Question 838603
<pre>
Here are all the possible dice rolls:

(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)  
  

 Events    Winnings X       P(X)   E(X)=X·P(X)    
----------------------------------------------
rolling 2   2*$6=$12        1/36    $12/36   
rolling 3   3*$6=$18        2/36    $36/36
rolling 4   4*$6=$24        3/36    $72/36
rolling 5   5*$6=$30        4/36   $120/36
rolling 6   6*$6=$36        5/36   $180/36
rolling 7   7*$6=$42        6/36   $252/36
rolling 8   8*$6=$48        5/36   $240/36
rolling 9   9*$6=$54        4/36   $216/36
rolling 10  10*$6=$60       3/36   $180/36
rolling 11  11*$6=$66       2/36   $132/36
rolling 12  12*$6=$72       1/36    $72/36 
------------------------------------------
     Total expectation  = &#8721;(E(X) = $1512/36 = $42
</pre>
what is the expected value of the game? 
<pre>
$42
</pre>
Is it a fair game?
<pre>
Yes because the cost to play is $42, which is the same as the  
expectation.

Thus if you play this game many times, you will average breaking
even. 

 (Games at fairs are NEVER fair. So no 'fair games' are fair games! hahaha!) 
 
Edwin</pre>