Find the expectation of the game.

Question

If a player rolls two dice and gets a sum of 2 or 12, she wins $20. If the person gets a 7, she wins $5. The cost to play the game is $3. Find the expectation of the game.

Answer 1

For two “fair” dice (equal probability of rolling 1 to 6 each), that means there is an equal probability of rolling each the numbers 11 to 66.  We refer to this as “duplicates allowed” and you will often see problem where an item drawn from a set is put back in before the next draw.
 
Now, this problem is interesting because it is not the numbers, but their sum that determines the winnings.  So, we have to calculate the frequency distribution of the sums.

     Sum          Ways to get sum       The results
      2                       1                             (2+2)
      3                       2                             (1+2),(2+1)
      4                       3                             (1+3),(2+2),(3+1)
      5                       4                             (1+4),(2+3),(3+2),(4+1)
      6                       5                             (1+5),(2+4),(3+3),(4+1),(5+1)
      7                       6                             (1+6),(2+5),(3+4),(4+3),(5+2),(6+1)
      8                       5                             (2+6),(3+5),(4+4),(5+3),(6+2)
      9                       4                             (3+6),(4+5),(5+4),(6+3)
     10                      3                             (4+6),(5+5),(6+4)
     11                      2                             (5+6),(6+5)
     12                      1                             (6+6)
                          ----------
             Total        36

The “expected value” of a play is just the average of the 36 possible outcomes.  In this problem, we take the ways to get the sum and multiply by (the dollars won/lost minus the cost to play).   [note: negative result will indicate player loss].

There are 3 sums for which the player wins something.  There are 1+1=2 ways to win $20.  There are 6 ways to win $5.  Also, remember that each game costs $3 to play and there are 36-2-1=33 ways to spend $3 and win nothing.  Subtracting the $3 to play the game so that we have net winnings, we have an average (“expected value”) of:

      [2 ways to win $20; 6 ways to win $5; 33 ways to win $0] / 36 possible outcomes

      (2*17) + (6*2) + (33*(-3))  / 36 
       ( 34      + 12       -  99 )   / 36
                    -1.47        The player’s expected loss per game is $1.47.

 

Extra note:  There are formulas for this that you will want to learn as problems get more complex.