SOLUTION: 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?

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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?
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
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?
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) = ∑[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) = ∑[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
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