Quote: "Is that method not susceptible to less shuffling? When your random number is already chosen you just work up through the pack until you find an unchosen card."
Nope.
It doesn't matter at all whether the number is already chosen or not, because it's all about the xth number of the REMAINING cards, where x is the random number.
Quote: "Worst case scenario, your first random number is 1. The second is 1 or 2, the third 1-3 and so on. In this scenario no shuffling happens."
No.
There is only no shuffling if every random number was 1. The odds of this scenario to happen with a deck of 52 cards are 1 : 52!, which equals 1 : 52*51*50*.....*2 which equals 1 : 8.06581751709E+67. Which is quite a bit.
With 14 cards odds are already 1 : 87178291200
Quote: " There can be varying degrees of this issue which reduce the shuffling. The problem gets greater as you get nearer the end of the pack."
No. Getting to the end of the pack has no influence at all on the shuffling.
The system works as follows:
Let's assume we have a deck with cards labeled A, B, C, D and E.
First random number is 1 (1-5), so A is drawn.
Second random number is 2(1-4), so C is drawn.
Third random number is 3(1-3), so E is drawn.
Forth random number is 1(1-2), so B is drawn.
Fifth random number is 1(1-1), so D is drawn.
As mentioned, there is only no shuffling if every random number is 1.
When you swap cards, chances are pretty "high" that a swapped card eventually lands on its original spot, no matter how often you shuffle.
As I said, the system picks the xth number of the REMAINING cards. You can't pick more randomly. The only limit would be the random generator.
Quote: "Picking back to back positions repeatedly can happen in any method, it's a valid random occurrence, it's also relatively unlikely in any method across a full size deck. "
That is correct