Random Magic Paths - is it truly random?

BigDaddy · #1 February 17th, 2005, 02:28 PM

Arallen and Bummer_Duck, if you are actually seeing long strings of alternating air and earth mages, I would suggest that it is a bug, but with the sophistication of random number tables and RNG these days, I believe the flaw would be in the windows OS. If you're using a different OS, I almost can't comprehend the problem. Depending on the actual coding of course, it could be caused by an artifact in the memory of the computer, caused by windows clunky and inefficient memory management.

Reboot, and try again! But be warned random systems do do unexpected things, and thats the beauty of them.

BigDaddy · #2 February 17th, 2005, 03:54 PM

Restarting the computer incessently is a dogmatic method of course, if you're not seeing a real pattern.

If the system wasn't random you could use and expert system (thats the engineers term for artificial intelligence) to predict the outcome. Of course with the turing machine's (a computer is a turing machine)inability to actually produce random numbers, with enough samples the expert system may eventually learn how predict the supposed randomness, but it would take quite a few samples and a very sophisticated expert system.

Has anyone learned anything from reading these posts?

(Corrected spelling)

Kristoffer O · #3 February 17th, 2005, 04:38 PM

I learned that Bone Daddy is a nerd

BigDaddy · #4 February 17th, 2005, 04:39 PM

Thats funny because i'm not bone daddy and Ivan's posts where nerdier. No offense intended Ivan.

Ivan Pedroso · #5 February 19th, 2005, 04:11 PM

Quote:

BigDaddy said:
Thats funny because i'm not bone daddy and Ivan's posts where nerdier. No offense intended Ivan.

Hehehe none taken

"20 sages in a row with exactly 1 path missing":
(7/8)^20 *8 = 0.554

Explanation:
the (7/8) is the chance of getting one of the 7 "good" paths when you hire a sage.
^20 is just 20 such sages in a row.
*8 stems from the fact that you can choose your pool of 7 "good" paths from the 8 possible paths in 8 different ways.

General formula:
(chance of one good)^(number in a row) * (ways to make a good pool)

Getting 20 in a row with exactly 2 paths missing:
(6/8)^20 * 28 = 0.0888
(28=(8!)/(6!2!) is the number of ways of taking 6 from a sample of 8)

Getting 20 in a row with 1 or more paths missing:
~65%

Surprisingly high ? or ?

alexti · #6 February 19th, 2005, 04:56 PM

Quote:

Ivan Pedroso said:
"20 sages in a row - exactly 1 path missing":
(7/8)^20 *8 = 0.554

Explanation:
the (7/8) is the chance of getting one of the 7 "good" paths when you hire a sage.
^20 is just 20 such sages in a row.
*8 stems from the fact that you can choose your pool of 7 "good" paths from the 8 possible paths in 8 different ways.

7/8 probability means that you will get one of a "good" paths, but repeating the roll 20 times does not guarantee that you'll get each of "good" paths at least once. Thus 7/8 ^ 20 is probability of missing "bad" path (and maybe some of the "good" paths), not one path exactly. If you write your 20 random picks, using digits from 0 to 7 to indicate different paths, 8^20 is a total number of possible numbers (outcomes). Not getting one particular path means getting number without particular digits. There's total of 7^20 such digits. However, there're numbers like 111...1, 222...2 etc amongst them. Those numbers indicate outcomes where 7 paths are missing. Other numbers will represent outcomes with various number of paths missing.

Quote:

Ivan Pedroso said:
General formula:
(chance of one good)^(number in a row) * (ways to make a good pool)

Getting 20 in a row without exactly 2 paths:
(6/8)^20 * 28 = 0.0888
(28=(8!)/(6!2!) is the number of ways of taking 6 from a sample of 8)

Same problem as above, it's probability of missing 2 paths or more.

Quote:

Ivan Pedroso said:
Getting 20 in a row with 1 or more paths missing:
~65%

Surprisingly high ? or ?

BigDaddy · #7 February 19th, 2005, 06:11 PM

Based on his knowledge of statistics he made an estimate, thats what the ~ is for.

The exact value is given by
((7/8)^20 * 8c7) + ((6/8)^20 * 8c6) + ((5/8^20 * 8c5) + . . . + ((1/8^20 * 8c1) = .647163

If he didn't figure it out exactly he's a pretty good guesser.

And he did say it properly:

Probability of not getting 1 or more paths after recruiting 20 random paths.

*8c5 = 8 combinataion 5 (ways to get 5 out of 8 with replacement) = 8!/5!3!

Ivan Pedroso · #8 February 19th, 2005, 06:28 PM

@Alexti

Ahhhh yeah - that was sloppy.

(7/8)^20*8 is NOT the probality of getting exactly one path missing. You are absolutely right about it not garanteeing that the other 7 paths are present.

Back when I have had time for a better look (sorry 'bout posting too hastilly

)

Turms · #9 February 17th, 2005, 04:46 PM

You reasoning is invalid BigDaddy. Johan Osterman already posted that Dominions II uses a pregenerated random number table. That means it's random number generator works exactly the same in any computer with any OS. If it didn't the battle replays wouldn't work.

I have still a question though:
How is the seed for a new turn generation chosen?
The normal way is to take it from the clock or something. That is the only place I could think that could somehow affect the randomness of picks (very unlikely).

Turms

BigDaddy · #10 February 17th, 2005, 04:53 PM

OK, I didn't notice Johan's post (I was wondering what method was used). Pre-generated number tables are very common (they are more efficient), but I think an artifact could still cause problems.

I don't think the seeding technique is likely to be the problem. I believe the seeds are used to select a starting point on the table and a "movement" pattern around the table, and it can be made even more comples. Repeats are highly unlikely in any case.