Question 838603
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=1/36}}}
Sum=3 or 11 {{{P=2/36}}}
Sum=4 or 10 {{{P=3/36}}}
Sum=5 or 9 {{{P=4/36}}}
Sum=6 or 8 {{{P=5/36}}}
Sum=7 {{{P=6/36}}}
The expected value is then,
{{{E=2*6*(1/36)+3*6*(2/36)+4*6*(3/36)+5*6*(4/36)+6*6*(5/36)+12*6*(1/36)+11*6*(2/36)+10*6*(3/36)+9*6*(4/36)+8*6*(5/36)+7*6*(6/36)}}}
{{{E=(1512/36)}}}
{{{E=42}}}
Is it fair? The expected value equals the cost to play.