Topic: 🚀 Sportsbet.io - Proud partner of Southampton FC 🚀 - page 1048. (Read 410277 times)
Although Darkstar has come up with a nice and simple method,there are still a few problems with it.Let me update this post later.I'm thinking of implementing one of the known and already proved Random Number Generator Algorithms.
Everyone is good with this method from DarkStar_?
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
Shouldn't you apply modulus (number of tickets)? Why - 1?
Without -1: Range of 0 - # Of Tickets - 1
With -1: Range of 1 - # Of Tickets
Edit: nvm, just add one the the answer. My bad.
Sportsbet.io, have you decided whether you are going to use multiple block hashs, more digits or a combo of both?
Everyone is good with this method from DarkStar_?
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
Shouldn't you apply modulus (number of tickets)? Why - 1?
The final promotional Lottery draw of 2016 Olympics is now over. Too bad I did not win anything this time.
Now I am waiting for sportsbet to settle my bets, because I still have 2 Olympic bets set as 'Pending'.
Now I am waiting for sportsbet to settle my bets, because I still have 2 Olympic bets set as 'Pending'.
All pending bets are now settled. Can you please let me know if you are all set?
I assume you meant "now" processed and settled
Quote
I have a buddy who could build this as long as he gets to publish it open source on Github, send me a PM and we'll talk
Perfect! I'll shoot you a message
The final promotional Lottery draw of 2016 Olympics is now over. Too bad I did not win anything this time.
Now I am waiting for sportsbet to settle my bets, because I still have 2 Olympic bets set as 'Pending'.
Now I am waiting for sportsbet to settle my bets, because I still have 2 Olympic bets set as 'Pending'.
All pending bets are now settled. Can you please let me know if you are all set?
The final promotional Lottery draw of 2016 Olympics is now over. Too bad I did not win anything this time.
Now I am waiting for sportsbet to settle my bets, because I still have 2 Olympic bets set as 'Pending'.
Now I am waiting for sportsbet to settle my bets, because I still have 2 Olympic bets set as 'Pending'.
All pending bets are now settled. Can you please let me know if you are all set?
Everyone is good with this method from DarkStar_?
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
Here is an example of the method I outlined:
For this example, I will be using the block that was mined right before this post (block 426230). Normally, a block that hasn't been mined would be used, but for convenience, I will use an already mined one.
The has of the block 426230 is
Code:
000000000000000000dd8a7bb73190bd8574df5f18966347a2094049e28c8cab
The last 7 characters are 28c8cab, and when converted to a decimal gives 42765483. Here is my imaginary ticket list (published before the block is mined)
Code:
1: darkstar
2: person2
3: person3
4: person4
5: person5
42765483 % 4 gives 3. 3+1 = 4, so person 4 would have won the 1 BTC prize in this case.
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
Here is an example of the method I outlined:
For this example, I will be using the block that was mined right before this post (block 426230). Normally, a block that hasn't been mined would be used, but for convenience, I will use an already mined one.
The has of the block 426230 is
Code:
000000000000000000dd8a7bb73190bd8574df5f18966347a2094049e28c8cab
The last 7 characters are 28c8cab, and when converted to a decimal gives 42765483. Here is my imaginary ticket list (published before the block is mined)
Code:
1: darkstar
2: person2
3: person3
4: person4
5: person5
42765483 % 4 gives 3. 3+1 = 4, so person 4 would have won the 1 BTC prize in this case.
I have a buddy who could build this as long as he gets to publish it open source on Github, send me a PM and we'll talk
Everyone is good with this method from DarkStar_?
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
Here is an example of the method I outlined:
For this example, I will be using the block that was mined right before this post (block 426230). Normally, a block that hasn't been mined would be used, but for convenience, I will use an already mined one.
The has of the block 426230 is
Code:
000000000000000000dd8a7bb73190bd8574df5f18966347a2094049e28c8cab
The last 7 characters are 28c8cab, and when converted to a decimal gives 42765483. Here is my imaginary ticket list (published before the block is mined)
Code:
1: darkstar
2: person2
3: person3
4: person4
5: person5
42765483 % 4 gives 3. 3+1 = 4, so person 4 would have won the 1 BTC prize in this case.
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
Here is an example of the method I outlined:
For this example, I will be using the block that was mined right before this post (block 426230). Normally, a block that hasn't been mined would be used, but for convenience, I will use an already mined one.
The has of the block 426230 is
Code:
000000000000000000dd8a7bb73190bd8574df5f18966347a2094049e28c8cab
The last 7 characters are 28c8cab, and when converted to a decimal gives 42765483. Here is my imaginary ticket list (published before the block is mined)
Code:
1: darkstar
2: person2
3: person3
4: person4
5: person5
42765483 % 4 gives 3. 3+1 = 4, so person 4 would have won the 1 BTC prize in this case.
Just a typo. You ought to say 'get your English right'. Grammar is to do with structure of sentences and stuff as far as I know. This puts you in the same position. 
Now,I can see personal hate coming out of that post.Did I hurt your emotions intentionally/unintentionally ?huh
I have heard that accusation before.
I don't think the lottery is provably fair or it can ever be made provably fair.
Well,you really need to stop living under the rock and get your facts right.Provable/Verifiable lotteries do exists.What have you known about the RNG's ? There are more than 100 algos precisely crafted out by our so called "Mathematicians".You ought to prove all of them wrong by saying "it can ever be made fair".Do research about the RNG's,website devs can implement one of the existing algos to propose fairness.I don't think the lottery is provably fair or it can ever be made provably fair.
I am sorry, I believe you are wrong there. (Advance apology. not hurting your feelings
)Not possible because you cannot verify a username is real or fake. Making this truly provably fair requires a very high level of transparency.
You can have some fancy blocks and hashes and any other thing you like, it won't make the lottery 100% provably fair.
Well something is better than nothing. Any attempt on making it better is appreciated.
I am sure Sportsbet offers better odds than Directbet.
Why are you backing up the site without concluding the facts ? Are you associated with them ?Their spokesperson ? Or just the urge to have a negative opinions against whatever is doubted about the website ? My statement is correct, isn't it?
I see the EPL leaderboard is updated already, but seems that it is only updated for the top positions only. I have won some mbtc last Saturday but my point is still the same with the last week update. Means that my winning last Saturday is not count towards the leaderboard? Or perhaps the update is not finished yet?
Should be all updated
The funny is, people who won the raffle still complaining about the raffle and me, who never win a single raffle not complaining, rofl.
I never win in the raffle, but not complain too
maybe I'm not enough luck in this lottery.
Anyone saw the olympic 17-21 august lottery redraw in 5 hours or so in the site? I cant figure out how to get any ticket to join these since all of tickets has been used on the previous lottery and Im not sure why is there a redraw again this time. Well with the olympic ended there is no way that somone could get ticket anymore but I saw around 170+ tickets there
edit : yep monsterv beat me to it
edit : yep monsterv beat me to it
I think the redraw lottery is for those who bet yesterday but did not receive the tickets for the last lottery (17-21). I received one ticket for the redraw, and I believe I received it for my bets on the Final Basketball match.
Anyone saw the olympic 17-21 august lottery redraw in 5 hours or so in the site? I cant figure out how to get any ticket to join these since all of tickets has been used on the previous lottery and Im not sure why is there a redraw again this time. Well with the olympic ended there is no way that somone could get ticket anymore but I saw around 170+ tickets there
edit : yep monsterv beat me to it
edit : yep monsterv beat me to it
The funny is, people who won the raffle still complaining about the raffle and me, who never win a single raffle not complaining, rofl.
Maybe they only give a suggestion I also never won the lottery in this period and I feel fine
By the way lottery still available ? I see in sportsbet.io (21:08 Olympics Redraw) Prize 1500 mbtc
How to participate in the lottery is no games available in Olympics
The funny is, people who won the raffle still complaining about the raffle and me, who never win a single raffle not complaining, rofl.
I see the EPL leaderboard is updated already, but seems that it is only updated for the top positions only. I have won some mbtc last Saturday but my point is still the same with the last week update. Means that my winning last Saturday is not count towards the leaderboard? Or perhaps the update is not finished yet?
The final promotional Lottery draw of 2016 Olympics is now over. Too bad I did not win anything this time.
Now I am waiting for sportsbet to settle my bets, because I still have 2 Olympic bets set as 'Pending'.
Now I am waiting for sportsbet to settle my bets, because I still have 2 Olympic bets set as 'Pending'.
All pending bets are now settled. Can you please let me know if you are all set?
- snip-
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
- snip -
Here is an example of the method I outlined:
For this example, I will be using the block that was mined right before this post (block 426230). Normally, a block that hasn't been mined would be used, but for convenience, I will use an already mined one.
The has of the block 426230 is
The last 7 characters are 28c8cab, and when converted to a decimal gives 42765483. Here is my imaginary ticket list (published before the block is mined)
42765483 % 4 gives 3. 3+1 = 4, so person 4 would have won the 1 BTC prize in this case.
Here is an example of the method I outlined:
For this example, I will be using the block that was mined right before this post (block 426230). Normally, a block that hasn't been mined would be used, but for convenience, I will use an already mined one.
The has of the block 426230 is
Code:
000000000000000000dd8a7bb73190bd8574df5f18966347a2094049e28c8cab
Code:
1: darkstar
2: person2
3: person3
4: person4
5: person5
Am ok with this method.
In my opinion now it's needed a clear specification on how this method will go on for all the prizes paid (26 places if I remember correctly)
It has the features needed to silence all the controversy and speculations that have been raised (also by me to be honest) : fairness and trasparency.
This is way better than the rudimental one I suggested (video recording of random.org draw).
How does everyone else feel about this method?
- snip-
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
1. Publish the list of tickets and who has which ticket, a bit before the actual draw. Make sure that tickets are numbered.
2. Wait for a specific block hash for the drawing. Once the block comes, take the last few digits of the block hash (maybe 7?) and convert it to a decimal, by using it as a hexadecimal. Next, use the number that you get and apply modulus (number of tickets - 1) to that number. You should get a number between 0 and the number of tickets, minus one. Add one to the number, and you have your first winner. You can get the other winners in other blocks, or if that takes too long, use the last 8 digits, than 9, e.t.c repeating the entire process I just outlined.
- snip -
Here is an example of the method I outlined:
For this example, I will be using the block that was mined right before this post (block 426230). Normally, a block that hasn't been mined would be used, but for convenience, I will use an already mined one.
The has of the block 426230 is
The last 7 characters are 28c8cab, and when converted to a decimal gives 42765483. Here is my imaginary ticket list (published before the block is mined)
42765483 % 4 gives 3. 3+1 = 4, so person 4 would have won the 1 BTC prize in this case.
Here is an example of the method I outlined:
For this example, I will be using the block that was mined right before this post (block 426230). Normally, a block that hasn't been mined would be used, but for convenience, I will use an already mined one.
The has of the block 426230 is
Code:
000000000000000000dd8a7bb73190bd8574df5f18966347a2094049e28c8cab
Code:
1: darkstar
2: person2
3: person3
4: person4
5: person5
Am ok with this method.
In my opinion now it's needed a clear specification on how this method will go on for all the prizes paid (26 places if I remember correctly)
It has the features needed to silence all the controversy and speculations that have been raised (also by me to be honest) : fairness and trasparency.
This is way better than the rudimental one I suggested (video recording of random.org draw).
Jump to: