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Drawing Aces

Date: 06/12/97 at 21:27:46
From: Anonymous
Subject: Math-probability

52 card deck of playing cards? 

I have asked my parents, my teacher, my class and have even searched 
the entire Web for the answer. I don't understand how you would write 
the answer out. I have no area what I need help with specifically - 
I need to know the whole answer. It seems like the hardest question I 
have ever heard. I am hoping you can help me figure it out!


Date: 06/13/97 at 05:38:10
From: Doctor Anthony
Subject: Re: Math-probability

In fact the answer can be written down in a few seconds if you are 
familiar with the topic in mathematics known as permutations and 

I will give the answer in the notation of combinations, and then 
consider a more basic approach.

You have to select 2 aces from the 4 available, and this can be done 
in 4_C_2 ("four choose two) ways = 6 ways. However the number of ways 
for selecting any two cards from 52 is 52_C_2 = 1326 ways. So the 
probability that you will select two aces when you remove two cards 
from the pack is given by:

          4_C_2       6         1
         -------- =  ----  =  -----
          52_C_2     1326      221

                                         4 x 3
The expression 4_C_2 is calculated from  -----  = 6
                                         1 x 2 

              52 x 51
So  52_C_2 = ---------  =  1326
               1 x 2 

The more complicated the problem, the more the use of combinatorial 
methods becomes necessary. However, the problem you asked about can be 
done without the use of combinations.

If you have a pack of 52 cards and you draw one card out, the 
probability that it will be an ace is 4/52 because there are four aces 
in a deck of 52 cards. Now you will be down to 51 cards, and of these 
3 will be aces, so if you draw a second card the probability that this 
one is also an ace is 3/51 (since there are only three aces and a 
total of 51 cards left). To get the probability that both events will 
happen, i.e. an ace on the first draw and an ace on the second draw, 
we must multiply the probabilities.

                            4        3       1
Probability of two aces =  ----  x  ---  =  --- 
                            52       51     221

This is same result we got before.

-Doctor Anthony,  The Math Forum
Associated Topics:
High School Probability
Middle School Probability

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