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#math
<seb-_> % 3
<tjcoder_> well, 1 ace is set. so it'd just be 1*{48 choose 4} right?
<asphyxia> he isnt here
<seb-_> Steve | Office: why is mbot dead?
<seb-_> Steve|Office: i mean you
<Steve|Office> You'd have to ask Cale.
<asphyxia> tjcoder_: Ehm, I dont know.
<seb-_> Cale!
<asphyxia> you have 48 remaining cards to draw from in order not to have second ace
<asphyxia> seems reasonable.
<tjcoder_> yeah...
<asphyxia> but dont take my word on it, my prob101 was a passed not passed course
<asphyxia> a mandatory one
<tjcoder_> nah, it sounds right. better than anything else i can think of anyways
<tjcoder_> which is... good enough, when it's due tomorrow and i'm about ready to fall asleep at work.
<asphyxia> gl though
<tjcoder_> crap. 1-P(1 ace)/P(1 ace)
<tjcoder_> :-S
<asphyxia> not true
<tjcoder_> haha yeah. i've obviously lost it tonight.
<asphyxia> wait
<asphyxia> P(1 ace) is your dependency. You know you have already drawn an ace.
<tjcoder_> right
<asphyxia> Now, P(1 ace only is a whole other story)
<asphyxia> ow
<asphyxia> P(1 ace only) is a whole other story
<asphyxia> given 1 ace, no other aces has been drawn.
<tjcoder_> asphyxia it's at least 1
<asphyxia> ah, it cant be. there is always a prob that you draw another ace
<tjcoder_> yeah.
<tjcoder_> oh
<tjcoder_> the 1-p(|) thing isn't necesarily true, i don't think.
<asphyxia> it is
<asphyxia> if m is one event and q is the exact opposite, the p(m) = 1 - p(q)
<asphyxia> tjcoder_: As I get it, you have already drawn an ace.
<tjcoder_> right, but p(r|q) + p(r|-q) = 1, not necesarily p(r|q) + p(-r|q) =1, i don't think
<tjcoder_> well, no, not even that, i guess
<asphyxia> That means you can draw another 51 cards.
<asphyxia> I see it as {48 choose 4}/{51 choose 4}
<tjcoder_> that's.... prob of no aces given 1 ace?
<asphyxia> I dont know
<asphyxia> Im kind of guessing. I may very well be totally off.
<tjcoder_> i think i should just sum the probabilities of picking 2, 3, or 4 aces
<tjcoder_> no that's totally wrong too
<tjcoder_> grr.
<asphyxia> but in my book, i have an example which states that the prob of drawing 13 cards without any aces is {52-4 choose 13}\{52 choose 13}
<tjcoder_> well, (p(2 aces) + p(3 aces) + p(4 aces))/p(at least 1 ace)
<tjcoder_> because p(2 aces and >=1 ace) should be the same as p(2 aces), i think.
<tjcoder_> well, more generally, P(>= 2 aces ^ >= 1 ace) should imply P(>= 2 aces)
<asphyxia> no
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