Based on a joke going around in Video Game randomizers, where items that are appearing to be “not randomized” are thought to be broken, I was interested in the probabilities related to this.

To simplify, we have *n* items, which are randomized to appear in *n* positions.

This can be considered to be similar to be having a bag of *n* tokens, numbered *1* to *n*, and we draw them sequentially, so that in a bag of 3, a draw of 3-2-1 is counted as a draw in sequential order, as 2 is drawn in position 2.

Now, the probability of drawing them all in order is simple, $frac{1}{n!}$.

But what is the probability of none of them being in order? (Because, obviously, none of them being in their original order is “truly” random…..) Based on a question and comment here it seems that Rencontres numbers fit my empirical test at low numbers, where

n | No sequential draw occurrences | Probability |
---|---|---|

3 | 2 | 2/6 = 1/3 |

4 | 9 | 9/24 = 3/8 |

5 | 44 | 44/120 = 11/30 |

So am I correct in saying that we can use the formula for Rencontres numbers here, which is

$$D_{n,0} = leftlceilfrac{n!}{e}rightrfloor$$

?

Which interestingly enough, without looking too hard on expanding that formula, seems to tend to around 0.3678 as n gets bigger.