[FREE RAFFLE][ENDED]MX12 BITCOIN ART - ERROR PIECE 715155 <-- Bitcoin Block

[Darker45]

00- Darker45

Please and thank you! Merry Christmas!

[noorman0]

37 - noorman0

ty

[owlcatz]

29 - owlcatz

Thanks!

[LitLitBit]

Wow great work!

09 - LitLitBit

Thanks!

[PontiusTP]

32 - PontiusTP

[mx12.levins]

Wow great work!

09 - LitLitBit

Thanks!

Number 9 already taken
22 - LitLitBit

[m2017]

#39 - m2017
Let's try our luck.

[mortare]

47 - Mortare
thnx

[examplens]

24 - examplens

[Frank AlpenCoin]

26 - Frank AlpenCoin
thank you

[Yabes]

28-Yabes, Last spot?

Thank you

[mx12.levins]

"IN 10 MINUTES WILL ANNOUNCE WINNER"

[mx12.levins]

Congratulations number 08 - heisty BTCBTCBTC You are the winner!!!

For information I used Python to find the winner:

[heisty]

I'll gladly accept the win but agree that the random number generator is a bit unconventional.  Thank you for hosting this raffle, I've replied to your PM.   

Happy holidays all.

[Cryptogreatdane]

Is it just me or does the random number generator only pick a number between 1 and 10.. not trying to be a bitter loser but 40 people were excluded?

[heisty]

Just eyeballing the script it might be the case that only numbers 1-10 were considered...  If this is what happened I'd be fine with picking a future block to ensure a fair re-roll.

[sam00]

If you put the following Code into a Python interpreter

import random
print(random.randrange(1,10))

then you will print a random integer (a full number) from 1 to 9.
The 1 is included and the 10 isn't.
I also think that's a bit "unconventional" for a raffle with 50 people

You could go ahead and use the following code:

import random
print(random.randrange(1,51)

This will include 1-50.
As others have mentioned before. You could also pick a bitcoin block in the future and let the last 2 digits of the hash of the block decide the winner. If its above 50, just go with the next block.
Just make sure to announce the block you chose before it is mined

[MoparMiningLLC]

worth noting is that the spot "00" was also included so a range of 1-51 would exclude that individual as well.  This was really 51 participants and only 9 were actually in the drawing.

[mx12.levins]

Congratulations number 08 - heisty BTCBTCBTC You are the winner!!!

For information I used Python to find the winner:

In general, the rule of good form in such raffles is to determine the winner using a blockchain or a hash, with a predetermined condition, and not with the help of an incomprehensible screenshot, the reliability of which no one can verify.

Not that I was upset about the loss, since I always lost in such lotteries, but this technique, to put it mildly, raises doubts.

Oh,
I will create another raffle as well and use blockchain this time. I never had exp with using hash/blockchain for raffles. Can you tell me how the raffle like this work? We have 50 or 100 numbers how you will pick up the winner? Let's say it was Bitcoin Block 715,145 00000000000000000008cfd241b914ea2e441a0529dd47c549a33220ec852a82

worth noting is that the spot "00" was also included so a range of 1-51 would exclude that individual as well.  This was really 51 participants and only 9 were actually in the drawing.

I have recorded video as well, the screenshot is only example.

import random
print(random.randrange(0,50))

[mx12.levins]

Just eyeballing the script it might be the case that only numbers 1-10 were considered...  If this is what happened I'd be fine with picking a future block to ensure a fair re-roll.

If you are okey with it, we can re-roll picking up one of the next blocks, your choice BTC