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Vessel of Hatred

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@Luhkoh Yeah, I wondered about the 0,95% increase :D

I did not do this using the actual values to calculate the expected value but rather I treated the number of picks n, the max number of drops k and the chance to drop each pick p as variables in order to avoid making any mistakes along the way. Here's the formula for the expected value:

Code: 
exp. # drops = n*p*(1-p^(n-1))/(1-p^n)

Assuming I made no mistakes on my way there of course ;)

It works for Andariel too, her p is different from Mephisto's but she has k=6 and n=7 just like him.

I also have the chances for each # of drops per kill in this general format, let me know if you want those. Obviously they will yield the same results as malawi's calculation because everything interesting that Mephisto (or anyone for that matter) can drop is rare enough.

@drmalawi Lesson potentially learned :p
 
You guys should always include a TL;DR version for us less technical minded people ...
 
@art_vandelay this is how you do expected number of drops method.

Chance of drop per pick p1 is 81.5%
We have 6.707 effective picks on p1 (with the optional 7th pick)
so 6.707 * 0.815 = 5.466 drops per kill

And on p3 chance of drop per pick is 97.06%
We have 6.164 effective picks on p3 (with the optional 7th pick)
so 6.164 * 0.9706 = 5.983 drops per kill

Gain = 5.983/5.466 = 1.0945 i.e. 9.5% higher which is the same number @Luhkoh gave and that I gave in the longer alternative derivation above (which kinda shows that higher player setting gives higher chance per pick but also fewer picks in general, which is kinda interesting to see how those two factors interact so to say)
 
@T72on1 are the bullet points in my first post good enough? That was my attempt to give easy numbers to use. TLDR for me personally is p1 mephy should be "generally" recommended over p3.

edit: @art_vandelay I'm still not sure what 0.95% increase you mean
 
drmalawi said:
@art_vandelay this is how you do expected number of drops method.

Chance of drop per pick p1 is 81.5%
We have 6.707 effective picks on p1 (with the optional 7th pick)
so 6.707 * 0.815 = 5.466 drops per kill

And on p3 chance of drop per pick is 97.06%
We have 6.164 effective picks on p3 (with the optional 7th pick)
so 6.164 * 0.9706 = 5.983 drops per kill

Gain = 5.983/5.466 = 1.0945 i.e. 9.5% higher which is the same number @Luhkoh gave and that I gave in the longer alternative derivation above (which kinda shows that higher player setting gives higher chance per pick but also fewer picks in general, which is kinda interesting to see how those two factors interact so to say)
Click to expand...

What I did is I looked at all possible outcomes for a drop from Mephisto and assigned them a weight Probability(drop)^#drops*Probability(nodrop)^#nodrops and then I added all up and normalized each weight by that value. This is how I got all the probabilities I'm using. Right now I'm not sure wether the normalizing was correct... I'll have to look at it again.

I calculate expected value pretty much the only way I know how: by definition adding up all the probabilites*values

Can you tell me which step was wrong?


drmalawi said:
There is no one claiming 0.95% increase. 1.095 is 9.5% increase.
Click to expand...

I'm very much aware of that, no need to be condescending ;)
 
@art_vandelay I havent investigated the specifics, but I would guess it has to do with the "weighted averaging" you mentioned. It often doesn't work in situations when I would think it would. This was something I kept running into doing the countess rune odds and ubeogesh kept showing me that was wrong. I don't know if it's the same situation here, but I hope that general point is helpful.
 
@Luhkoh That's maybe it. Hard to say right now as I didn't exactly make a lot of notes back when I did those calculations. :confused:

And I know that his way works. I would rather know the exact reason why mine doesn't. Being presented the right answer without a reasoning why mine is wrong is not really helpful.

Top down education sucked a decade ago and it doesn't feel much nicer today :D


drmalawi said:
No I do not have time for that unfortunately.

Have a study of my two methods and I am sure you can figure it out yourself eventually.
Click to expand...

Thanks for the confidence Dr. Malawi, maybe I do have it in me to understand it after all! :):):)


***


Anyway in an effort not to derail this: Only very recently did I find out that the statues by the RoF waypoint actually turn their heads towards you as you pass them o_O
 
art_vandelay said:
And I know that his way works. I would rather know the exact reason why mine doesn't. Being presented the right answer without a reasoning why mine is wrong is not really helpful.

Top down education sucked a decade ago and it doesn't feel much nicer today :D
Click to expand...

To be fair, I do not get paid for explaining maths on a Diablo forum :p

And sometimes, the best reply to "why is this not correct" is simply "just because it isn't".
Like why isn't the derivative of f(x) = e^(-3x) equal to sin(x)? Well, because it is not correct.

This is the way of doing it: Expected number of successes = number of trials times the chance for success per trial.
I.e. basic probability, just as you would have treated expected number of "6" when you roll a regular dice x number of times.
 
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@art_vandelay It's just a lot of effort to use your formula and figure out exactly what it assumes and where the incorrect reasoning is, at least for me. I may get interested and dig into it at some point (but probably not today), and if I do I will definitely message you.

Anyways sorry if it got condescending. For me, i was generally confused and frustrated with the 1% thing, and I legitimately thought it was unclear that you were no longer claiming I used that number when you responded that you were wondering about it. It's frustrating to get corrected for saying something you didn't say.

But anyways I hope the ratios are generally helpful, they gave me a serious new conclusion (that I'd wasted a fair amount of time on p3 mephy :rolleyes:)

And very cool on the statues. I often find I have not been appreciating the scenery in the game as much as it deserves. It's a really good looking game!
 
drmalawi said:
To be fair, I do not get paid for explaining maths on a Diablo forum :p

And sometimes, the best reply to "why is this not correct" is simply "just because it isn't".
Like why isn't the derivative of f(x) = e^(-3x) equal to sin(x)? Well, because it is not correct.

This is the way of doing it: Expected number of successes = number of trials times the chance for success per trial.
I.e. basic probability, just as you would have treated expected number of "6" when you roll a regular dice x number of times.
Click to expand...

And no one forces you to do so ;)

I'm of the opinion that wrong answers are a valuable tool for understanding how something works. For example I just learned why my model didn't work out: I took something like 'drop drop drop drop drop drop nodrop' as a possible drop from Mephisto and wrongly thought the probability for this drop should be P(drop)^6*P(drop) which was wrong because the drop algorithm stops after 6 drops and so the last factor needs to be omitted.

And true to my namesake I'm going to redo my way out of spite until I get the same numbers as you. I'm sorry though that I was being sarcastic above! :)
 
art_vandelay said:
And no one forces you to do so ;)

I'm of the opinion that wrong answers are a valuable tool for understanding how something works. For example I just learned why my model didn't work out: I took something like 'drop drop drop drop drop drop nodrop' as a possible drop from Mephisto and wrongly thought the probability for this drop should be P(drop)^6*P(drop) which was wrong because the drop algorithm stops after 6 drops and so the last factor needs to be omitted.

And true to my namesake I'm going to redo my way out of spite until I get the same numbers as you. I'm sorry though that I was being sarcastic above! :)
Click to expand...

Probability for number of items dropped per kill btw is calculated with the Binomial formula plus clever inclusion of the optional 7th pick. It is actually a nice problem in its own right and not as trivial as one might expect.

Much easier just to simulate this with a RNG if you know some basic programming.
 
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@drmalawi I just finished redoing my way and the formula I got agrees with your values for p1/p3 Mephisto. I got this result:
Code: 
E(#drops)=6p+p(1-p^6)
where p is chance to drop per pick.
This is the same thing as (your approximation of effective picks)*(chance to drop per pick). So I guess it doesn't matter at all wether chance of nodrop is small. (because I didn't use any approximations and assumed p was an arbitrary probability)


Can you say which terms you dropped from the Taylor expansion in this part:
Chance of no drop per pick is calculated as P_0 = No Drop /(No Drop + Prob sum) = 18.5% (p1), 2.94% (p3), 0% (p5+)

Chance of 7th pick is calculated as "chance getting at least one no drop in 6 picks" : P_7 = 1-(1-P_0)^6 = 70.7% (p1), 16.4% (p3), 0% (p5+)

Effective number of picks is 6 + P_7 = 6.707 (p1), 6.164 (p3), 6 (p5+)
(can be derived using Taylor expansion since we assume the drop chance P_0 is small)
Click to expand...


***

In any case I'm frustrated that doing it without approximations takes three times as long and literally yields the same result...
 
art_vandelay said:
Can you say which terms you dropped from the Taylor expansion in this part:
Click to expand...

(1-x)^y = 1-y*x if x is small.

But I put the stuff about taylor expansion trick in the wrong place actually.
It should only be where I put in under "II)" I forgot to remove it from the place you quoted it ;)
I have changed my original text now.

your formula is correct.

I give you C+ on this assignment ;)

But in general, using Taylor is kinda neat when one deals with very small probabilities and want to have compressed expressions.
E.g. we can calculate drop chance per pick of say Jah rune from Meph.
We wanna know the drop chance per kill to get at least one Jah rune.
But, in practice, getting 2 or more Jah runes is gonna be insanely small compared to dropping one.
Let's use 6 picks now only, we want to calculate 1-(1-p_Jah)^6, where p_Jah is drop chance per pick of Jah rune, but it is easier just to give/use 6*p_Jah which comes from "Tayloring" the previous expression.

Let's take use p_Jah = 0.000003 (think its in the correct "ball park")

1-(1-p_Jah)^6 = 1.79998*10^-5
6*p_Jah = 1.8*10^-5

The error we make by "Tayloring" is 2*10^-15 ...
 
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Luhkoh said:
Nice! Sounds very promising. Even more so if lvl 2 starts coughing up more packs. Afaik level 1 and 2 are two completely different areas.
Click to expand...

Did 50 runs in total counting monsters. 7.7 boss packs on average, compact map. Yep, definitely a keeper. Only got 5 packs as lowest one time, but maybe I just missed a pack there. Highest was 1x 11 and 2x 10. TooMuchEnergy will stick to Pits for quite some time I guess :).
 
@drmalawi I messed around a bit with Duriel's drops and I'd like to hear your input on it. To calculate item chances, like you did, the chance for at least one item to drop is needed. So I calculated the chance for Duriel to drop 0 items from his 'Duriel Base' TC for a given kill. Here are the details:

So Duriel's TC looks like this:
View attachment 14395
For simplicity I'm going to write b(k) for the event that 'Duriel - Base' is picked and dropped k items (k is a number from 0,1,2,3,4,5,6) and 't' for 'tsc' (scroll of town portal). I also write p for the chance to drop for each pick performed by 'Duriel - Base'.

So the chance to drop 0 items from the 'Duriel Base' TC is given by:
P(0 items from his base TC)=Sum_(k=0 to 5) P(t occurs 5-k times AND b(0) occurs k times)

Now I represent any one Duriel kill by writing down an ordered list consisting of t's and k's. For example:
t 0 t t 0
means three tsc's and two 'Duriel - Base' picks with 0 drops in that order.

Since we are only interested in the case where he drops 0 items from his base TC we won't have to deal with any case where the 'Duriel' TC terminates before the 5th step. So something like:
t t 3 t
can't happen. So the Duriel kills in our case (0 items) will be a list of five 0's and t's.

For a given number k of there's only binom(5,k) many possibilities to make such a list as above consisting of k times '0' and 5-k times 't'. So we get:
P(t occurs 5-k times AND b(0) occurs k times)=binom(5,k) * (1/3)^(5-k) (2/3)^k * (1-p)^(7*k) = (1/3)^5 * binom(5,k) * (2*(1-p)^7)^k

Summing these up for k=0,...,5 and using the binomial formula I arrive at:
P(0 items from his base TC) = ((1+2*(1-p)^7)/3)^5

Finally the actual chances:
View attachment 14397
The chance of 0 items becomes 1 over 243 as p becomes 1.

Throughout all of this I treated the tsc from the 'Duriel' TC and the scrolls you can get from 'Duriel - Base' as separate items. So if you multiply the 1-P(0 items) chance by the chance to drop a tsc once you're in the base TC you will not get the correct chance to drop at least one tsc for Duriel. This is pretty much irrelevant as the true prize is quest Duriel who can't drop tsc from his base TC.

The expected number of drops per kill coming from his base TC is still eluding me. However I was able to compute P(1 item from his base TC).
 
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Before I bank all my gold (and potentially lose it all), I just wanted to make sure that the gold stored in the Gomule 'bank' off to the right under the 'Clipboard' area will not disappear. It seems straightforward (it's called bank, after all :rolleyes:) but I just wanted to confirm. I've never had to bank gold before, and I was assuming that I would have to put it in a stash...but I guess not?

Thanks.
 
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