Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives

Random Card Shuffling Probabilities

Date: 6/11/96 at 20:37:33
From: Anonymous
Subject: Probability of two cards being....

I was wondering the other day what the probability of say at least two 
cards.  However a more interesting problem is the probability of at 
The answer is 1-P(no similar cards are next to each other), but how do 
you count that?

Date: 6/12/96 at 6:10:30
From: Doctor Anthony
Subject: Re: Probability of two cards being....

If you want the probability that no 'eights' are next to each other, 
you can model the problem in the following way:

Let 'eights' be represented by stars, *, and non-eights by strips |.  
if the cards are laid out in a row at random, you get something like 

  |||*||||**||||....... and so on.

You will notice that the condition for no two stars to be together is 
that a single star must occupy a gap between two strips.  Suppose 
therefore that we first lay out the 48 strips (non-eights) in a row.  
Now there are 49 gaps available (49 because there is a space before 
the first strip and after the last one), and we must choose 4 out of 
the 49 in which to place a star.  
This can be done in  49_C_4 ways = 211876 ways.
However the number of ways of laying out 48 strips and 4 stars in any 
order is 
       52!/(48!.4!)  = 52_C_4 = 270725

The probability that no eights are together is therefore given by:

       (49_C_4)/(52_C_4) = 0.782624

The probability that at least two eights are together is therefore 
 1 - 0.782624 = 0.217376

These are also the a priori probabilities for any other kind of card, 
say kings or sixes. The fact that eights ARE together would however 
affect the number of gaps available when considering other face-value 
cards, so the probabilities are not independent, if, for example, you 
wanted probabilities that no eights and no queens are together.

I will give the matter further thought, but I think your general 
request for the probability that NO two cards of the same face-value 
are together will prove very complex indeed.

-Doctor Anthony,  The Math Forum

Date: 6/13/96 at 23:16:30
From: Doctor Peter

This problem involves counting to number of permutations, since we are
dealing (no pun intended) with the order of the cards in the deck.  
There are 52! permutations of the deck, but the calculation of the
number of permutations with at least two like cards together involves
a fairly lengthy recursive formula.  There is not enough space to give
it here, but it involves the number of permutations of the 4 together
plus 3 together plus 2 together all times the rest of the decks' 
permutations, which also needs to incorporate this.  I made a spread-
sheet model based on this, and got that the P(at least 2 together) is
about equal to 96.35%.  Contact me for additional explanation.

-Doctor Peter,  The Math Forum
 Check out our web site!
Associated Topics:
College Probability
High School Permutations and Combinations
High School Probability

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:


parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]


Ask Dr. MathTM
© 1994-2013 The Math Forum