Poker Hand with At Least One Ace

Date: 04/07/2003 at 17:55:35
From: Matti
Subject: A poker hand that contains at least 1 Ace

Dr. Math,

There is an exercise in my textbook that says, "What is the 
probability that a five-card poker hand contains at least 1 ace?"

The textbook gives the answear to this probability: 

   1-C(48,5)/C(52,5)

I do not agree; I feel that this would be a more accurate answer: 

   (C(4,1)*C(51,4))/C(52,5)

I first choose 1 ace out of 4 and then I have "at least one ace." Then 
I can choose the next 4 out of the 51 remaining cards. Then I've got 
C(4,1) * C(51,4) possible correct outcomes. I divide this by all 
possible poker hands or: C(52,5)

Could anyone tell me which is correct?

Kind regards,
Matti

Date: 04/07/2003 at 18:13:13
From: Doctor Roy
Subject: Re: A poker hand that contains at least 1 Ace

Hi,

Thanks for writing to Dr. Math.

That's not correct. Since it is "at least" one ace, you have to 
account for drawing 1, 2, 3, or 4 aces. It is easier to calculate the
following: Probability of drawing 0 aces. The probability of drawing
"at least" 1 ace is:

  Pr[drawing at least 1 ace] = 1 - Pr[drawing 0 aces]

The probability of drawing 0 aces is:

    C(48,5)/C(52,5)

The numerator is the ways we can draw 5-card hands out of 48 cards
(all the cards except for the aces). The denominator is the ways we
can draw any 5-card hands. So the ratio is the number of ways of
drawing no aces. If we subtract it from 1, we get the probability of
drawing "at least" 1 ace.

Does this help?  Please feel free to write back with any questions you
may have.

- Doctor Roy, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 04/08/2003 at 08:57:37
From: Matti
Subject: Thank you (A poker hand that contains at least 1 Ace)

Your explanation was different from what we had and I can see 
now that this approach is correct.

Thank you.