Got my BFL Single today and I'm raffling it away for 0.5BTC! - page 7.

DeathAndTaxes

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Gerald Davis

Quote from: Matthew N. Wright on March 16, 2012, 03:32:37 PM

I totally agree. I want to learn this, but it still doesn't make sense to me completely, and that bothers me since I'm usually a fast mover on things. When the block comes around can you show me hands on what has to happen again?

I can tell you now to avoid it being after the fact.

Step 1) Take the last 10 digits of the block hash (in hex)
Step 2) Convert that to a decimal number.
Step 3a) If the number is 1099511622840 or larger the draw is void. Repeat steps #1 to #3 on the next block.
Step 3b) If the number is < 1099511622840 then you have a valid random number. Continue to step #4
Step 4) Winning Ticket = (random number) mod 40270

The only change is "Voiding" random # 1099511622840 and up because they are "uneven". The odds of that happening are ~ 1 in 222 million.

Like I said if you do something else I won't mind as I trust you. I just think it is useful example of using the blockchain to avoid needing implicit trust. Kinda an extension of the rational for creation Bitcoin in the first place.

Epoch

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Quote from: Matthew N. Wright on March 16, 2012, 03:32:37 PM

Quote from: DeathAndTaxes on March 16, 2012, 03:31:08 PM

Your call Matt. I only have 2 tickets so I don't really care. It would be useful excercise in using the block chain to ensure no ability to cheat. Random.Org is only random if you actually make it random not pick a buddy as the winner.

Not saying you would but using the blockchain as a double blind picking method reduces the need for implicit trust.

I totally agree. I want to learn this, but it still doesn't make sense to me completely, and that bothers me since I'm usually a fast mover on things. When the block comes around can you show me hands on what has to happen again?

Excellent decision on using the blockchain. It is a natural, and full transparent, random number generator.

wogaut

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Quote from: Matthew N. Wright on March 16, 2012, 03:32:37 PM

Quote from: DeathAndTaxes on March 16, 2012, 03:31:08 PM

Your call Matt. I only have 2 tickets so I don't really care. It would be useful excercise in using the block chain to ensure no ability to cheat. Random.Org is only random if you actually make it random not pick a buddy as the winner.

Not saying you would but using the blockchain as a double blind picking method reduces the need for implicit trust.

I totally agree. I want to learn this, but it still doesn't make sense to me completely, and that bothers me since I'm usually a fast mover on things. When the block comes around can you show me hands on what has to happen again?

Fair enough, I am pretty sure there are plenty of people who will rave over it!

It will make the random number more special! Grin

DeathAndTaxes

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Gerald Davis

Quote from: wogaut on March 16, 2012, 03:29:50 PM

Just want to add, if the block hash is < 1099511622840 it is good, which means that 1099511622840 is repeating ticket number 0, (a consequence of the mod operation).

Good catch. Will update.

Matthew N. Wright

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Quote from: DeathAndTaxes on March 16, 2012, 03:31:08 PM

Your call Matt. I only have 2 tickets so I don't really care. It would be useful excercise in using the block chain to ensure no ability to cheat. Random.Org is only random if you actually make it random not pick a buddy as the winner.

Not saying you would but using the blockchain as a double blind picking method reduces the need for implicit trust.

I totally agree. I want to learn this, but it still doesn't make sense to me completely, and that bothers me since I'm usually a fast mover on things. When the block comes around can you show me hands on what has to happen again?

wogaut

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Quote from: Matthew N. Wright on March 16, 2012, 03:23:22 PM

The raffle rules stated that:

Your email address would be published
The winner would be chosen by yesterday
Random.org would be used to choose the winner

So far I've done none of those. I'm all for learning and making this a better experience for everyone, but understand that every modification makes me feel like I'm straying even further from keeping my word.

I appreciate sticking around and helping everyone, and on that note, keep it as simple as possible for those who cant 2 math.

Matt, I know there's no voting on a raffle that you made, so it's really up to you.

But IMO for many of us it would be more fun if the block-chain is used for determination of the outcome.

It's still simple enough to replicate

DeathAndTaxes

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Gerald Davis

Your call Matt. I only have 2 tickets so I don't really care. It would be useful excercise in using the block chain to ensure no ability to cheat. Random.Org is only random if you actually make it random not pick a buddy as the winner.

Not saying you would but using the blockchain as a double blind picking method reduces the need for implicit trust.

wogaut

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Quote from: DeathAndTaxes on March 16, 2012, 03:21:13 PM

Quote from: Matthew N. Wright on March 16, 2012, 03:15:11 PM

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.

Don't do that.

From above:

Matt a "fix" is simple and fair.

The only "unfairness" comes from "remainder" greater than the max multiple. 10 hex digits = 16^10 = 1099511627776 but there are 40,270 tickets which doesn't divide into all the random number range evenly.

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

So max valid random number is 40270 * 27303492 = 1099511622840

If the value of last 10 digits of the block hash is > 1099511622840 then it is a "void" and just use the next block to find the winner.

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%. Even if you don't the "unfairness" is very small. The first 4936 tickets have 27,303,493 chances to win. The rest of the tickets have 27,303,492 Still given wow easy it is to do it fair just do it fair.

Just want to add, if the block hash is < 1099511622840 it is good, which means that 1099511622840 is repeating ticket number 0, (a consequence of the mod operation).

But otherwise I like it

Matthew N. Wright

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The raffle rules stated that:

Your email address would be published
The winner would be chosen by yesterday
Random.org would be used to choose the winner

So far I've done none of those. I'm all for learning and making this a better experience for everyone, but understand that every modification makes me feel like I'm straying even further from keeping my word.

I appreciate sticking around and helping everyone, and on that note, keep it as simple as possible for those who cant 2 math.

DeathAndTaxes

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Activity: 1218

Merit: 1079

Gerald Davis

Quote from: Matthew N. Wright on March 16, 2012, 03:15:11 PM

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.

Don't do that.

From above:

Matt a "fix" is simple and fair.

The only "unfairness" comes from "remainder" greater than the max multiple. 10 hex digits = 16^10 = 1099511627776 but there are 40,270 tickets which doesn't divide into all the random number range evenly. Some of the tickets have 1 more chance of winning.

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

The first 4936 tickets have 27,303,493 chances to win. The rest of the tickets have 27,303,492. Very small but a slight bias towards the earlier tickets. To be even the max valid random number is 40270 * 27303492 - 1 = 1099511622839. Just exclude numbers larger than that.

Simple version: If the value of last 10 digits of the block hash is >= 1099511622840 then it is a "void drawing". Use the last 10 digits of the next block to find the winner fairly. Don't worry the odds of that happening are very very very small. Only 1,099,511,622,840 to 1,099,511,627,776 are "bad". 4936 unfair numbers out of 1099511627776 or 0.0000004%. The odds of two bad drawings in a row are 1 in 46 quadrillion.

On edit: fixed on small error (1099511622840 is the first "unfair" number not the last "fair" number).

danieldaniel

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Oh wow, stupid me! Nevermind!

theymos

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Quote from: DeathAndTaxes on March 16, 2012, 03:14:08 PM

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.

Matthew N. Wright

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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.

DeathAndTaxes

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Gerald Davis

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%

theymos

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Quote from: deepceleron on March 16, 2012, 01:52:06 PM

(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.)

Epoch

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Quote from: Matthew N. Wright on March 16, 2012, 03:08:19 PM

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

Set!

Matthew N. Wright

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

wogaut

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Quote from: DeathAndTaxes on March 16, 2012, 03:05:51 PM

Son what is the magic block #?

I need to set my block clock.

Same here

filharvey

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Quote from: Epoch on March 16, 2012, 03:00:59 PM

Quote from: filharvey on March 16, 2012, 02:57:38 PM

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

DeathAndTaxes

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Gerald Davis

Son what is the magic block #?

I need to set my block clock.

Topic: Got my BFL Single today and I'm raffling it away for 0.5BTC! - page 7. (Read 29329 times)