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I was recently running tests to get items to drop randomly in an old-school computer RPG. I wanted to verify that all items in a range, say, 1-80 would drop in a certain dungeon. But I couldn't do things all at once.

After one test run with the 80-item dungeon, here might be example output from my array of item drops:

Items dropping in D: 1-5, 7-12, 15-21, ...., 73-80

In other words, my program doesn't just print out "yes no yes no" but lumps together a range of consecutive numbers that randomly turned up, where a range is defined as any set of $x \ge 1$ continuous integers that have been rolled. So 1 alone would count as a range, and so would 40-78.

I noticed that this expanded for a while, then it shrank. But I'm curious what m would give the maximum number of ranges. This necessitates a formula for the expected number of ranges in terms of $m$ and $n$, and I have to admit I don't know where to start there. Obviously I could run Monte Carlo simulations to give a guess, but I'm interested in the general formula/derivation.

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  • $\begingroup$ A side question would be, what would give the longest text output? For instance, if the output was like 01, 02-03, 06-11, 14, (etc.) how would weighting by range size affect n as a function of m? $\endgroup$ – aschultz 2 days ago
  • $\begingroup$ Related. $\endgroup$ – hilberts_drinking_problem 2 days ago
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Following the derivation described here, we can obtain an explicit probability of getting $k$ intervals when selecting $m$ out of $n$ available objects by extracting coefficients of the following generating function:

$$\frac{1-w+uw}{(1-z)(1-w)-uwz}$$

Here, $w$ counts objects that do appear, $z$ counts the objects that don't and $u$ counts runs/intervals.

Using Mathematica, we can explicitly compute the expected number of runs for n=numItems and a given m as follows:

numItems = 30; 
f[z_, w_, u_] := (1 - w + u*w)/((1 - z) (1 - w) - u*w*z); 
countNRuns[m_, r_] := 
 SeriesCoefficient[
  f[z, w, u], {z, 0, numItems - m}, {w, 0, m}, {u, 0, r}];
expectedRuns[m_] := 
  N[Sum[r*countNRuns[m, r], {r, 0, m}]/Binomial[numItems, m]];
DiscretePlot[expectedRuns[m], {m, 0, numItems}]

yielding the following for n=30:

enter image description here

Explicit computation for large $n$ is costly, but it seems that the maximal number of runs is achieved when the number of selected items if approximately $n/2 = 40$.

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