Got my BFL Single today and I'm raffling it away for 0.5BTC!

[Matthew N. Wright]

Okay so what block number are we on right now?

[wogaut]

171436
...so I'd choose like two in the future whenever the spreadsheet is finished

[DeathAndTaxes]

171436
...so I'd choose like two in the future whenever the spreadsheet is finished

I would make it further out.  The further the block is out the more people who can review the list and possibly spot errors.  Would suck if there is an error in the list, someone wins and then people realize the error(s).  Do you have it stand, go again, etc.

I would say once list is published announce block # and make it 24 blocks out (~ 4 hours).

[wogaut]

171436
...so I'd choose like two in the future whenever the spreadsheet is finished

I guess I could wait one hour more too *sohardtowait*

[kakobrekla]

171436
...so I'd choose like two in the future whenever the spreadsheet is finished

I would make it further out.  The further the block is out the more people who can review the list and possibly spot errors.  Would suck if there is an error in the list, someone wins and then people realize the error(s).  Do you have it stand, go again, etc.

I would say once list is published announce block # and make it 24 blocks out (~ 4 hours).

I support, at least 24. Makes a round number. 171460

[Matthew N. Wright]

171436
...so I'd choose like two in the future whenever the spreadsheet is finished

Spreadsheet is finished.

https://docs.google.com/spreadsheet/ccc?key=0Ajtx05YrHtIydFpNdjFuQ0diUTkxV3ZTcnczb2RJN0E

The original, unhashed version (which includes invoices for verification at Bit-Pay has been sent to Theymos for records purposes and future scammer investigations, which I'm sure some sore loser will try to instigate at one point or another).


EDIT: Forgot to mention this, but I converted all of the email addresses to lowercase (wtf@people using different casing in email addresses anyway) so keep that in mind when searching for your address. Since I've already received a dozen complaints that someone couldn't find their email address and I've had to hold their hand and show them how, I won't be responding to complaints of that nature in the future.

[wogaut]

171436
...so I'd choose like two in the future whenever the spreadsheet is finished

Spreadsheet is finished.

Great, maybe we could agree on a block, like 171450, also a very round number in the future. Plenty of time to check the sheet until then.

[filharvey]

You sorted the list by the SHA, not sure I like that. I had 2 purchases sperated by quite a bit of time, I like the idea of the tickets for those not being in 1 block and 2 blocks of numbers.

Phil

[SgtSpike]

You sorted the list by the SHA, not sure I like that. I had 2 purchases sperated by quite a bit of time, I like the idea of the tickets for those not being in 1 block and 2 blocks of numbers.

Phil

It's exactly random and the same number of chances regardless of where they end up in the list...    

[Epoch]

You sorted the list by the SHA, not sure I like that. I had 2 purchases sperated by quite a bit of time, I like the idea of the tickets for those not being in 1 block and 2 blocks of numbers.

Phil

You do realize that it makes absolutely no difference statistically, right? Your odds of winning (or losing) are identical regardless of where or how your tickets are split within the total range.

[Matthew N. Wright]

You sorted the list by the SHA, not sure I like that. I had 2 purchases sperated by quite a bit of time, I like the idea of the tickets for those not being in 1 block and 2 blocks of numbers.

Phil

I sorted the list by email addresses actually. The SHA was after-the-fact.

[DeathAndTaxes]

Son what is the magic block #?

I need to set my block clock.  

[filharvey]

You sorted the list by the SHA, not sure I like that. I had 2 purchases sperated by quite a bit of time, I like the idea of the tickets for those not being in 1 block and 2 blocks of numbers.

Phil

You do realize that it makes absolutely no difference statistically, right? Your odds of winning (or losing) are identical regardless of where or how your tickets are split within the total range.

I realize that, I just generally prefer to have different blocks of tickets / numbers. Just my personal view.

Lets not stop the process. Lets move ahead of how it is currently.

Phil

[wogaut]

Son what is the magic block #?

I need to set my block clock.  

Same here 

[Matthew N. Wright]

The official block to determine the winning ticket number will be 171,450.

[Epoch]

The official block to determine the winning ticket number will be 171,450.

Set!

[theymos]

(last-10-digits-of-block-hash) % (number-of-tickets) = winning ticket number

The results of this aren't evenly distributed in most cases.

Say the random number is between 0 and 7, and there are 5 tickets from 0-4. Then the possible results are:
0 mod 5 = 0 (0 is the winner)
1 mod 5 = 1
2 mod 5 = 2
3 mod 5 = 3
4 mod 5 = 4
5 mod 5 = 0
6 mod 5 = 1
7 mod 5 = 2
As you can see, tickets 0-2 have a better chance than 3-4

For proper randomness, you actually need to re-roll when the random number is greater than your range of winners. This could be done by hashing the block hash again or choosing a different block. (There are probably other algorithms for properly distributing the randomness, but AFAIK they're always less efficient when generating random numbers and I don't know any such algorithms off-hand.)

[DeathAndTaxes]

Good point therymos although you just need to re-roll when the random number is > the largest even MULTIPLE of the ticket #.

i.e. there are 40270 tickets.  So for 0-40270 it is even.  0 to 2x(40270) is even 0 to n x (40270) is even.

The only issue is the "remainder" greater than the max multiple.

10 hex digits = 16^10 = 1099511627776


1099511627776 / 40270 = 27303492.12  (that last 0.12 is unfair)

So max valid random number is 40270 * 27303492 = 1099511622840

Matt a "fix" is simple and fair.

If the value of last 10 digits of the block hash is > 1099511622840 then it is a "no-go" and you use the next block.

Don't worry the odds of that happening are very small.  Only 1099511622841 to 1099511627776 are "bad".  4936 unfair numbers out of 1099511627776 or 0.0000004%

[Matthew N. Wright]

If you guys complicate this any further I swear I'm going into PHP, typing

Code:

echo rand(0,41269);

and finishing this like a boss.

[theymos]

Good point therymos although you just need to re-roll when the random number is > the largest even MULTIPLE of the ticket #.

Right. I was assuming that the max random number would be chosen to be the smallest power of two larger than the largest ticket.