Question 432850
Suppose we roll two ordinary, 6-sided dice. What is the expectation of the sum
of the two values showing? What is the expectation of the maximum of the two
values showing?
<pre>
Expectation of sum:

Here are all 36 possible rolls with a pair of dice:

(1,1)  (1,2)   (1,3)  (1,4)   (1,5)  (1,6)
 
(2,1)  (2,2)   (2,3)  (2,4)   (2,5)  (2,6)
 
(3,1)  (3,2)   (3,3)  (3,4)   (3,5)  (3,6)
 
(4,1)  (4,2)   (4,3)  (4,4)   (4,5)  (4,6)
 
(5,1)  (5,2)   (5,3)  (5,4)   (5,5)  (5,6)
 
(6,1)  (6,2)   (6,3)  (6,4)   (6,5)  (6,6)


Their sums are

  2      3       4      5      6      7

  3      4       5      6      7      8

  4      5       6      7      8      9

  5      6       7      8      9     10

  6      7       8      9     10     11

  7      8       9     10     11     12

There is 1 2, 2 3's, 3 4's, 4 5's, 5 6's 6 7's, 
5 8's, 4 9's, 3 10's, 2 11's and 1 12.  And there'
are 36 possible rolls, so the probabilities are the
number of ways to roll the sum over 36.

So we list the probablity distribution function:

Sum of roll   Prob. of sum     
    x            P(x)          x·P(x) 
-------------------------------------
    2            1/36           2/36
    3            2/36           6/36
    4            3/36          12/36 
    5            4/36          20/36
    6            5/36          30/36
    7            6/36          42/36
    8            5/36          40/36
    9            4/36          36/36
   10            3/36          30/36 
   11            2/36          22/36
   12            1/36          12/36
-------------------------------------
TOTALS          36/36=1       252/36 = 7

Expectation = E(x) = &#8721;[x·P(x)] = 7

So if you rolled two dice many many times and averaged
up all the sums, you would expect to get an average of about 7.

===================================================

Expectation of maximum:

Here are all 36 possible rolls with a pair of dice:

(1,1)  (1,2)   (1,3)  (1,4)   (1,5)  (1,6)
 
(2,1)  (2,2)   (2,3)  (2,4)   (2,5)  (2,6)
 
(3,1)  (3,2)   (3,3)  (3,4)   (3,5)  (3,6)
 
(4,1)  (4,2)   (4,3)  (4,4)   (4,5)  (4,6)
 
(5,1)  (5,2)   (5,3)  (5,4)   (5,5)  (5,6)
 
(6,1)  (6,2)   (6,3)  (6,4)   (6,5)  (6,6)


The maximums are

  1      2       3      4      5      6

  2      2       3      4      5      6

  3      3       3      4      5      6
 
  4      4       4      4      5      6

  5      5       5      5      5      6

  6      6       6      6      6      6      

There is 1 1, 3 2's, 5 3's, 7 4's, 9 5's and 11 6's.  
And there are 36 possible rolls, so the probabilities are the
number of ways to roll the maximum over 36.

So we list the probablity distribution function:

Max of roll   Prob. of max     
    x            P(x)          x·P(x) 
-------------------------------------
    1            1/36           1/36
    2            3/36           6/36
    3            5/36          15/36 
    4            7/36          28/36
    5            9/36          45/36
    6           11/36          66/36
-------------------------------------
TOTALS          36/36=1       161/36 = 4.47222···

Expectation = E(x) = &#8721;[x·P(x)] = 161/36

==============================================

So if you rolled two dice many many times and averaged
up all the maximums, you would expect to get an average of about 4.47222···

Edwin</pre>