chance to succeed as "average" in other cases, so I thought it was a nice number to use here.
Finally, what if I want to know the time to find or cube 2 Ists? Or 3 Ists? Or 6 for that pitzerker phase blade? Well you COULD use malawi's complementary probability method for more runes, but the number of combinations you have to calc gets very big. At some high number of Ists the "long term average runs" becomes very accurate, but I don't know how many Ists that requires. I also don't plan to try and figure it out
. I would say to myself "well we know it's between 176 and 136 for expectation." And I would be fine with that. Maybe if you want 6 Ists for the phase blade, shoot for something like 150x6 = 900 runs. After all, it is only an average or 63.2% CL. It very well may take many more runs, or many less. But I do like seeing the numbers and getting a practical idea of what to expect. If someone else digs further into that math, I would of course enjoy seeing it, but don't deem it of enough value to warrant a lot of effort.Anyways that's it! Hope the sheet and tables are helpful and/or interesting. And of course any discussion or questions are welcome!
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