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Kdice Sandbox odds
jakestone wrote
at 1:22 PM, Monday January 1, 2007 EST
Yay for Statistics class last year and some quick programming and a good polynomial library.

Here are the odds for 16 dice kdice in a nice little png. The numbers are to a high degree of accuracy since I wanted that 3 offense versus 16 defense to actually show a number other than 0. It's <b>possible</b>, but don't bet any money on it.

<a href="http://img237.imageshack.us/img237/3218/16kdiceoddstn3.png"; target="_blank"><img src="http://img237.imageshack.us/img237/3218/16kdiceoddstn3.th.png"; border="0" alt="kdice odds for 16 dice" title="kdice odds for 16 dice" /></a>

Just in case the HTML gets filtered, you can find the image at http://img237.imageshack.us/img237/3218/16kdiceoddstn3.png , message me if you want raw numbers.

Replies 1 - 10 of 10
jakestone wrote
at 1:27 PM, Monday January 1, 2007 EST
Just in case someone doesn't get it, that itty bitty pic is a thumbnail to the big pic with all the numbers readable. Paranoia, I'll admit, but I just had this thought where people would start griping about not being able to read the numbers in the thumbnail.
Tech wrote
at 2:56 PM, Monday January 1, 2007 EST
Trust me, it would have happened...
MadWilly wrote
at 2:58 PM, Monday January 1, 2007 EST
hey great idea. But i cant read the numbers in those post of yours would you mind posting a larger image please?

SCNR.
fuzzycat wrote
at 3:23 PM, Monday January 1, 2007 EST
"I wanted that 3 offense versus 16 defense to actually show a number other than 0. It's possible, but don't bet any money on it. "

I'm just wanting for the day it happens, and somebody complaining the hell out of the games, how random is tinkered, and it is "realistical" impossible to happen.

---
I find it interesting that 16v16 has a higher chance of winning then 8v8. hmm, why is this so? And is this approaching exactly 0,50% for infinite<->infinte ?
Tech wrote
at 6:49 PM, Monday January 1, 2007 EST
The more dice, the smaller the chance of a tie, which is what skews it off from 50%. Yes, it approaches 50%.
the brain wrote
at 7:59 PM, Monday January 1, 2007 EST
"I find it interesting that 16v16 has a higher chance of winning then 8v8. hmm, why is this so? And is this approaching exactly 0,50% for infinite<->infinte ?"

The reason for this is that the winning odds for a X-vs-X fight are completely based on the number of combinations of draws (chance to win = ((total combinations - draw combinations)/2) / total combinations). The proportion of draw combinations decreases with the number of dice (1vs1 = 6:36 ~= 16%, 2vs2 = 146:1296 ~= 11%, etc.), and so the winning odds will be closer to 50% for an increasing numer of dice.

It's hard to actually proof that the limit is precisely 50% though, it may be the case that the rate of ascent of the number of draw combinations will approach the rate of ascent of the total combinations (36^n), resulting in a limit close to, but not exactly, 50%.

I've evaluated it to a 166-vs-166 dice situation (best I could do numerically :p), which resulted in a 49.359% win percentage.
Tech wrote
at 8:43 PM, Monday January 1, 2007 EST
Well...What's the chances of tying with an infinite number of dice against an infinite number of dice? It's lower than any positive number, so it's zero. That's the way I think of it. not exactly a rigorous proof...
the brain wrote
at 9:38 AM, Tuesday January 2, 2007 EST
With infinitely many dice there are infinitely many possibilities to tie, so it's hard to say.
But it doesn't matter that much anyway, either it's 50% or a number very very close to 50% ;).
Tech wrote
at 6:43 PM, Tuesday January 2, 2007 EST
Bump. :p
MadWilly wrote
at 8:40 PM, Saturday December 13, 2008 EST
those numbers are still true, reliable sources pretended to know.
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