Randomize a List problem

[Albert Chan]

(09-14-2020 08:51 PM)Hlib Wrote:  FOR n t RAND ... then after many clicks we may get all set of lists (6 for #3 and 24 for #4).
Assuming that RAND can take any value from the interval [0,1] and taking into account the specifics of the ROLLD command, I calculated all possible options for the shown results for both versions of the program using the full search method. There is no error here. But let`s say we can`t be sure. So we need to get the formula.

For a list of 3 items, we have 3! = 6 permutations.
But, your suggested shuffling can have 3*3*3 = 27 ways.

By the pigeonhole principles, some will appear more often, some less (*)
Starting from {3,2,1}, and gone thru all 27 possible ROLLD shuffling, we have:

{3,2,1} × 5
{2,3,1} × 4
{3,1,2} × 5
{1,3,2} × 5
{2,1,3} × 4
{1,2,3} × 4

Shuffling will not be fair. I did a simulation, and confirmed this.

(*) even if one divides the other, there is still a possibility of unfair shuffling, depending on shuffling details.