[Closed]

[ezeminer]

Closed

[cryptoheadd]

I am planning to do this raffle soon. It's an everyone wins kind of raffle. It will involve roughly 15BTC in prizes.

Thanks to TheNewAnon135246 for the idea.

How much are you planning to charge per ticket?

[ezeminer]

Somewhere around 1BTC depending on how big the prize pool is. total goodies I estimated at 17BTC currently.

[owlcatz]

I'm with minerjones on this one - 1 btc is a lot to many of us - I am still trying to figure out if I want to buy one of those new BTCC titanium coins as well, so now I'm on hold for a bit, until I see what your prizes may be... Looking forward to that part!? 

[ezeminer]

Yes. I would play, but maybe make the tickets a little lower or more tickets...

thanks

I can probably get it down to 0.5 per ticket with 32 tickets, I'll just need to find more prizes.

[dazedfool]

Yes. I would play, but maybe make the tickets a little lower or more tickets...

thanks

I can probably get it down to 0.5 per ticket with 32 tickets, I'll just need to find more prizes.

I can supply some prizes if you would like....

I think if you broke up the Cas set and added some other fun prizes, you'd get more interest

[chronicsky]

I will play too

So everyone wins

[MRBONG411]

As always love the raffles but I will probably be priced out of this one.

Will be awesome to watch though.

[ezeminer]

Yes. I would play, but maybe make the tickets a little lower or more tickets...

thanks

I can probably get it down to 0.5 per ticket with 32 tickets, I'll just need to find more prizes.

I can supply some prizes if you would like....

I think if you broke up the Cas set and added some other fun prizes, you'd get more interest

I will likely do that, just need to get more prizes to fill the last slots.

I will play too

So everyone wins

That is what I would like to do.

[chronicsky]

Great

Will be know the prices of the raffle or will it be a mystery??
I'll  prefer to know them

[ezeminer]

Great

Will be know the prices of the raffle or will it be a mystery??
I'll  prefer to know them

I'll release a list of all the prizes.

[chronicsky]

Great

Will be know the prices of the raffle or will it be a mystery??
I'll  prefer to know them

I'll release a list of all the prizes.

Great!
look forward to it!

Also , i hope that since ticket cost will be high and it might take time to fill all tickets - so probably just register name and then when all filled - take payment ??

Just what i think

[ezeminer]

Great

Will be know the prices of the raffle or will it be a mystery??
I'll  prefer to know them

I'll release a list of all the prizes.

Great!
look forward to it!

Also , i hope that since ticket cost will be high and it might take time to fill all tickets - so probably just register name and then when all filled - take payment ??

Just what i think

Registering names wouldn't be a bad idea.

I'm also trying to find other fair ways to generate numbers other than blockchain. At the moment I could do 25 prizes varying from 0.06BTC to 4.0BTC. But it wouldn't evenly distribute chances among block hashes.

For example (Block hash ending in):
0-8 0 - entrant A - meaning 00, 10, 20, 30, 40, 50, 60, 70, 80
9-f 0 - entrant B - meaning 90, a0, b0, c0, d0, e0, f0
0-8 1 - entrant C
9-f 1 - entrant D

This would generate 32 tickets with even probabilities of winning.

[miffman]

Great

Will be know the prices of the raffle or will it be a mystery??
I'll  prefer to know them

I'll release a list of all the prizes.

Great!
look forward to it!

Also , i hope that since ticket cost will be high and it might take time to fill all tickets - so probably just register name and then when all filled - take payment ??

Just what i think

Registering names wouldn't be a bad idea.

I'm also trying to find other fair ways to generate numbers other than blockchain. At the moment I could do 25 prizes varying from 0.06BTC to 4.0BTC.

Personally, I wouldn't spend 1BTC for a chance to get a 0.06 prize, even if I could win a 4BTC prize.

[ezeminer]

Great

Will be know the prices of the raffle or will it be a mystery??
I'll  prefer to know them

I'll release a list of all the prizes.

Great!
look forward to it!

Also , i hope that since ticket cost will be high and it might take time to fill all tickets - so probably just register name and then when all filled - take payment ??

Just what i think

Registering names wouldn't be a bad idea.

I'm also trying to find other fair ways to generate numbers other than blockchain. At the moment I could do 25 prizes varying from 0.06BTC to 4.0BTC.

Personally, I wouldn't spend 1BTC for a chance to get a 0.06 prize, even if I could win a 4BTC prize.

Yea, I can understand that. So far with 25 prizes, tickets would be 0.64, if I had a fair way of generating numbers. Meaning I cannot alter the results.

I tried to run raffles for BTC in the games and rounds section, my way was to upload a small video of each drawing using a bitcoin gambling site for number generation and hashes as proof.

[chronicsky]

Great

Will be know the prices of the raffle or will it be a mystery??
I'll  prefer to know them

I'll release a list of all the prizes.

Great!
look forward to it!

Also , i hope that since ticket cost will be high and it might take time to fill all tickets - so probably just register name and then when all filled - take payment ??

Just what i think

Registering names wouldn't be a bad idea.

I'm also trying to find other fair ways to generate numbers other than blockchain. At the moment I could do 25 prizes varying from 0.06BTC to 4.0BTC. But it wouldn't evenly distribute chances among block hashes.

For example (Block hash ending in):
0-8 0 - entrant A - meaning 00, 10, 20, 30, 40, 50, 60, 70, 80
9-f 0 - entrant B - meaning 90, a0, b0, c0, d0, e0, f0
0-8 1 - entrant C
9-f 1 - entrant D

This would generate 32 tickets with even probabilities of winning.

To tell you i didn't understood it at all

[ezeminer]

Great

Will be know the prices of the raffle or will it be a mystery??
I'll  prefer to know them

I'll release a list of all the prizes.

Great!
look forward to it!

Also , i hope that since ticket cost will be high and it might take time to fill all tickets - so probably just register name and then when all filled - take payment ??

Just what i think

Registering names wouldn't be a bad idea.

I'm also trying to find other fair ways to generate numbers other than blockchain. At the moment I could do 25 prizes varying from 0.06BTC to 4.0BTC. But it wouldn't evenly distribute chances among block hashes.

For example (Block hash ending in):
0-8 0 - entrant A - meaning 00, 10, 20, 30, 40, 50, 60, 70, 80
9-f 0 - entrant B - meaning 90, a0, b0, c0, d0, e0, f0
0-8 1 - entrant C
9-f 1 - entrant D

This would generate 32 tickets with even probabilities of winning.

To tell you i didn't understood it at all

I'll use real blocks as an example:

Say the raffle used block 413332 then 413333 and so on and so on.
Block #413333, the hash ended in "4d"
Block #413332 ended in "50"

So if you picked "0-8 0" you would get the first prize.
If you picked "0-8 d" you would get the second prize.

The entry list looks like this:

Code:

0-7 0 - Person A
8-f 0 -
0-7 1 -
8-f 1 -
0-7 2 -
8-f 2 -
0-7 3 -
8-f 3 -
0-7 4 -
8-f 4 -
0-7 5 -
8-f 5 -
0-7 6 -
8-f 6 -
0-7 7 -
8-f 7 -
0-7 8 -
8-f 8 -
0-7 9 -
8-f 9 -
0-7 a -
8-f a -
0-7 b -
8-f b -
0-7 c -
8-f c -
0-7 d - Person B
8-f d -
0-7 e -
8-f e -
0-7 f -
8-f f -

[chronicsky]

Great

Will be know the prices of the raffle or will it be a mystery??
I'll  prefer to know them

I'll release a list of all the prizes.

Great!
look forward to it!

Also , i hope that since ticket cost will be high and it might take time to fill all tickets - so probably just register name and then when all filled - take payment ??

Just what i think

Registering names wouldn't be a bad idea.

I'm also trying to find other fair ways to generate numbers other than blockchain. At the moment I could do 25 prizes varying from 0.06BTC to 4.0BTC. But it wouldn't evenly distribute chances among block hashes.

For example (Block hash ending in):
0-8 0 - entrant A - meaning 00, 10, 20, 30, 40, 50, 60, 70, 80
9-f 0 - entrant B - meaning 90, a0, b0, c0, d0, e0, f0
0-8 1 - entrant C
9-f 1 - entrant D

This would generate 32 tickets with even probabilities of winning.

To tell you i didn't understood it at all

I'll use real blocks as an example:

Say the raffle used block 413332 then 413333 and so on and so on.
Block #413333, the hash ended in "4d"
Block #413332 ended in "50"

So if you picked "0-8 0" you would get the first prize.
If you picked "0-8 d" you would get the second prize.

The entry list looks like this:

Code:

0-8 0 - Person A
9-f 0 -
0-8 1 -
9-f 1 -
0-8 2 -
9-f 2 -
0-8 3 -
9-f 3 -
0-8 4 -
9-f 4 -
0-8 5 -
9-f 5 -
0-8 6 -
9-f 6 -
0-8 7 -
9-f 7 -
0-8 8 -
9-f 8 -
0-8 9 -
9-f 9 -
0-8 a -
9-f a -
0-8 b -
9-f b -
0-8 c -
9-f c -
0-8 d - Person B
9-f d -
0-8 e -
9-f e -
0-8 f -
9-f f -

Looks awesome!!

Look forward to what the prices will be

[TookDk]

For example (Block hash ending in):
0-8 0 - entrant A - meaning 00, 10, 20, 30, 40, 50, 60, 70, 80
9-f 0 - entrant B - meaning 90, a0, b0, c0, d0, e0, f0
0-8 1 - entrant C
9-f 1 - entrant D

This would generate 32 tickets with even probabilities of winning.

Your hex calculations are broken.
There would be significant higher probability to hit entrant A and C versus entrant B and D.

Instead use
0-7 and 8-F for the second last nibble.

[ezeminer]

For example (Block hash ending in):
0-8 0 - entrant A - meaning 00, 10, 20, 30, 40, 50, 60, 70, 80
9-f 0 - entrant B - meaning 90, a0, b0, c0, d0, e0, f0
0-8 1 - entrant C
9-f 1 - entrant D

This would generate 32 tickets with even probabilities of winning.

Your hex calculations are broken.
There would be significant higher probability to hit entrant A and C versus entrant B and D.

Instead use
0-7 and 8-F for the second last nibble.

Thanks, I just noticed that when you mentioned it. Will update