Conquer Club

It's mathematically impossible for the distribution of the dice to be even, since the random parts are taken from a bit stream (base 2), and dice are in base 6. If we had 8-sided dice to roll, it would be possible. However, there is bound to be a slight unbalance in actual distribution. However, how unbalanced it is, and which dice is favors (there will be 2 dice or 4 dice it favors, most likely), is determined by the actual equations used.

Discuss.

Also, lack, how exactly do you determine the value of each die, given the bit stream from random.org?

If you try this out, it might show you any byist dice. http://www.conquerclub.com/forum/viewtopic.php?t=5655
It's called a dice analyser.

@kc-jake: your post makes no sense at all. How did you come up with that?

I assume he's talking about a binary setup 2,4,8,etc. He's saying that the number 7 and 8 options would be rolled over and counted with 2 previous options.

There's quite a simple solution to that. 8 possibles and every time number 7 or 8 shows up you disregard it. Thus you get a random selection of 6.

zip_disk wrote:I assume he's talking about a binary setup 2,4,8,etc. He's saying that the number 7 and 8 options would be rolled over and counted with 2 previous options.

There's quite a simple solution to that. 8 possibles and every time number 7 or 8 shows up you disregard it. Thus you get a random selection of 6.

Hmmm, good point. I always forget about throwing results away. I did calculate that if you determined a number from 4 or more bytes (using the modulus operator), the "percentage of error" would be so small that it would have absolutely no effect on strategy.