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Probabilities and Expected Value in Seating Arrangements


Date: 05/12/98 at 08:01:18
From: Ian Peters
Subject: Probability

Dear Dr. Maths,

Three couples are seated at random around a round table. Let M denote 
the number of husbands seated next to their wives. Find:

   (a) the probability that M = m for m = 0,1,2 and 3
   (b) the expected value of M using part(a)

I know that for 1 couple siting together, the probability would be 2/
5: say the husband sat down, then there are only 5 seats left, and his 
wife can sit on either side of him, so it would be 1/5 + 1/5. But then 
as I work to 2 couples, I don't know how to state the situation, and I 
got nonsense answers (sometimes the probability is greater than 1, 
which is impossible).

And also, what does it mean by the expected value of M using part (a)? 
Can you give me one example, say, to find the E(M) as m = 2, please?  
Thanks for your time, Doctor.

Ian


Date: 05/12/98 at 19:36:23
From: Doctor Anthony
Subject: Re: Probability


For part (a), with a little systematic placing, we have:

     P(0) = 32/120

     P(1) = 48/120

     P(2) = 24/120

     P(3) = 16/120
   ------------------
     Total = 1           

The denominator 120 is the total number of permutations 5! = 120.

Working out above results is done in the following way. For example: 
with P(3), we consider

       (aA)

   (bB)    (cC)

where a = husband, A = wife, and so on.

Keep (aA) in fixed position. Then we can swap A and a, b and B, c and 
C, giving 2*2*2 different arrangements; and finally we could swap the 
b pair with the c pair, giving a further factor 2. All together, there 
are:

   2*2*2*2 = 16   

arrangements with all three seated together.

In part (b), we find the expected value of M using part (a):

   Expected value = 0*P(0) + 1*P(1) + 2*P(2) + 3*P(3)

                  = 0 + 1*48/120 + 2*24/120 + 3*16/120

                  = (1/120)[48 + 48 + 48]

                  = 3*48/120
  
                  = 1.2

-Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   


Date: 05/12/98 at 21:28:53
From: ian peters
Subject: Re: Probability

Dear Dr. Anthony,

I still don't understand that why P(2) is not same as P(3). If P(2) is 
correct, that should mean that there are only 2 seats and 1 couple 
left -- so doesn't that mean the third couple will sit together as 
well?

Also, why is P(1) not 2/5? Can you show me how you work P(1) and P(2) 
out, please? Thanks, Doctor Anthony.

Ian


Date: 05/13/98 at 11:35:15
From: Doctor Anthony
Subject: Re: Probability

Two pairs could separate a third pair:

    (aA)
   b    B
    (cC)

You can choose which two pairs in 3 ways, and then the a pair can be 
swapped, the b pair can be swapped, and the c pair can be swapped, 
giving:

   3*2*2*2 =  24 ways

With 1 pair together, you could have:

    (aA)
   c    b
    B  C

You could choose the pair to be together in 3 ways. Then the a pair 
can be swapped, the b pair can be swapped, the c pair can be swapped. 
Also, keeping the the a pair together, you could swap the b pair with 
the c pair. This would give:

   3*2*2*2*2 =  48 ways

We have already shown that the three pairs together can occur in 16 
ways.

You could go through another exercise with no pairs together, but the 
number of ways is clearly 120 - 16 - 24 - 48 = 32 ways.

I have checked this result, but it is somewhat tedious, requiring 4 
lots of situations each giving 8 arrangements, totalling 32:

      A             A            A            A
   c     C       c     B      C     b      c     B
   b     B       b     a      a     c      b     c
      a             C            B            a

In these, we keep A fixed and move a through three possible positions. 
Then for each diagram, we can swap the b pair, swap the c pair, and 
swap the b pair and c pair, giving 2*2*2 = 8 ways of arranging the 
letters.

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

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