Question

If a person rolls a sum of 8 when she rolls two dice, she wins $10. If the game is to be fair, how much (to the nearest penny) should the person pay to play the game?

Answer 1

Let c be the cost of the game. then the expected winnings is

(10-c)*P(roll an 8)+(-c)P(you don't get an 8)

Answer 2

is there some way to simplify it down?

Answer 3

what if i did this? Is this the same?
10cP+8-c

Answer 4

(10-c)*P(roll an 8)+(-c)P(you don't get an 8)

=10*P(roll and 8)-c

set equal to zero (fair game)

then c=10*P(roll an 8)

Answer 5

so wouldnt the answer be $1.39

Answer 6

yes, that looks good.

to review

the amount you get if you roll an 8 is 10-cost
the amount you "get" is -cost if you roll a "not 8"

the expected "earnings" is
(10-cost)* P(8) + -cost*P(not 8) = 0 (which means in the long run, you don't win or lost any money)

distribute:
10* P(8) - cost*P(8) - cost* P(not 8) = 0

factor out -cost:
10* P(8) - cost * (P(8) + P(not 8) ) =0

notice the probability of getting an 8 plus the probability of not getting 8 is 1 (100% one or then other happens)
so the equation simplifies to
10*P(8) - cost *1 = 0
or
cost = 10*P(8)

you now need to find the probability of getting an 8
there are 5 ways (use a table to find the combinations)
out of 36 different rolls. P(8)= 5/36

cost = 10*5/36