# What is the probability that a five-card poker hand contains the ace of diamonds and the ace of hearts?

I am stuck in this question. Any help would be appreciated.

### 3 Answers

- PopeLv 73 years agoFavorite Answer
number of possible hands = C(52, 5) = 52! / (47! 5!)

number of possible successful hands = C(50, 3) = 50! / (47! 3!)

probability of success

= C(50, 3) / C(52, 5)

= [50! / (47! 3!)] / [52! / (47! 5!)]

= (50! 5!) / (52! 3!)

= (5)(4) / [(52)(51)]

= 5 / [(13)(51)]

= 5 / 663

≈ 0.00754

- 3 years ago
1 * 1 * 50 * 49 * 48 = number of hands with Ace of Diamond and the Ace of Hearts

52! / (5! * 47!) = number of possible hands

(50! / 47!) / (52! / (5! * 47!)) =>

50! * 5! * 47! / (47! * 52!) =>

50! * 5! / 52! =>

50! * 5! / (50! * 51 * 52) =>

5! / (51 * 52) =>

5 * 4 * 3 * 2 / (51 * 52) =>

5 * 3 * 2 / (51 * 13) =>

5 * 2 / (17 * 13) =>

10 / 221

- Anonymous3 years ago
I think I'm figuring this right. You have five chances out of 52 for the first ace and 4 chances out of 51 for the second ace. 5/52 x 4/51= 0.753%, so about three-quarters of one percent.

Damn, I was right and you chose the Johnny come lately. Thanks a lot.

2 different answers. I don't know which of you is right. :(