The winner will be determined by the hash function of Block #333933 and #333394, unless the desired resulting number gleaned is higher than 19 (fuck me, now I have Bruce Peterson on the brain
![Roll Eyes](https://bitcointalk.org/Smileys/default/rolleyes.gif)
).
The Hash for Block #333922 is 00000000000000000b2dde107d49f8b98eb9e1947cf87d3bb66af9fd2d7c82c0 as depicted here:
https://blockchain.info/block-index/511618/00000000000000000b2dde107d49f8b98eb9e1947cf87d3bb66af9fd2d7c82c0![](https://ip.bitcointalk.org/?u=http%3A%2F%2Fs1.postimg.org%2F8nhfkbe3j%2Fhash.jpg&t=665&c=LHButu2geeYJxQ)
I will disregard all the letters up to the first letter, in this case the letter "b", thus: b2dde107d49f8b98eb9e1947cf87d3bb66af9fd2d7c82c0
Then, I will isolate the numbers from the letters, thus:
b 2 dde 107 d 49 f 8 b
98 eb 9 e 1947 cf 87 d 3 bb 66 af 9 fd 2 d 7 c 82 c 0
I will glean the fifth [element], in this case the number 98, as highlighted above.
98 is larger than 19, thus I head on over to the next Block/hash to see if the resulting number can be had.
<guess I really thought this through, for now I have to wait for it to further illustrate this part, but I'll detail the second stage of the determination process while waiting>
Once a number between 1 and 19, inclusive, is realized, I'll head on over to both ordered lists and count down, i.e., if the number is 12, then Light and jparsley will be the two ready for the next round. I will put the resulting names in alphabetical order, in this case jparsley (first) and Light (second). jparsley will be consider odd (think number 1) and Light will be considered even (think number 2). I will then look at the very next Block's hash function and see what the very first digit of the third number is. If it's odd, Light would've won, and if it's even, then jparsley would've won, example purposes only.
Using the same hash function as above, I'll illustrate how that odd/even number will be obtained:
b 2 dde 107 d
49 f 8 b 98 eb 9 e 1947 cf 87 d 3 bb 66 af 9 fd 2 d 7 c 82 c 0
2 is the first digit of the first number 2
1 is the first digit of the second number 107
4 is the first digit of the third number 49
4 is even (thus Light would've won this round and the overall contest)
I believe that what I've devised is fair. What I can't believe is that we're still on Block #333922, thus the results may be a while in the making, but I see no problem there.
Let me know if I failed somewhere in my reasoning/fairness, for there's no way I can game the results at this point, even in essence giving entrants two chances to win.
~Bruno Kucinskas