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#math
<gzl> so 1 goes to 2
<gzl> !
<gzl> similarly you'll find that 2 goes to 1
<gzl> and 3 goes to 3
<gzl> which is why it's equal to (1 2)
<asphyxia> ok. if I have (4 5) (5 6) (4 5)
<asphyxia> 4 goes to 5, 5 goes to 6 and stays
<gzl> yes
<asphyxia> and 6 goes to 5 and 5 goes to 4 and stays
<asphyxia> so I have (4 6)
<asphyxia> is equal to the statement above.
<gzl> yes
<gzl> see now?
<asphyxia> I guess
<gzl> ...
<asphyxia> Yes, I do
<asphyxia> Well, Im confused by the notation. But I guess it'll be gone in a week.
<gzl> the notation is just a simple way of writing down functions
<asphyxia> Is it a common practice to write (i j), where i < j?
<gzl> you can succinctly write down where each thing is sent
<gzl> it's used in group theory, when you manipulate transpositions and cycles and stuff
<asphyxia> ok.
<gzl> you don't necessarily need to have i < j, after all (i j) and (j i) are the same
<asphyxia> nice, then
<asphyxia> Thanks a bunch on banging my head against the wall, till I realize stuff :)
<asphyxia> I should go to class instead of being such a menace.
<gzl> I concur
<asphyxia> rofl
<Llamanade> does anybody have a graph of different rates of growth
<Llamanade> for funtions
<asphyxia> Infact, I have class in like 3 hours and I am totally spiked on caffeine. I should be able to live through to the weekend at 12 'o clock. Then I can crash be able to avoid the parties of friday night
<tjcoder_> so... anyone wanna tackle a probability problem? well, help me, at least.
<tjcoder_> http://www.mathbin.net/8291
<asphyxia> tjcoder_: You have one ace, right. What is the probability of having one ace in 5 cards out of 52 cards?
<asphyxia> Thats your denominator.
<Llamanade> is there a maple channel
<tjcoder_> yeah, that's just 4 choose 1 * 48 choose 4 (/52 choose 5 for the probability
<tjcoder_> but... i don't get what to do with the numerator.
<asphyxia> ehm
<asphyxia> That is the event that you have 2 aces
<asphyxia> out of 5 cards. Thats how I read it
<asphyxia> It could though also be the prob that you have 2 or more aces
<tjcoder_> i asked my prof, and he said 2 or more
<asphyxia> ok
<tjcoder_> do i just need to use the sum rule to get 2 aces, 3 aces, 4 aces?
<asphyxia> then you are better of doing something like P(2 or more|given one) = 1-P(none of the remaining 4 is ace| given one ace)
<tjcoder_> ahh
<tjcoder_> that kinda makes sense
<asphyxia> yeah
<asphyxia> I dont know how to express it though, I'm kind of trying to forget all this prob of finite distributions
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