View Full Version : lol what are the odds?
Eppitheblackguitar
04-08-2009, 03:45 AM
http://i520.photobucket.com/albums/w327/marky1man/map3.jpg
http://i520.photobucket.com/albums/w327/marky1man/map4.jpg
http://i520.photobucket.com/albums/w327/marky1man/map4.jpg
http://i520.photobucket.com/albums/w327/marky1man/map1.jpg
plz notice the upper right hand corner to see the different servers lol
Static_Fang
04-08-2009, 04:25 AM
Hahahahahaha, Ignis rocks on 3 servers!!!!
ArcticWolf
04-08-2009, 05:25 AM
I'll go mathematical:
You want identical situations on all the servers. You know that the possible combination of forts is 3^(9*4). You want identical combinations for every server, so we have 3^9 possible combinations.
The statistical ratio shows that the odds are 3^9 / 3^36
Please, don't tell me I'm wrong. I'm having my Algebra exam next week.
EDIT: That's 1 over 7625597484987
Miraculix
04-08-2009, 06:07 AM
I'll go mathematical:
You want identical situations on all the servers. You know that the possible combination of forts is 3^(9*4). You want identical combinations for every server, so we have 3^9 possible combinations.
The statistical ratio shows that the odds are 3^9 / 3^36
Please, don't tell me I'm wrong. I'm having my Algebra exam next week.
EDIT: That's 1 over 7625597484987
Per server, you want 1 combination out of 3^9 (3 values per building, 9 buildings) possible combinations ( 1/(3^9) ).
Then you want this to happen at 4 independent servers, so you take that number to the power of 4.
So that would be 1 over 150094635296999121, if calculations didn't fail me.
But that is irrelevant, as the probability is 100% after a patch since all servers copy amun status :p
_dracus_
04-08-2009, 08:59 AM
So that would be 1 over 150094635296999121, if calculations didn't fail me.
Don't try to even impress me now if you don't come with a number where you need a PhD Thesis to even write it.
voyager_3
04-08-2009, 10:35 AM
Some clarifications:
Suppose chances of taking any fort/castle by any realm are equal. Then, probability of specific map situation on a server is 3^(-9).
Probability that on all 4 servers we have the specific map situation (eg. the one from the pictures) is 3^(-36) (which is ( 3^(-9) )^4 ) or otherwise ~1.5*10^(-17).
However, there are possible 3^9 specific situations, therefore probability that on all 4 servers we have the same map situation (no matter which specific one) is
3^(-36) * 3^9 = 3^(-27) or otherwise ~7.6*10^(-12).
Mattdoesrock
04-08-2009, 10:38 AM
But that is irrelevant, as the probability is 100% after a patch since all servers copy amun status :p
Mira won. :P
Angel_de_Combate
04-08-2009, 12:09 PM
/me head is spinning
ArcticWolf
04-08-2009, 09:02 PM
Some clarifications:
Suppose chances of taking any fort/castle by any realm are equal. Then, probability of specific map situation on a server is 3^(-9).
Probability that on all 4 servers we have the specific map situation (eg. the one from the pictures) is 3^(-36) (which is ( 3^(-9) )^4 ) or otherwise ~1.5*10^(-17).
However, there are possible 3^9 specific situations, therefore probability that on all 4 servers we have the same map situation (no matter which specific one) is
3^(-36) * 3^9 = 3^(-27) or otherwise ~7.6*10^(-12).
That's right, because you may think of each fort as wooden blocks, and each kingdom as paint. You have 9 wooden blocks and you can paint them in three different colours, so you take a block, paint it and repeat until you have painted them all. That's 3*3*3*3*3*3*3*3*3, or 3^9.
Suppose you want to repeat this for each server, so we have to multiply it four times... Which translates as 3^36.
Now you want all the possible combnations where the wooden blocks are the same all over the servers. You just need to know all the possible combinations you can make of a single state or model, so that's 3^9.
Now you want to know the probability. Probability is "what I want over what I may get", so it is:
(3^9) over (3^36)
That's the same as:
(3^9) times (3^-36)
we can use a property of exponential calculus in multiplication to simplify it: if both numbers are the same base, add them. 9-36 = -27.
So it is 3^-27.
/me <3 math :D
Nightchill
04-09-2009, 12:27 PM
errr... get a life (?) :D
SPARTISH
05-14-2009, 01:38 AM
I'll go mathematical:
You want identical situations on all the servers. You know that the possible combination of forts is 3^(9*4). You want identical combinations for every server, so we have 3^9 possible combinations.
The statistical ratio shows that the odds are 3^9 / 3^36
Please, don't tell me I'm wrong. I'm having my Algebra exam next week.
EDIT: That's 1 over 7625597484987
i think thats wrong because u have to consider all servers are in teh same situation, and are probably doing the same thing (alsius always trying to take meni and algaros on all servers) ignis on RA is probably just like ignis on german server. i cba opening screeshots so if this did not happen on teh same time and the person was just waiting around for the same thing to happen then its different.
btw u have to consider the many different combinations they could have had? like ignis could have had 1 fort anyware on the field, thats 9 combinations alraedy then u have to do the different combinations of the other realms what forts they could have had?....
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