# 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?

### 2 Answers

- TonyLv 67 years agoFavorite 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

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- bubbicoLv 43 years ago
difficulty-free. the prob to entice all of the comparable in advantageous condition is (13c5*39c0)/52c5 it is to %. one suite. you have 4 suites, so the respond situations 4 that provides you with 0.001981 compliment it is 0.998

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