Probability of drawing one spade AND one of either 10,9 or 8 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 But this is for drawing 2 cards, how shall I input the third card drawing possibility into this?

In this case I find it a little easier to calculate the probability of failing to get what you want; then you can subtract that from $1$ to get the desired probability. You fail if you draw no spade or no middle card. There are $\binom{39}3$ ways to draw three non-spades, and there are $\binom{40}3$ ways to draw no middle card. However, some sets of three cards have been counted twice, because they contain neither a spade nor a middle card. There are $13$ spades and $9$ non-spade middle cards, so there are $30$ cards that are neither spades nor middle cards and therefore $\binom{30}3$ sets of three cards that are ‘bad’ in both ways. Altogether, then, there are

$$\binom{39}3+\binom{40}3-\binom{30}3$$

bad draws out of the total of $\binom{52}3$ possible draws, and the desired probability is

$$1-\frac{\binom{39}3+\binom{40}3-\binom{30}3}{\binom{52}3}\;,$$

which is a little less than $\frac13$.