# Among the 2598960 possible 5 card poker hands how many will contain at least one ace?

Among the 2598960 possible 5 card poker hands how many will contain at least one ace?

If I do n(number of 5 card hands) - n(no ace): The only way to obtain a no ace would be 2598960C0 = 1 Is that right?

So the answer is 2598959?

### 5 Answers

- 4 years agoBest Answer
This is a typical combinatorial problem. Generally, you have two (or more) groups taken from a population and you want to know the number of ways a certain number can be taken from each population.

Your problem is too tuff. Consider a bag of 10 marbles, 2 red, 8 blue. Draw 4. What is the probability (or number) of 1 red and 3 blue?

The theory says you must choose 1 from the 2 red and 3 from the 8 blue. If you want the probability, you must divide by the number of ways you can draw 4 from 10.

Hence, # ways of getting 1 red is 2C1; of getting 3 blue is 8C3; number of total ways; 10C4

probability is:

2C1 x 8C3

--------------

10C4

note how the 2 8 = 10 and the 1 3 = 4

Now, to YOUR problem.

Choose 1 ace from 4 cards, and 4 "others" from 48 cards = 4 C 1 x 48 C 4

Choose 2 ace from 4 cards, and 3 "others" from 48 cards = 4 C 2 x 48 C 3

Choose 3 ace from 4 cards, and 2 "others" from 48 cards = 4 C 3 x 48 C 2

Choose 4 ace from 4 cards, and 1 "others" from 48 cards = 4 C 4 x 48 C 1

add them up

Alternatively, you could approach it as you did:

Choose 0 Ace from 4 cards and 5 "others" from 48 cards = 4 C 0 x 48 C 5

and subtract from 2598959 (if you have confidence in your instructor that

259859 is the total number of hands.

always,

tony

- Scarlet ManukaLv 74 years ago
No, I have no idea why you think there is only one hand with no aces. Common sense should tell you this is wildly wrong.

In the deck there are 48 non-aces and 4 aces. To get a hand with no aces, we need to pick 5 cards from the 48 non-aces, so there are 48C5 = 1,712,304 ways to do it.

So there are 2,598,960 - 1,712,304 = 886,656 hands with at least one ace.

- Elizabeth MLv 74 years ago
The correct answer is 52C5 - 48C5 since 52C5 different hands are available, without restriction but there are 48 non-aces so 48C5 hands

with no ace. Either you have a hand with no ace or you have a hand

with at least one ace which gives 52C5 - 48C5 which I make 2443296.

- How do you think about the answers? You can sign in to vote the answer.