PatrickHarnett


March 09, 2012, 12:34:31 AM 

I will not post anything inflammatory for 72 hours.
Can we raffle this? (going into a coma would not count)








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cablepair


March 09, 2012, 12:40:36 AM 

Thread updated. Rules for normal participation (IMPORTANT!) Click the image above
 Purchase any number of raffle entries you would like.
 Provide a valid email address that you do not mind sharing publicly.
Rules for referrals (IMPORTANT!) Provide your email address to a referral
 When they purchase their raffle credits, they will receive an additional 5% credit for each credit they purchase.
 You will receive an additional 25% credit for each credit purchased through the referral.
Referral example:
You purchase 1 credit. You now have 1 raffle credit. Friend 1 purchases 1 credit through your referral email. You now have 1.25 raffle credits. Friend 1 now has 1.05 raffle credits. Friend 2 purchases 3 credits through your referral email. You now have 2 raffle credits. Friend 2 now has 3.15 raffle credits.
Deadline (IMPORTANT!)A winner will be chosen 7 days from today's date and contacted through the email address the provided at the time of entry.The winner be chosen in conjunction with random numbers provided by random.org across a spreadsheet of raffle credits. the evil genius strikes again, well done Matt




LoupGaroux


March 09, 2012, 12:48:19 AM 

I will not post anything inflammatory for 72 hours.
Can we raffle this? (going into a coma would not count) Sure if I win this raffle, I will raffle off my silence for 72 hours. Although that is going to cost Matthew the glowing review... I'm not a cheap whore you know, just a whore.

▄████▄ ▄████████▄ ▄████████████▄ ▄████████████████▄ ████████████████████ ▄█▄ ▄███▄ ▄███▄ ▄████████████████▀ ▄██████████ ▄▄▄▀█████▀▄▄▄▄▀█████▀▄▄▄ ▀██▄ ▄██▀ ▀██▄ ▄██▀ ▀██▄ ▄██▀ ██ ▄█████▄▀▀▀▄██████▄▀▀▀▄█████▄ ▀██▄ ▄██▀ ▀██▄ ▄██▀ ▀██▄ ▄██▀ ▄█▄ ▀██████████████▄ ████████████████████████████ ▀██▄ ▄██▀ ▀██▄ ▄██▀ ▀██▄ ▄██▀ ▀█▀ ██ ▀████████████████████████▀ ▀██▄ ▄██▀ ▀██▄ ▄██▀ ▄█▄ ▀██▄ ▄██▀ ██ ▀████████████████████▀ ▀███▀ ▀███▀ ▀█▀ ▀███▀ ▄███████████████████████████████████▀ ▀████████████████▀ ▀████████████▀ ▀████████▀ ▀████▀
 ║║ ║█ ║█ ║║  .
 .
║║ ██ ║║
 .
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║║ ██ ║║
 .
 ║║ █║ █║ ║║  



cyberlync


March 09, 2012, 12:53:55 AM 

aaaand I'm out again. If it's too good to be true, It'll probably get edited later.

Giving away your BTC's? Send 'em here: 1F7XgercyaXeDHiuq31YzrVK5YAhbDkJhf



Red Emerald


March 09, 2012, 01:14:10 AM 

Are you sure the winner isn't going to be announced in 46 weeks?




Matthew N. Wright


March 09, 2012, 01:16:15 AM 

aaaand I'm out again. If it's too good to be true, It'll probably get edited later.
Quote it for truth then. Nothing will change. I'm not Atlas.




deepceleron


March 09, 2012, 01:28:53 AM 

 A winner will be chosen 7 days from today's date and contacted through the email address they provided at the time of entry.
 The winner will be chosen in conjunction with random numbers provided by random.org across a spreadsheet of raffle credits.
This method cannot be independently verified.  There can be false unpaid entries (even with payment verified on blockchain, they can be paid in by the organizer to himself).
 The winning ticket selection method cannot be seen or verified by all outside parties.
I can at least solve the second part. The bitcoin blockchain contains lots of completely random hashes that are NSArecommended as unreversible, not predetermined, outside control of any one party, and published publicly. Here is my recommended method to declare winning ticket numbers of any forum lottery: 1. Publish block# to determine winner. For this lottery's timeline, block 171420 (~8 days from now) after 120 confirmations could be the one to determine the winner ( block 170267 used in example below), 2. Disclose ticket numbers (spreadsheet) beforehand (start at 0, list forum IDs, etc), allow all entrants to see they are present in list and that there are no mystery entries nobody can account for. 3. Determine total number of tickets at entrance cutoff time (lets say 711 in this example), 4. The random data to be used from the block will be the last 10 digits of the hash, 5. The winning ticket number will be determined by the modulo (remainder) function [python = math.fmod(int("hash_last_10", 16), num_of_entries)]. Winning number as output by Win7 x32 calculator if any disputes of solution. Here is how to independently determine the winner with no programming on Windows calculator (fixed):  Verify the block hash from two independent sources (ideally one being your local blockchain),
 Open Windows Calculator. Select View>Programmer. Select "Hex" radio button,
 Paste last 10 digits of hash into calculator (73ef1acb71 in block 170267 in this example),
 Press "Dec" radio button to convert to decimal (497932749681)
 Press "Mod",
 Enter number of tickets (711)
 Press =
 (If the list of ticket numbers started at ticket #1 instead of ticket #0, add 1)
 See winner (example winning ticket number is #276)




Matthew N. Wright


March 09, 2012, 01:30:44 AM 

 A winner will be chosen 7 days from today's date and contacted through the email address they provided at the time of entry.
 The winner will be chosen in conjunction with random numbers provided by random.org across a spreadsheet of raffle credits.
This method cannot be independently verified.  There can be false unpaid entries (even with payment verified on blockchain, they can be paid in by the organizer to himself).
 The winning ticket selection method cannot be seen or verified by all outside parties.
I can at least solve the second part. The bitcoin blockchain contains lots of completely random hashes that are NSArecommended as unreversible, not predetermined, outside control of any one party, and published publicly. Here is my recommended method to declare winning ticket numbers of any forum lottery: 1. Publish block# to determine winner. For this lottery's timeline, block 171420 (~8 days from now) after 120 confirmations could be the one to determine the winner ( block 170267 used in example below), 2. Disclose ticket numbers (spreadsheet) beforehand (start at 0, list forum IDs, etc), allow all entrants to see they are present in list and that there are no mystery entries nobody can account for. 3. Determine total number of tickets at entrance cutoff time (lets say 711 in this example), 4. The random data to be used from the block will be the last 10 digits of the hash, 5. The winning ticket number will be determined by the modulo (remainder) function [python = math.fmod(int(hash_last_10, 16), num_of_entries)]. Winning number as output by Win7 x32 calculator if any disputes of solution. Here is how to independently determine the winner with no programming on Windows calculator:  Verify the block hash from two independent sources (ideally one being your local blockchain),
 Open Windows Calculator. Select View>Programmer. Select "Hex" radio button,
 Paste last 10 digits of hash into calculator (73ef1acb71 in block 170267 in this example),
 Press "Mod",
 Enter number of tickets (711)
 Press =
 See winner (example winning ticket number is #141)
Anyone have any objections to this?




notme
Legendary
Offline
Activity: 1736


March 09, 2012, 01:32:22 AM 

 A winner will be chosen 7 days from today's date and contacted through the email address they provided at the time of entry.
 The winner will be chosen in conjunction with random numbers provided by random.org across a spreadsheet of raffle credits.
This method cannot be independently verified.  There can be false unpaid entries (even with payment verified on blockchain, they can be paid in by the organizer to himself).
 The winning ticket selection method cannot be seen or verified by all outside parties.
I can at least solve the second part. The bitcoin blockchain contains lots of completely random hashes that are NSArecommended as unreversible, not predetermined, outside control of any one party, and published publicly. Here is my recommended method to declare winning ticket numbers of any forum lottery: 1. Publish block# to determine winner. For this lottery's timeline, block 171420 (~8 days from now) after 120 confirmations could be the one to determine the winner ( block 170267 used in example below), 2. Disclose ticket numbers (spreadsheet) beforehand (start at 0, list forum IDs, etc), allow all entrants to see they are present in list and that there are no mystery entries nobody can account for. 3. Determine total number of tickets at entrance cutoff time (lets say 711 in this example), 4. The random data to be used from the block will be the last 10 digits of the hash, 5. The winning ticket number will be determined by the modulo (remainder) function [python = math.fmod(int(hash_last_10, 16), num_of_entries)]. Winning number as output by Win7 x32 calculator if any disputes of solution. Here is how to independently determine the winner with no programming on Windows calculator:  Verify the block hash from two independent sources (ideally one being your local blockchain),
 Open Windows Calculator. Select View>Programmer. Select "Hex" radio button,
 Paste last 10 digits of hash into calculator (73ef1acb71 in block 170267 in this example),
 Press "Mod",
 Enter number of tickets (711)
 Press =
 See winner (example winning ticket number is #141)
You forgot: add 1 to mod result unless you numbered tickets starting with 0




deepceleron


March 09, 2012, 01:39:19 AM 

You forgot: add 1 to mod result unless you numbered tickets starting with 0
I just updated the post, the example result 141 was in hex, you must convert it back to decimal = 321. I do specify the first ticket is #0. Another oops with the calculator procedure is that you'd be entering the number of tickets in hex, which gives wrong results if you input decimal, I'll fix that too.




notme
Legendary
Offline
Activity: 1736


March 09, 2012, 01:40:44 AM 

I see that now... at least I helped you catch something .




bittenbob


March 09, 2012, 01:41:22 AM 

 A winner will be chosen 7 days from today's date and contacted through the email address they provided at the time of entry.
 The winner will be chosen in conjunction with random numbers provided by random.org across a spreadsheet of raffle credits.
This method cannot be independently verified.  There can be false unpaid entries (even with payment verified on blockchain, they can be paid in by the organizer to himself).
 The winning ticket selection method cannot be seen or verified by all outside parties.
I can at least solve the second part. The bitcoin blockchain contains lots of completely random hashes that are NSArecommended as unreversible, not predetermined, outside control of any one party, and published publicly. Here is my recommended method to declare winning ticket numbers of any forum lottery: 1. Publish block# to determine winner. For this lottery's timeline, block 171420 (~8 days from now) after 120 confirmations could be the one to determine the winner ( block 170267 used in example below), 2. Disclose ticket numbers (spreadsheet) beforehand (start at 0, list forum IDs, etc), allow all entrants to see they are present in list and that there are no mystery entries nobody can account for. 3. Determine total number of tickets at entrance cutoff time (lets say 711 in this example), 4. The random data to be used from the block will be the last 10 digits of the hash, 5. The winning ticket number will be determined by the modulo (remainder) function [python = math.fmod(int(hash_last_10, 16), num_of_entries)]. Winning number as output by Win7 x32 calculator if any disputes of solution. Here is how to independently determine the winner with no programming on Windows calculator:  Verify the block hash from two independent sources (ideally one being your local blockchain),
 Open Windows Calculator. Select View>Programmer. Select "Hex" radio button,
 Paste last 10 digits of hash into calculator (73ef1acb71 in block 170267 in this example),
 Press "Mod",
 Enter number of tickets (711)
 Press =
 See winner (example winning ticket number is #141)
Anyone have any objections to this? The issue I see is that the bonus program you laid out issues fractions of entries. Are only full entries counted towards the draw? (ie 1.05=1, 1.85=1, 2.05=0, etc)




notme
Legendary
Offline
Activity: 1736


March 09, 2012, 01:44:54 AM 

The issue I see is that the bonus program you laid out issues fractions of entries. Are only full entries counted towards the draw? (ie 1.05=1, 1.85=1, 2.05=0, etc)
This could be solved by multiplying all ticket amounts by 100.




draco49


March 09, 2012, 01:55:14 AM 

The issue I see is that the bonus program you laid out issues fractions of entries. Are only full entries counted towards the draw? (ie 1.05=1, 1.85=1, 2.05=0, etc) Good point. I think a user's final entry count will be turned into a weighted number for the purpose of the drawing. Some clarification on this would be appreciated Use me as your referral, draco49@tormail.net, and if I win I'll pay everyone who used my referral their 0.50BTC back.

Coming Soon  bitmax.co (http://bitmax.co)  the weekly bitcoin lottery
** I use BitVPS.com (http://BitVPS.com)  You should too! **



rjk


March 09, 2012, 01:57:45 AM 

Aww piss, I post the first response, step out for a meeting, and then Matthew turns it into a raffle. Fuck this shit, I'm out.




notme
Legendary
Offline
Activity: 1736


March 09, 2012, 02:02:47 AM 

Use me as your referral, draco49@tormail.net, and if I win I'll pay everyone who used my referral their 0.50BTC back. How are you going to verify that?




deepceleron


March 09, 2012, 02:04:28 AM 

Anybody want to make a javascript page to do the ticketpicker math I specify above? I'll try it if no genius can whip one up in minutes, I'd like to make sure many/any method can arrive at same answer for winner. Input 1: Last 10 of hash (10 digits hexidecimal) Input 2: Number of entries (base10 decimal) Input 3: Do ticket numbers start at 0? (yes = 0, no = 1) Output 1: Input 1 % Input 2 + Input 3 Aww piss, I post the first response, step out for a meeting, and then Matthew turns it into a raffle. Fuck this shit, I'm out.
Right?? I was ready to hit send and buy that 0.5BTC BFL single!




Epoch
Legendary
Offline
Activity: 917


March 09, 2012, 02:06:35 AM 

following

BTC: 1DJVUnLuPA2bERTkyeir8bKn1eSoRCrYvx NMC: NFcfHSBBnq622pAr1Xoh9KtnBPA5CUn6id



Ferroh


March 09, 2012, 02:14:40 AM 

You're not giving them away, you're selling raffles for them in a pyramid gambling scheme, where we don't know how many raffles are sold and therefore what our odds of winning are. It is mildly depressing that the majority of the bitcoin community:  Jumps on this idea with wild excitement, with apparently no consideration of why it might be a bad idea to buy into this.
 Discusses stuff like this in a forum where the "im in LETS DO THIS LOL!" posts can't be downvoted so that the "um guys, this is a pyramid/gambling scheme" comments get buried. (I'm looking forward to new forum software.)




notme
Legendary
Offline
Activity: 1736


March 09, 2012, 02:15:08 AM 

Anybody want to make a javascript page to do the ticketpicker math I specify above? I'll try it if no genius can whip one up in minutes, I'd like to make sure many/any method can arrive at same answer for winner. Input 1: Last 10 of hash (10 digits hexidecimal) Input 2: Number of entries (base10 decimal) Input 3: Do ticket numbers start at 0? (yes = 0, no = 1) Output 1: Input 1 % Input 2 + Input 3 Aww piss, I post the first response, step out for a meeting, and then Matthew turns it into a raffle. Fuck this shit, I'm out.
Right?? I was ready to hit send and buy that 0.5BTC BFL single! http://pastebin.com/6LqsUgUgAddress is in signature if you want to say thanks.




