I use the randomized unlock screen where it changes the positioning of the numbers 0-9 each time you unlock the phone. Assuming you use 4 unique numbers, no repeats, I want to calculate the odds of those 4 numbers appearing in their usual spots on a standard 0-9 keypad, i.e. the zero at the bottom, and the 1 beginning in the upper left and counting down--but not concerning ourselves with the placement of numbers which aren't used in your keycode.

So let's say your passcode is 7843. What are the odds that the 7, the 8, the 4, and the 3 appear where they would typically appear on a non-randomized keypad?

The first position, the one-place, must be any one of six different numbers out of the total of ten as it's not in your code:
10/6 = 1.667
The second position must be any one of five remaining numbers not in your code out of nine total possibilities:
1.667 x (9/5) = 1.667 x 1.8
The third position must be only one number out of the remaining eight numbers, as it is unique to your code:
1.667 x 1.8 x 8
The fourth position must be only one number out of the remaining seven numbers, unique to your code:
1.667 x 1.8 x 8 x 7
The fifth position must be any one of four remaining numbers not in your code, out of six total:
1.667 x 1.8 x 8 x 7 x (6/4) = 1.667 x 1.8 x 8 x 7 x 1.5
The sixth position must be any one of three remaining numbers not in your code, out of five total:
1.667 x 1.8 x 8 x 7 x 1.5 x (5/3) = 1.667 x 1.8 x 8 x 7 x 1.5 x 1.667
The seventh position must be only one of the remaining four numbers, unique to your code:
1.667 x 1.8 x 8 x 7 x 1.5 x 1.667 x 4
The eighth position must be only one of the remaining three numbers, unique to your code:
1.667 x 1.8 x 8 x 7 x 1.5 x 1.667 x 4 x 3
The ninth, non-unique:
1.667 x 1.8 x 8 x 7 x 1.5 x 1.667 x 4 x 3 x 2
The tenth:
1.667 x 1.8 x 8 x 7 x 1.5 x 1.667 x 4 x 3 x 2 x 1

Meaning the odds of this occurring are about 1 in 10,000. To be exact, 1 in 10,084. Not a bad feature.
This assumes the numbers populate from top-left to bottom-right and I imagine it would vary depending on the code used. This also assumes the expected degree of randomization methodology.

All this was on my mind tonight so just thought I'd create an account and share :)