In a game of Open Face Chinese poker, I encounter this problem quite often. One draws three cards at a time and one of these cards must be discarded, so two of three drawn cards can be used. A typical problem and question in this game would be: What is the probability of drawing one spade AND one of either 10,9 or 8 (call it Middle cards)with drawing three cards? We could have 10,9,8 of spades but we will need either another spade or another of the same group the draw to be successful. My approach was following - Success scenarios: 1) Spade and Middle Card - 13/52*10/51 2) Spade and Middle Spade 10/52*3/51 3) Middle and Middle Spade 9/52*3/51 4) Middle Spade and Middle Spade 3/52*2/51 If I sum up all the results, would that bring me to the correct answer?