wheelz1200
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July 08, 2016, 04:09:33 PM |
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The spreadsheet associates all addresses with a book number (by row), so this is indistinguishable for your request. I haven't posted the spreadsheet yet as I haven't gotten addresses from everyone. Purchasers should be able to see their own address via their browsers "find" function. All the addresses will receive counterparty tokens and the winning address holder will also get the gold bar shipped to them.
That does not answer my question, so I'm asking you again, what happens when the winning number is a number of a book that hasn't been sold. I know what you are going to do, but if a book isn't sold have nowhere to ship the bar to as nobody can claim it. Mitchell brings up a good point, the idea of unclaimable tickets. This was a problem in the original plan that we didn't consider, so the solution is that sales will simple have to stay open and even though some market participants will know the number of the book eligible for the gold, even if it didn't sell, some lucky person in the future will get it instead 1 day till the halving! Soooooo. You could possibly have to sell 5000 books until you might have a winner, at which point you know what the winning book is....dont know if you want to stick to that "plan"
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photonresearch (OP)
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July 08, 2016, 04:20:21 PM |
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The spreadsheet associates all addresses with a book number (by row), so this is indistinguishable for your request. I haven't posted the spreadsheet yet as I haven't gotten addresses from everyone. Purchasers should be able to see their own address via their browsers "find" function. All the addresses will receive counterparty tokens and the winning address holder will also get the gold bar shipped to them.
That does not answer my question, so I'm asking you again, what happens when the winning number is a number of a book that hasn't been sold. I know what you are going to do, but if a book isn't sold have nowhere to ship the bar to as nobody can claim it. Mitchell brings up a good point, the idea of unclaimable tickets. This was a problem in the original plan that we didn't consider, so the solution is that sales will simple have to stay open and even though some market participants will know the number of the book eligible for the gold, even if it didn't sell, some lucky person in the future will get it instead 1 day till the halving! Soooooo. You could possibly have to sell 5000 books until you might have a winner, at which point you know what the winning book is....dont know if you want to stick to that "plan" I evaluated other solutions to that problem, but those solutions had the side effect of revealing exactly how many books have sold which is useful to our competitors (Now that the spreadsheet is out, later buyers can extrapolate the same information), unless there is a version I didn't consider This has to further be balanced with the 24 hours remaining before the halving
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photonresearch (OP)
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July 08, 2016, 04:21:26 PM |
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digicoinuser
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July 08, 2016, 05:50:41 PM Last edit: July 08, 2016, 06:16:12 PM by digicoinuser |
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That's rough, if 500 people (high end estimate on my analysis) purchased the book before the raffle it would be a 10% chance that someone wins who has a book currently.
Once you know the number of the book that is to win I don't think there is anything stopping you from revealing it to a friend so they can purchase the winning combo. I'd see it as 100% fair if you only entered the purchasers who already bought the book, although I did mention my worries to purchasers on IRC immediately after the raffle was announced.
I didn't purchase it so I don't really have a say in how it's run, but I can provide my reasoning that selecting from book numbers that aren't sold doesn't make me want to buy a book since the odds are horrible in my eyes.
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photonresearch (OP)
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July 08, 2016, 07:05:51 PM Last edit: July 08, 2016, 09:19:24 PM by photonresearch |
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That's rough, if 500 people (high end estimate on my analysis) purchased the book before the raffle it would be a 10% chance that someone wins who has a book currently.
Once you know the number of the book that is to win I don't think there is anything stopping you from revealing it to a friend so they can purchase the winning combo. I'd see it as 100% fair if you only entered the purchasers who already bought the book, although I did mention my worries to purchasers on IRC immediately after the raffle was announced.
I didn't purchase it so I don't really have a say in how it's run, but I can provide my reasoning that selecting from book numbers that aren't sold doesn't make me want to buy a book since the odds are horrible in my eyes.
Hmmmmmmmmmmm Good points, then the only thing I could do to alleviate the businesses own concerns is to have a less probably fair raffle with me as the intermediary It simply involves changing the random number generator to a lower number than 5,000 but not saying what that number is Thinking about this more that seems like a decent compromise, but makes the whole spreadsheet thing not necessary, thought I would be able to avoid seemingly arbitrary last minute changes
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monkeynuts
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July 08, 2016, 10:02:40 PM |
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I am sure it was previously stated that the book would only be sold until the halving
I also am sure that it was stated that not all 5000 books had been printed yet, indeed far from it, and that 5000 was the absolute max that would be done. So inference is that only as many books would be printed, as had indeed been sold
I struggle to see how it can be justified to base the raffle on anything other than actual books sold pre halving. Anything else has the appearance of shifting the goal posts
Will the mystery lower number for the raffle be the real number of books sold pre halving ?
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photonresearch (OP)
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July 09, 2016, 12:13:36 AM |
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Will the mystery lower number for the raffle be the real number of books sold pre halving ?
Would be theoretical possible to guess that number by trying any number from 2 to 5000 in the rng function I posted, and getting the same winning number that my server posts Unless multiple max numbers return the same winning number using the block hash. A little technical but does that make sense? From my perspective it doesn't seem possible for me to Introduce concerns
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photonresearch (OP)
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July 09, 2016, 01:53:42 AM Last edit: July 09, 2016, 02:34:54 AM by photonresearch |
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I am sure it was previously stated that the book would only be sold until the halving
I also am sure that it was stated that not all 5000 books had been printed yet, indeed far from it, and that 5000 was the absolute max that would be done. So inference is that only as many books would be printed, as had indeed been sold
I struggle to see how it can be justified to base the raffle on anything other than actual books sold pre halving. Anything else has the appearance of shifting the goal posts
Will the mystery lower number for the raffle be the real number of books sold pre halving ?
This would be solved with a 3-character hash draw. You have 3360 possible selections, which is far greater than the books sold thus far I would imagine. Considering that the halvening is in less than 100 blocks and the books are not being sold after this, it would be an easy, fair solution. I will even handle all the entries if needed. Thanks I like having a little mystery to it with what the max number is
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Chainsaw
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July 09, 2016, 01:54:02 PM |
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I don't have a horse in this race, but I would strongly suggest defining this unambiguously before the end of raffle, or you may have a giant mess on your hands.
The draw need only be done once. So long as the algorithm is defined and not changed after the hash is announced, there is no rush. Manual intervention could take place.
If I had purchased a raffle entry for any raffle, and later found out that 'not enough tickets were sold so we'll leave it open if/until they are closed', I would likely be contacting authorities for fraud. You have a fair number of people with money on the line stating this.
Your algorithm could simply perform a modulus of the generated chosen number from 1 to 5000, against the number of entrants. That way if a number over the number of tickets sold occurred, it would just wrap back into the number of tickets actually sold.
For example, of 5000 tickets, let's say 215 copies sold. The random number is 3,183. Rather than wait for almost 3,000 more copies of the book to be sold, simply take 1 + (3183 % 215) and get 174.
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photonresearch (OP)
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July 09, 2016, 03:22:19 PM |
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I am sure it was previously stated that the book would only be sold until the halving
I also am sure that it was stated that not all 5000 books had been printed yet, indeed far from it, and that 5000 was the absolute max that would be done. So inference is that only as many books would be printed, as had indeed been sold
I struggle to see how it can be justified to base the raffle on anything other than actual books sold pre halving. Anything else has the appearance of shifting the goal posts
Will the mystery lower number for the raffle be the real number of books sold pre halving ?
This would be solved with a 3-character hash draw. You have 3360 possible selections, which is far greater than the books sold thus far I would imagine. Considering that the halvening is in less than 100 blocks and the books are not being sold after this, it would be an easy, fair solution. I will even handle all the entries if needed. Thanks I like having a little mystery to it with what the max number is Mystery is fine, but we are asking for transparency. I would take the advice from above before this turns into a crap-storm of unhappy people You have about 3 hours to decide. I decided to change the max number down from 5000, increasing the odds for everyone and removing the possibility of unclaimed tickets and removing the possibility of keeping sales open Everyone will be able run the code and change the RNG function max to get an idea of what the max number was by trying every max number from 2 to 5000. This should actually create a distribution of possible winning book #s that reveal a lot about sales, and simultaneously prove there were no improprieties. Let me know your perspective, it is possible to retrieve block 420,000 at any time (especially in event the first block 420,000 gets orphaned and replaced by a longer chain) From my perspective: greater odds for everyone, nobody unhappy = good. Let me know if your perspective is different, or why the 3 letter hash / modulus is better
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coin@coin
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July 10, 2016, 09:03:40 AM |
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Fortify
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July 10, 2016, 09:21:49 AM |
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Asking for advice on how the drawing should proceed and now someone is already claiming a win? Seems like some terrible planning. It makes way more sense to settle on an algorithm first. I evaluated other solutions to that problem, but those solutions had the side effect of revealing exactly how many books have sold which is useful to our competitors (Now that the spreadsheet is out, later buyers can extrapolate the same information), unless there is a version I didn't consider
Competitors? This looks like a one time art piece with a quirky selling point (Casascius raffle). Who exactly would benefit from knowing the amount of books you sold? The major selling point that seemed to attract a lot of people to your overpriced book is the possibility of winning a rare item and the assumption you are unlikely to have another.
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Zepher
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July 10, 2016, 12:35:23 PM |
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Correct me if I'm missing something here, but a lot of this isn't adding up. According to the link provided by proton for the draw, the winning book number is #13.. Now.. How do we know who has book 13 if buyers have not received their counter party tokens? Edit: How is this "Provably Fair"?
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Mitchell
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July 10, 2016, 08:00:30 PM |
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Correct me if I'm missing something here, but a lot of this isn't adding up. According to the link provided by proton for the draw, the winning book number is #13.. Now.. How do we know who has book 13 if buyers have not received their counter party tokens? Edit: How is this "Provably Fair"? #13? How did you get #13? This is the result I got from the the script: { 'block height': 420000, hash': '000000000000000002cce816c0ab2c5c269cb081896b7dcb34b8422d6b74ffa1', book number': 3291 } The picking is fair (not sure if it's provably fair) as it uses a blockchain hash which cannot be known until it exists. As the random function always uses the same seed, the first number will always be the same (allowing us to verify the result). I just ran the script three times to proof this:
![](https://ip.bitcointalk.org/?u=https%3A%2F%2Fi.imgur.com%2FngTKudK.jpg&t=665&c=PvuvfAOXpzDlrw) This is my (very very slightly edited script): var request = require('request'); var RNG = require('rng-js'); var blockNumber = 420000; request({ url : 'https://bitcoin.toshi.io/api/v0/blocks/' + blockNumber, method : 'GET', }, function (error, response, body) { if (response.statusCode == '404') { console.log({ "block height" : "block 420,000 is not yet available", "block hash" : "please check in the future", "winning book number" : "n/a" }); } else { if (error) { console.log('Error sending message: ', error); } else if (response.body.error) { console.log('Error: ', response.body.error); } else { var blockInfo = JSON.parse(response.body); var rng = new RNG(blockInfo.hash); console.log({ "block height" : blockNumber, "block hash" : blockInfo.hash, "winning book number" : rng.random(1, 5000) }); } } });
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Zepher
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July 10, 2016, 09:56:04 PM |
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Hmmm, apologies. Thanks Mitchell. The problem is that no one has received counterparty tokens from what I've heard, so they don't even know what book number they have. This is an error in my opinion.
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My only payment address: 1ZephertJThxkHih7XcaUHBkMSnvkTt5u
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