EPTOn June 11 2011 05:53 scorch- wrote:
It's not random vs random. It's zerg vs random for 2 games. The question is how often does random roll zerg both games when your friend tells you that he saw an awesome ZvZ?
It's not random vs random. It's zerg vs random for 2 games. The question is how often does random roll zerg both games when your friend tells you that he saw an awesome ZvZ?
If it's zerg vs random then the only combinations for all the outcomes of the games are
Game 1:
(I'll use a and b for players a and b a is nestea who always plays zerg)
(Za,Tb), (Za,Pb), (Za,Zb)
Game 2:
(Za,Tb), (Za,Pb) (Za,Zb)
Because we know in one game there was a ZvZ we can infer from that that while there are still 5 possibilities remaining in our random series only the possibilities of the remaining game matter. Thus proving what the possibility of one out come in one game is to be 1/3rd.
If it's detecting the probability that both games are ZvZ (under no prior conditions) it's 1/9th. Under the assumption that 1 game has already been played and it was stated to be a ZvZ it would seemingly eliminate one of the possibilities out of the 6 but this isn't true. (see above paragraph)
Think of it this way:
A player rolls 2 die. One of them is a 6, what is the possibility of the other being a 6 thus both of them being a 6. Well since there are 6 sides it's going to occur 1/6th of the time. Because one outcome is guaranteed to occur you can eliminate that from the experiment.
