What is the probability of a five-card poker hand containing the ace of hearts?

Hello there,

The probability of getting a five-card poker hand containing the ace
of hearts is 1 / 52.

This kind of probability question requires you to multiply individual
probability odds to obtain the total number of cases.
First, the probability of obtaining ace of hearts is 1 in 52
Then the probability of the next card with no restrictions is 51/52
(since 51 cards are remaining)
By similar logic, the next 3 values are 50, 49 and 48.
Thus the total number of cases of getting a hand containing the ace of
hearts is 1x51x50x49x48 = 5997600.
The total number of five card poker hands is 52x51x50x49x48 =
311875200.
Thus the probability of getting a fivecard poker hand containing the
ace of hearts is 5997600 / 311875200 = 1 / 52.

Thank you for your question,
tox-ga

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Clarification of Answer by tox-ga on 06 Nov 2002 19:55 PST

Correction to the answer, the probability is 5/52.  Ignore the first
two comments given on this question as they're incorrect.

A simple way to look at it is looking for the probability of NOT
getting the ace of heart, which is (51x50x49x48x47)/(52x51x50x49x48) =
47/52
Thus the chances of getting the ace of heart is 5/52.

Researcher:
According to your logic, it is just as likely that a person would
select the ace of hearts after drawing just one card which obviously
has a probability of 1/52.  I believe the correct answer is 1/52 +
1/51 + 1/50 + 1/49 + 1/48.

Tox, your logic is severely flawed, as elipc points out above.

Elipc has the correct answer, as each time the pool from which to draw
the cards is reduced by one.

The correct answer is 5/52.

If you insist on reducing the pool by 1 for each subsequent card, then
you also have to account for NOT getting the HeartAce as a previous
card.  So:

1/52 + (51/52)*(1/51) + .....

As you can see, the extra stuff gets factored out.


Paddlequeen

answer given is incorrect.

Paddlequeen:
Thanks.  I need to think about it a bit more, but I believe you are correct.

way to go Darthcow, your logic is "severely flawed" as well.
Elipc had the wrong answer.
Something more humorous than a flawed answer is a flawed criticism.
Tox-ga's clarification is correct.
His original answer did not account for the ace of heart being able to
be drawn as 2nd, 3rd, 4th or 5th card, which is why it was 1/5 of the
correct answer.

This is the same question as:
https://answers.google.com/answers/main?cmd=threadview&id=100825 ,
what's the probability of a random day being in April, just with
different constants.

Mathee:
Please take a look at the dates of the posts.  The researcher's
corrected answer was only posted once paddlequeen came up with the
correct answer (which I promptly agreed to in my second post).  My
original problem with the researcher's logic was valid, as was
darthcow's agreement.  The fact that I did not arrive at the correct
answer in no way makes the researcher's original answer correct.

Sorry about that... that was actually my initial thought as well, but
then I thought that maybe elipc was right... arg... need to rethink
this one :P.

It's a trick question.  

"What is the probability of a five-card poker hand containing the ace
of hearts?"

Everybody assumed that we are using 52 cards, but most decks used in
casinos for poker have more than 52 cards so people don't cheat.

Casinos?  Who said anything about casinos!?  :-)

good call.

So far as I know, this is a very basic question of high school level.

I don't think it worth so much discussing and I wonder why nobody mention about

 the basic rule of combinations, that is , C(n, r)= n!/(r!(n-r)!)

 all combinations of 5 hands is C(52, 5)
 the combinations of one heart ace and four other cards is 1* C(51, 4)

 so the probability is C(51,4)/ C(52, 5)= (51*50*49*48/ 4!)/(52*51*50*49*48/ 5!)
 = 5!/(52*4! ) = 5/ 52 (cancel 51*50*49*48 first)

 So the 2nd answer of researcher is correct, even though,
 in my opinion, the idea is important not the answer. 

 But I prefer to solve it directly, also for me, I never multiply

 such sequence of product as 52*51*50*49*48

 Anyway, always, simplify before we do (boring) calculations.

 khwang

So far all answers and comments refer to the initial deal, and this is
not required parameter of the question. The question would need to
adress five card stud poker for these answers to be correct.
If this were a stud game with one deck and no jokers, there are 10.4
five card hands, so the probability would be 1 in 10.4 or 10 in 104.
If this is draw poker and all 5 cards can be discarded the probability
becomes 1 in 5.2 or 10 in 52.