SOLUTION: Suppose it cost $42 to roll a pair of dice. You get paid $6 times the sum of the numbers that appear on the dice. If the cost of the game is not taken into consideration, what is t

Algebra ->  Probability-and-statistics -> SOLUTION: Suppose it cost $42 to roll a pair of dice. You get paid $6 times the sum of the numbers that appear on the dice. If the cost of the game is not taken into consideration, what is t      Log On


   

Question 838603: Suppose it cost $42 to roll a pair of dice. You get paid $6 times the sum of the numbers that appear on the dice. If the cost of the game is not taken into consideration, what is the expected value of the game? Is it a fair game?
Found 2 solutions by Fombitz, Edwin McCravy:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Look at the possible outcomes,
11=2
12=3
13=4
14=5
15=6
16=7
21=3
22=4
23=5
24=6
25=7
26=8
31=4
32=5
33=6
34=7
35=8
36=9
41=5
42=6
43=7
44=8
45=9
46=10
51=6
52=7
53=8
54=9
55=10
56=11
61=7
62=8
63=9
64=10
65=11
66=12
Calculate the probability for each sum,
Sum=2 or 12 P=1%2F36
Sum=3 or 11 P=2%2F36
Sum=4 or 10 P=3%2F36
Sum=5 or 9 P=4%2F36
Sum=6 or 8 P=5%2F36
Sum=7 P=6%2F36
The expected value is then,

E=%281512%2F36%29
E=42
Is it fair? The expected value equals the cost to play.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
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  = ∑(E(X) = $1512/36 = $42

what is the expected value of the game? 
$42

Is it a fair game?
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