| ▲ | pmarreck 2 days ago | |
If, instead of doing a binomially-distributed cut, we do a 2-step cut which consists of first a random cut (then take the bottom portion and place it on the top, which in essence shifts all the cards by a random amount), and THEN do a perfect half-and-half split to perform the actual shuffle with? Would that get us closer to "random enough", quicker? | ||