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the possibility you win
191078819 wrote
at 7:49 PM, Sunday October 19, 2008 EDT
hi all!

i calculated the possibility that attackers and defencer(?) will win. ill show below.

(if someone already calculated, sorry)

a = the number of dices that attacker owns.
d = the number of dices that defencer owns.
p(a,b) = the possibility that attacker will win.


p(8,8) = 0.470
p(8,7) = 0.673
p(8.6) = 0.843
p(8,5) = 0.947
p(8,4) = 0.989
p(8,3) = 0.999 (0.9990)
p(8.2) = 0.999 (0.99997)
p(8,1) = 1.000

p(7,8) = 0.274
p(7,7) = 0.468
p(7,6) = 0.685
p(7,5) = 0.862
p(7,4) = 0.961
p(7,3) = 0.994
p(7,2) = 0.999 (0.9998)
p(7,1) = 1.000

p(6,8) = 0.121
p(6,7) = 0.259
p(6,6) = 0.466
p(6,5) = 0.699
p(6,4) = 0.883
p(6,3) = 0.975
p(6,2) = 0.998
p(6,1) = 0.999 (0.99995)

p(5,8) = 0.036
p(5,7) = 0.103
p(5,6) = 0.242
p(5,5) = 0.463
p(5,4) = 0.717
p(5,3) = 0.909
p(5,2) = 0.987
p(5,1) = 0.999 (0.9998)

p(4,8) = 0.006
p(4,7) = 0.025
p(4,6) = 0.083
p(4,5) = 0.220
p(4,4) = 0.459
p(4,3) = 0.742
p(4,2) = 0.939
p(4,1) = 0.997

p(3,8) = 0.000 (0.0004)
p(3,7) = 0.003
p(3,6) = 0.014
p(3,5) = 0.060
p(3,4) = 0.191
p(3,3) = 0.453
p(3,2) = 0.778
p(3,1) = 0.973

p(2,8) = 0.000 (0.000005)
p(2,7) = 0.000 (0.00007)
p(2,6) = 0.000 (0.0007)
p(2,5) = 0.006
p(2,4) = 0.035
p(2,3) = 0.152
p(2,2) = 0.443
p(2,1) = 0.837

(needless to say, for example, p(7,8) + p(8,7) is not equal to 1)

what do you feel about these numbers??


Replies 1 - 10 of 26 Next › Last »
£1000000 wrote
at 8:09 PM, Sunday October 19, 2008 EDT
I like numbers
kam|k2 wrote
at 8:43 PM, Sunday October 19, 2008 EDT
agreed, nice topic.
NeoconRuler wrote
at 8:50 PM, Sunday October 19, 2008 EDT
p(funny) = 0
moondust wrote
at 6:15 AM, Monday October 20, 2008 EDT
Thank you very much for your work...

It's interesting to see the "raw" likelihood".

This also shows that it is actually comparatively risky to do a 4v3 in round one. (compared to a 2v1)

Also: interesting how comparatively far away the likelihood is from 50% when it comes to 2v2/3v3/4v4 etc...


WELL DONE (your name)
Thraxle wrote
at 8:29 AM, Monday October 20, 2008 EDT
http://kdice.wikispaces.com/



Hope you didn't spend TOO much time doing that...........
191078819 wrote
at 8:58 AM, Monday October 20, 2008 EDT
to moondust;
the possibility of 3vs3 and 2vs2 was much higher than i expected...

and to Thraxle;


i tried these calculation for my homework of programming class
(making the question oneself and solve that).
It took one hour and a half.

but my approach was to count one million or 10millions times on
computer, so the likelihood of the page you showed is
more accurate than mine.

thanks for telling, i was convinced that my code was not wrong.
ChristianSoldier wrote
at 9:47 AM, Monday October 20, 2008 EDT
So you have an 8v2 as more likely to win than a 6v1.

The 6v1 is easy to calculate analytically. It would be 1 / 6^7 .. since all 7 dice have to have precisely 1 value: 1,1,1,1,1,1 v. 6 out of 6^7 possible combinations.

So this is: .000003572245.
vs. your .00005

If you ran your random generator 10M times then you had approximately 500 times where the 6v1 failed as compared to the most probable result of 35.7. One would expect a distribution around 35.7, and I suspect 500 is a highly unlikely sample estimate. Therefore, you have demonstrated a flaw in your code, either your pseudorandom number generator is insufficiently random (probably) and\or you are accumulating your statistics with significant roundoff error (owing to likely using a single or double floating point - try it with a decimal type (128 bits - massive precision) if it is available).
ChristianSoldier wrote
at 9:51 AM, Monday October 20, 2008 EDT
I challenge you to actually find the correct answer to the question of which is more likely to win!!! 8v2 or 6v1 !
Thraxle wrote
at 10:11 AM, Monday October 20, 2008 EDT
http://kdice.wikispaces.com/How+the+probability+table+is+calculated

CS:

I'm no mathematician, but the calculation used to compute the table in the link seems correct. This would indicate that a 6v1 loses 36/100,000 rolls and the 8v2 loses 166/100,000 rolls, making the 6v1 the safer attack.
ChristianSoldier wrote
at 10:14 AM, Monday October 20, 2008 EDT
Huh - I have the answer. It's actually what I suspected it would be.
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