# Combinatorics: Ways to pick 5 card poker hands to get the ace of spaces and another the ace of hearts.

Please someone explain to me how to do the following problem:

How many different 5 card poker hands can you get where one of the cards must be the ace of spades and another the ace of hearts?

My answer is $$\binom{52}{4} + \binom{51}{3}$$ because there are total of $52$ cards with $4$ aces. There are $\binom{52}{4}$ ways to choose the ace of spade. After we choose the ace of spade, there are total of $51$ cards left in the deck with $3$ aces. There are $\binom{51}{3}$ ways to choose the ace of hearts. Then we add them both.

• There is only one way to pick the ace of spades and only one way to pick the ace of hearts. May 24, 2017 at 0:30
• The Rule of Addition is that : the count of an union of disjoint events equals the sum of the count of the events. This is not applicable here. You want to find the count of an intersection of non-disjoint events. May 24, 2017 at 0:53
• There is not 4 ways to chose the ace of spades May 24, 2017 at 19:31

Once you have selected the ace of spades and the ace of hearts, you have 50 remaining cards from which you choose 3 to complete the 5-card hand. Thus the total number of ways is simply $$\binom{50}{3}.$$

• What is a hand in Poker? May 24, 2017 at 0:45
• It is 5-cards, Vada Curry, as specified in your opening post. May 24, 2017 at 0:52
• A hand is simply a collection of cards from the deck. A card player usually holds the cards in their hand when playing a game, hence the collection is called a hand. So another way to state this problem is "How many ways can you select 5 cards from a 52 card deck that contain both an ace of spades and an ace of hearts?" May 24, 2017 at 0:53

There is only one ace of spades and one ace of hearts
Take the other three from the 50 remaining

$$\frac{\binom{2}{2} \binom{50}{3}} {\binom{52}{5}} = 1 / 132.6 \approx 0.0075414781$$

• The question does not ask for a probability. May 24, 2017 at 21:37
• @N.F.Taussig How many different is not a probability? Good day. May 24, 2017 at 23:58
• It is not. What you are supposed to calculate is the number of hands (your numerator), not the probability of obtaining such a hand. May 25, 2017 at 0:05
• @N.F.Taussig Including a denominator makes it wrong? May 25, 2017 at 2:15
• Yes. The number of outcomes is the number of different hands that contain an ace of spades, an ace of hearts, and three other cards, which, as you found, is $$\binom{2}{2}\binom{50}{3}$$ The probability is the number of outcomes divided by the number of all possible outcomes, which, in this case, is $$\binom{52}{5}$$ since the hands are being selected from a deck with $52$ cards. May 25, 2017 at 8:15