Average AC of monsters per CR?

In D&D 4e, you could tell most monster’s AC just from its level and role (Brute, Skirmisher, Soldier).

This is not the case in 5e, but an estimation would be very useful when you try to compare builds.
My group uses 13 + proficiency bonus, this has the benefit that you can completely remove the proficiency bonus from the calculation. However, this is a bit high at the beginning, and low at the end.

  • Does someone have a better formula?
  • Better yet, has anyone found a compilation of existing monsters, that could be used to calculate this for myself?

Number of subarrays of size k or less are encountered during Quicksort on average?

I am reading the book Analysis of Algorithms. The focus of chapter 1 is to analyze the running time of the original Quicksort algorithm. The code of this algorithm is given in the book.

public class Quick {    private static int partition(Comparable[] a, int lo, int hi)    {       int i = lo, j = hi+1;       while (true)       {          while (less(a[++i], a[lo])) if (i == hi) break;          while (less(a[lo], a[--j])) if (j == lo) break;          if (i >= j) break;          exch(a, i, j);       }       exch(a, lo, j);       return j;    }     private static void sort(Comparable[] a, int lo, int hi)    {       if (hi <= lo) return;       int j = partition(a, lo, hi);       sort(a, lo, j-1);       sort(a, j+1, hi);    } }  

There is an exercise at the end of chapter 1 asking the number of subarrays with size k or less.

How many subarrays of size k or less (not counting subarrays of size 0 or less) are  encountered, on the average, when sorting a random file of size N with quicksort? 

For simplicity, I will only count the subarrays with the size exactly k. My approach to this problem is building a recurrence equation. Let $ C(n)$ be the average number of subarrays with the size k encountered during quicksort. We have $ C(n) = \frac{2}{n}\sum_{i=1}^{N-1}C(i)$ . Simplify this equation using the same technique in the book, we have $ C(n)=\frac{n+1}{n}C(n-1)$ . Because $ C(n)=0$ with $ n<k$ and $ C(n)=1$ with $ n=k$ . So by expanding the equation and canceling the factor, we have $ C(n)=\frac{n+1}{k+1}$ . So the answer of the original problem is $ (n+1)(\frac{1}{2}+…+\frac{1}{k+1})$ .
Is this a correct way to solve this problem? Did I miss something in my arguments above?

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Are there average damage-per-round guidelines?

Still being rather new to character-creation, I want to see if the character I create has the appropriate power for its class and level. Are there average damage-per-round per level tables to be found? Maybe per class/role?

Alternatively, if there are no such tables, could estimations be made?

I know that damage-per-round is not the only thing characters can be good at in a campaign, but in this instance I wanted to compare only DPR. I want to identify a fitting role for a character I am creating, and DPR plays a role in this process.

Accelerated and average position

I've been experimenting with the new AI on adwords. My website is for lead gen. I've tried Max. Conversions and CPA, and had some success.
I'm was hoping someone here would know of the newest course that might be out there. I want to learn how to use audiences, and demographics in my campaign.

Can someone point me in the right direction?

Can you take the average damage roll when changed into a Monster?

In this answer the idea of PCs using the following quote from the Monster Manual came up:

Any damage dealt or other effects that occur as a result of an attack hitting a target are described after the “Hit” notation.¬†You have the option of taking average damage or rolling the damage; for this reason, both the average damage and the die expression are presented.

I’m wondering whether this rule only applies to monsters or if PCs who have used Wild Shape or similar things to become monsters can also use this.