Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

The First Ace


Date: 06/12/97 at 15:42:49
From: Saul Klinow
Subject: Probability With Statistical Applications

I am having difficulty understanding the solution of an example 
problem provided by my textbook (Subject title by Mosteller, Rourke & 
Thomas). The problem is:

"The first ace.  An ordinary bridge deck of 52 cards is thoroughly 
shuffled.  The cards are then dealt face up, one at a time, until an 
ace appears.  What is the probability that the first ace appears at 
the kth card or sooner?"

The solution rendered by the text is as follows:

"Denote by F the event "first ace at the kth card or sooner".  Then 
the complementary event Fbar is the event "4 aces after the kth card".  
The first k symbols of every of every sample point in Fbar are all N's 
(non aces).  Therefore the number of sample points in Fbar is the 
number of ways of arranging 4 A's (aces) and 48-k N's in the remaining 
52-k places.  This number is

  (52-k)!/(48-k)!(4!)

Hence

  P(Fbar)=(52-k)!/(48-k)!(4!)  / 52!/(48!)(4!)

and 

  P(F) = 1 - P(Fbar). "

If the first ace is contained within the group of cards drawn up to 
and including the first ace, which describes the event F, how can 
Fbar, its complement include 4 aces?  Also, why are the first k 
symbols of every point in Fbar non aces?

Thank you for you  great service and wonderful cooperation.

Saul Klinow


Date: 06/12/97 at 20:10:06
From: Doctor Anthony
Subject: Re: Probability With Statistical Applications

Since F is the event "first ace at the kth card or sooner" then the 
converse of this (the complementary event), Fbar, must be that all 
four aces come after the kth card. We therefore have 52-k cards left, 
which must contain all 4 aces.  

The number of ways of arranging 4 A's and 48-k  N's is the number of 
ways of arranging 52-k letters, 4 being alike of one kind and 48-k 
being alike of a second kind. This number is given by

          (52-k)!
        -----------  =  (52-k)_C_4
         4! (48-k)!

                         (52-k)_C_4
So prob(Fbar)        =  ------------
                           52_C_4

and of course the required probability, that the first ace occurs at 
the kth card or earlier, is this probability subtracted from 1.

                             (52-k)_C_4
                P(F) =  1 -  -----------
                               52_C_4

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Probability
High School Statistics

Search the Dr. Math Library:


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

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/