Huh? I get 2,563,893 to 1. Of course, the odds are better still if you don't limit yourself to that exact vanity address, or exactly the first and last digits of the hash, or that exact hash function (indeed, the use of such an obscure hash function suggests that many were tried in the hope of finding one that works). Nothing to see here, folks.
Ok, lets see that math. Wrong is wrong in math. Lets see how good you are, including all the vectors presented here. Or is this a sock puppet account with ulterior motives trying to discount or distract people away from this post
In your equation analysis be sure to include the following.
1. How many alpha-numeric hash combinations are there in sha-256
2. How many md6-256 alpha-numeric combinations are there.
3. How unique are the hashes presented by hashing " v0.1 " in both sha-256/md6-256
4. What are the vectors of conversion between both hashes
5. How difficult is it to produce a random "five string letter" vanity address
6. How does this difficulty rise substantially when limited within the scope of only being allowed to manipulate 2 alphanumeric characters within hashing a random word?
7. How many two alpha numeric character combination changes are possible within a sha-256/md6-256 hash?
Ok Foxpup, I will give you a chance to correct your math; otherwise, you are just another keyboard cowboy speaking about something you know nothing about.