   Probability with Cards
  I'm trying to calculate the odds of getting at least a pair in a hand of Texas Hold'Em after the flop with no pair currently.Hold Cards: A KBoard: Q 2 7Ways to get a pair are to get at least 1 of (AKQ27), or a pair of cards not AKQ27 (eg. 8 8)Cards left to get: 2Cards left in deck: 47Cards of AKQ27 left in deck: 15 (5 * 3 remaining of that rank)Total remaining hands: 47 * 46 = 2162Hands not with (AKQ27): (47 - 15) * (46 - 15) = 32 * 31 = 992(this group includes a pair of the last 2 cards)I subtract 15 from the first of the 47 remaining cards and from the second card of the 46 reminaing (47-1) to get the number of cards left that are not (AKQ27).  Hands with an (AKQ27) : 15 * 46 = 690(this includes hands with 1 or 2 of (AKQ27))I use 15 to represent the number of (AKQ27) left and then 46 as the remianing cards, which includes the 14 other (AKQ27).This is where I break down.  The hands with an (AKQ27) plus the hands without an (AKQ27) should equal the total hands.  But I get 992 + 690 = 2162.  It isn't coming out right.  Someone please point out my probably easy mistake or tell me what I am not factoring in.  Thank you.   
  (47 - 15) * (46 - 15) = 992 Neither the first nor second card is from the 1515 * (47-15) = 480 The first card must be one of the 15 the second can't(47-15) * 15 = 480 The second card must be one of the 15 the first can't15 * 14 = 210 Both are from the 15992+480+480+210=2162You're missing the case when the first card isn't part of the 15, but the second is (or vice versa depending on how you came up with your numbers).   