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#1
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If I see the flop in EP at NL with the following hands:
AA KK QQ JJ TT 99 AKs AQs AJs KQs KJs QJs JTs AKo AQo KQ How do I work out the correct probability of me seeing the flop? I started writing down some numbers trying to do it systematically like this Pr(Ace first card) = 1/13 * Pr(Any A, K, Q, Js) = 12/51 Pr(K first card) = 1/13 * Pr(Any Q, Js) = 5/51 etc. Then I realised that I may be not accounting for an ace on the 2nd card, because AK and KA are different when you are doing probability this way...and is the 1/13 for a K falling wrong because I have already factored in that it isn't an ace...I'm getting way too confused here. Haven't done maths for a year now. I'm not asking for anybody to work it out for me, I'm just asking what is the correct way to systematically work out the probability of me seeing the flop with a certain group of hands. Is there an easier way using nCr or nPr? Thanks in advance. |
#2
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You are making it way too complicated.
There are 52 * 51 / 2 = 1326 distinct starting hands. Each pair can be made 6 ways, each suited hand 4 ways, and each offsuit hand 12 ways. so your answer would be Pairs: 6*6 = 36 ways Suited hands: 7*4 = 28 ways Offsuit hands: 12*3 = 36 ways Total = 100 hands out of 1326 total hands = 7.54% |
#3
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And that's it? Thanks a lot, saved me a lot of time. 7.54%? That's a small number. I'll work the other positions out myself, which shouldn't be too hard. [img]/forums/images/icons/crazy.gif[/img]
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#4
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Each offsuit hand can be made 16 ways.
So your % is actual higher. |
#5
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</font><blockquote><font class="small">In reply to:</font><hr />
Each offsuit hand can be made 16 ways. So your % is actual higher. [/ QUOTE ] Each non-pair hand can be made 16 ways. 12 of these are offsuit, 4 are suited. |
#6
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I think you are right. So the percentage I worked out is correct?
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