Attempting to dual boot Vista/Ubuntu(my distro). Started with Vista because it seemed simpler as GRUB just makes itself the bootloader after you install linux, rather then having to reinstall GRUB if you go linux first. So I go along with it, Vista loads smoothly. Go ahead and install linux: doesn't see either OS(no GRUB error). Shruged it off and formated/deleted both partitions via Vista boot cd. Reinstall Vista: "bootmgr is missing". A note although I can not predictively reproduce this, I know theres two instances of Vista in the MBR, one is bootable and one is not, as I said this is not predictively reproducible.

So right now my goal is to delete everything(including the MBR) off the drive and start all over again, problem is I don't know how to do this. Is there a better way?

Before you do anything, boot the Ubuntu liveCD, and from a terminal run this and post the results (it'll give us an idea of the Vista partition table)After that, you can scrub the disk like this (zero first 100 sectors - nothing scientific)It appears Vista allocates partitions differently than everyone else - including XP. Allocate your partitions from Ububtu liveCD, then do the Vista install.

(presumes /dev/sda as your disk - it may be /dev/hda)

66+0 records in
66+0 records out
66 bytes (66 B) copied, 0.00177635 seconds, 37.2 kB/s
00000000 80 20 21 00 07 fe ff ff 00 08 00 00 00 50 38 3a |. !..........P8:|
00000010 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................|
*
00000040 55 aa |U.|
00000042
(END)

(I think this is what your looking for?)

Yep - thanks.

So scrub it?

Ah sorry - yep, I would. I wanted that listing simply for info.
As a test I scrubbed a disk like I suggested, and gparted handled it fine - even formatted a NTFS partition. I expect Ubuntu will have gparted.