InLeo PREMIUM Lottery - how winner will be picked

PREMIUM subscriptions started on InLeo about ~10 days ago.

What are the benefits one can expect for his 10 HBD/month fee?
This can be seen very easy - just press the "Premium" MENU option on the left side.

This PREMIUM feature is very healthy to the whole LEO/HIVE ecosystem, as all it's subscription fees 100% go to the permanent Hive/LEO liquidity pool.

Yet I see a number of people saying, that this fee is to high, or not affordable to them. Or simply brings not enough value in all things what they will get back for their 10 HBD

To encourage new members, I decided to make a #premium100 LOTTERY for the FIRST 100 subscribers.
The winner of this #lottery will get a possibility to chose, which of the FOUR possible prize options he wants the most:

The winner will have a completely FREE choice, based exclusively on his own needs and preferences.

I am tracking 10 HBD credits to the @leopool wallet, and each consecutive payment gets a unique "Lottery ticket number". All in a strict chronological order.
As of today, the entries are like this:

(red-on-yellow entries show the winners of the secret surprises under certain seats)

But WHO will be the winner?

Or in other words - how one single account name will be picked, without any doubts of my personal preferences, sympathies, or anything which may spoil the transparency.
How to make drawing maximum transparent, and completely "random-like", completely UNPREDICTABLE ?
I was thinking about this task several days...

Until I found (I think) the right tool: HASHING.

As long time bitcoin-addicted person, I have learned years ago, what hashing is, and how it's results are unpredictable, hacker proof, easy & simple to achieve, and extremely secure.
Simply put - I AM big FAN of Hashing.

So what will be done?

1. I take all 100 participant accounts in a one-long-nobreak-text string, and calculate the SHA256 hash of it. The output of this hash (string of HEX characters) will be divided by the pairs-of-two, and one certain pair, representing the number in the range of 1-100, will be the wining ticket number.

Just like this example

SHA256(@khaleelkazi@yanbolludeniz@mrberenben@leofinance@[email protected]@beststart@brando28@taskmaster4450le@nonsowrites@idksamad78699@mistakili@mcoinz79@anderssinho@wealthwess@jongolson@katerinaramm@esmeesmith@l337m45732@solymi@neopch@alokkumar121@onealfa@[email protected]@dwayne16@crptogeek@ifarmgirl@jeffjagoe@bradleyarrow@forexbrokr@luckyali@selfhelp4trolls@adambarratt@muratkbesiroglu@niallon11@moretea@alex-rourke@shortsegments@randumb@bitcoinman@manniman@fokusnow@trumpman@haveyoursay@anomadsoul@efastromberg94@scaredycatguide@detlev@bitcoinflood@michupa@freecompliments@pouchon@gadrian@lisamgentile1961@coolguy222@mes@uyobong@walterjay@[email protected]@jk6276@elianaicgomes@idiosyncratic1@sapphirecrypto@dbooster@preparedwombat@biyimi@tsunsica@luchyl@lipe100dedos@vimukhti@pepetoken@silversaver888@melbourneswest@forkyishere@alessandrawhite@gargi@davchi2@sandymeyer@kingworldline@fifthspatial@mightpossibly@thorkellnft@phanty@steemflow@act87@act88@act89@act90@act91@act92@act93@act94@act95@act96@act97@act98@act99@act100) =

= 34b855e1feaf827cdba3c4783d234affec5ddb67dd8cb3160c5a8f380c1c364b

1. Then we divide this HEX output by pairs:
   34 b8 55 e1 fe af 82 7c db a3 c4 78 3d 23 4a ff ec 5d db 67 dd 8c b3 16 0c 5a 8f 38 0c 1c 36 4b

2. But which PAIR is now the winning?
   For that, I will take the SHA256 of the first 96 account names, and this first HEX-digit of the output hash will show, which pair must be used for picking the winner

Example: SHA256(@khaleelkazi@yanbolludeniz@mrberenben@leofinance@[email protected]@beststart@brando28@taskmaster4450le@nonsowrites@idksamad78699@mistakili@mcoinz79@anderssinho@wealthwess@jongolson@katerinaramm@esmeesmith@l337m45732@solymi@neopch@alokkumar121@onealfa@[email protected]@dwayne16@crptogeek@ifarmgirl@jeffjagoe@bradleyarrow@forexbrokr@luckyali@selfhelp4trolls@adambarratt@muratkbesiroglu@niallon11@moretea@alex-rourke@shortsegments@randumb@bitcoinman@manniman@fokusnow@trumpman@haveyoursay@anomadsoul@efastromberg94@scaredycatguide@detlev@bitcoinflood@michupa@freecompliments@pouchon@gadrian@lisamgentile1961@coolguy222@mes@uyobong@walterjay@[email protected]@jk6276@elianaicgomes@idiosyncratic1@sapphirecrypto@dbooster@preparedwombat@biyimi@tsunsica@luchyl@lipe100dedos@vimukhti@pepetoken@silversaver888@melbourneswest@forkyishere@alessandrawhite@gargi@davchi2@sandymeyer@kingworldline@fifthspatial@mightpossibly@thorkellnft@phanty@steemflow@act87@act88@act89@act90@act91@act92@act93@act94@act95@act96) = 4e90545c4c1a506974b361889472c7c9e217969a3fbebc7c577b9c17144f8465

As we see, first digit is 4, it shows that pair "4" must be used. But we see that pair "4" was a HEX number e1. e1(hex)= 225(dec)

Yes, two digit HEX numbers can be anything in the range 0-255 in decimal system. So we will use next HEX-pair to the right (starting from 4!), until we find something which brings dec <101

As we see, next such pair is way to the right, is "3d", which corresponds in decimal system 61. 3d= = 61

Voila ! #61 ticket is the winner, and its #premium holder is @drax.leo
Of course, in reality all numbers will be absolutely different. And the results are absolutely UNPREDICTABLE, until the last 100th ticket will be known.

One even smallest change in argument text string changes the output hash completely, from beginning to the end, every character. And it changes all in an absolutely unpredictable way.
Nobody knows yet which accounts will get tickets 87-96, and which accounts will come in seats 97, 98, 99 & 100.

There is absolutely no way to game-out this system, and somehow make that some preferred account could win.

I would be happy to hear, if somebody think this is not right, and that this algorithm can be spoofed somehow.

P.S.
In case we can not find a pair which gives decimal output <101 (highly unlikely, though), the longer output hash SHA512 will be used instead of SHA256 (while hashing all 100 accounts)

@onealfa

#premium100 #lottery #premium

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