Bash – How to delete all of a set of files in a random order

I have a set of files, all that are named with the convention file_[number]_[abcd].bin (where [number] is a number in the range 0-size of drive in MB). i.e there is file_0_a.bin, file_0_b.bin, file_0_c.bin and file_0_d.bin and then the 0 would become a 1 and so on.

The number of files is figured out at run-time based on the size of the partition. I need to delete all of the files that have been created, but in a pseudo-random manner. in blocks of size that I need to be able to specify, i.e where there is 1024 files, delete 512, then delete another 512.

I have the following function for doing it currently, which I call the required number of times, but it will get progressively less likely to find a file that exists, to the point where it might never complete. Obviously, this is somewhat less than ideal.

What is another method that I can use to delete all of the files in a random order?

deleteRandFile() #$1 - total number of files
{
    i=$((RANDOM%$1))
    j=$((RANDOM%3))
    file=""

    case $j in
    0)
        file="${dest_dir}/file_${i}_a.bin";;
    1)
        file="${dest_dir}/file_${i}_b.bin";;    
    2)
        file="${dest_dir}/file_${i}_c.bin";;
    3)
        file="${dest_dir}/file_${i}_d.bin";;
    esac

    if ! [[ -f $file ]]; then
        deleteRandFile $1
    else
        rm $file
    fi

    return 0;
}

Edit:
I'm trying to delete in random order so that I can fragment the files as much as possible. This is part of a script that begins by filling a drive with 1MB files, and deletes them, 1024 at a time, then fills the 'gap' with 1 1GB file. Rinse and repeat until you have some very fragmented 1GB files.