Guide Forgery Chest Mythic Probability

Motivation
Now, you may be thinking, hasn't this been done before? And you'd be right - there are (as far as I'm aware) two posts that address the question of the expected number of Mythics from the Forgery chest after a certain number of runs. However, both posts make the same error, so I thought I'd put my response to them in a new post.

For formality's sake, let n be the number of complete (all dungeons) Forgery runs, and p the probability of getting a Mythic from a complete Forgery run chest (p = 0.01).
In essence, they start by stating that the probability of not getting a Mythic in any chest is equal to (1-p)^n = 0.99^n, which is true. Then, the probability of getting one Mythic is defined as the complement of that, being 1 - 0.99^n. However, this is actually the probability of getting one Mythic or more.

So, what number of Mythics can you actually expect after n runs?

Method & Results
Let's denote the random variable associated with the number of Mythics obtained as X. We can gain some insight from the Probability Mass Function (PMF) of X. Basically, you can feed the number of Mythics (given a certain number of runs) to the PMF, and it will tell you how probable it is for that to occur.

X follows a Binomial distribution; its PMF takes the form of
.

For a Binomial distribution, the expected value (thus mean number of Mythics) can easily be found through n*p. Thus, after 100 runs, you'll have gained 1 Mythic on average. However, note that at 100 runs, there is still a 36.6% chance of 0 Mythics, and a 26.4% chance of 2 or more Mythics.

In the figure below, the probability of getting a certain number of Mythics is shown for the cases of 50, 100, 200 and 300 runs (as also considered by the most recent post).​

​

Conclusion
In short, the number of Mythics you can expect after n complete Forgery runs is equal to n*p, where p = 0.01. Based on this, you can estimate the cost per Mythic as (n*[run cost])/(n*p) = [run cost]/p and determine whether running the Forgery for Mythics is worthwhile, but that's somewhat outside of the scope of this post. In any case, thank you for reading :)

 

Hmmm interesting.

 

I can't be the only who saw the words Binomial Distribution, saw a graph and was "Idk what's going on but I'm sure he's correct"

 

Ah, Binomial Distribution, from calling Dream out to solving if Forgery is worth farming for Mythics, there's nothing it cannot do.

 

This was on the test I failed a few months ago

 

idk what this says but its telling me i can get mythics and thats good enough for me! very interesting though

 

Sorry I didn't really go into what a Binomial distribution is, though as has been mentioned -

There are quite some videos that explain what a Binomial distribution is and how it relates to the Dream cheating accusations (e.g. https://www.youtube.com/watch?v=rbhx3UNpV64), so feel free to check that out if you're interested.

 

I mean, in my post explaining how to get the chance on the one thread it wasn't an error, it was intentional. If I phrased it incorrectly that's on me, my bad (did I? I don't particularly remember) but I just didn't really think that most people would care about the specific chance of getting 2, 3 etc. so I just did the chance of getting one or more.
Plus, the chance of getting multiple are rather small especially when compounded with wynn's active player base (and considering how many are the available level, and will actually run forgery chest that many times) it likely won't happen to any player for a long time

also this video is epic would recommend. I used it as a guide for myself the other day cause I was pissed about some rates in a game

 

That’s a dream epic exposed word so I’m sure they’re fine

 

I actually know binomial distribution cuz Im taking stats but doing math outside of class hurts my brain

 

isnt 50 runs a lot?

 

Without dungeon bombs? Yes
With dungeon bombs? Yesish

 

50 runs of forgery chests be like 350 dungeon runs, assuming that you do a run per 6 mins you spend 2100 mins, which is 1.46 (1.458333...) days worth of time of doing dungeons. Of course we consider 200+ runs because more chance of mythic (and it is near statistically impossible for no mythic). You have to farm 5.833 days straight of it, with keys or dungeon bombs and the chance of sudden restarts.

With that absurd amount of time I could go ingredient/emerald farming with totems and loot gear for money just to buy an unrolled mythic (i believe it is a thing). It may be slower, however this method is more stable and at least less boring for me.

(thats why I asked for a forgery chest rework in the other thread)

 

Tell me you used a graphing calculator and didn’t actually use the whole formula cause u can find the info in 2-var stats I think

 

My math wasn’t perfect in all fairness, and the mythic portion was more of a panicked endnote for the Main goal of the XP sums to get to 106. I was more concerned about whether or not you’d get a mythic than how many considering most of us have maybe seen 1 in our total playtime, less so of it’s viability. But still, good on you for showing this might be a marginally competitive strategy compared to loot running

 

It wasn't my intention at all to call anyone out with this thread - sorry for coming across a bit harsh. The difference isn't that significant when the number of runs is small, and indeed you need to spend a lot of time on Forgery runs for a decent chance at a Mythic. There were simply a few misleading statements in the original posts which I wanted to address - e.g. "33 STACKS of LE for a 95% chance to find ONE mythic", when on average you'll have gotten 3 Mythics by 300 runs (still not very cost-effective, but that's beside the point).

I just used the whole equation in MATLAB since that also allows me to create figures easily :)