help me math problems!?

Question

We roll a pair of dice. If the sum of the dice is 7, you pay me $31. If the sum is not 7, I pay you the number of dollars indicated by the sum of the dice. What is the price that you should pay to play the game that would make the game fair?

a. $2.24 b. $0.67 c. $4.11 d. -$1.63

Answer 1

c) $0.67

I didn't do the precise calculation ...

but if 5 times I get paid (average) $7 each
and 1 time I pay $31

(35-31) / 6 = 0.67

[on average, 1/6 of rolls are 7 and 5/6 are non-7

under = 2,3,4,5,6 "psuedo-ave" = 4
over = 8,9,10,11,12 "psuedo-ave" = 10

psuedo ave-ave = (10+4)/2 = 7

5*7 - 1*31 ....

arggh ... it's right

in "psuedo-under, the weighting favors the higher and in psuedo-over, the weighting favors the lower ..
ave-ave "balances" the weightings

Answer 2

I have an answer, but I'm not 100% positive: Wat i did was i found the chance of rolling 7-
3(# of possibilities for 7) / 30(total # of possibilities). this gives you a 10%. so 90% chance that you will receive money.

then find the average amount of money you receive if you were to win, which is the sum of 2-6,8-12 because those are your possible rolls (you cant roll a 1 w/ 2 dice, and 7 doesn't earn you money) The average is 7$ or 70/10.

so then u have 10% to get (-31) and 90% to get (7)
so to be equal, (-31) should be 90% larger than 7, or $27.90. So for this to be fair you shouls have to pay an additional 4.11.

I realize this is off by a cent, not sure why but i think c. is your answer.

If you could, please tell me if i was right or wrong.

Answer 3

X + Y = 7 , +$31
X + Y not = 7, -$(X+Y)

to get 7, you'd need
1, 6 / 6, 1
2, 5 / 5, 2
3, 4 / 4, 3

that's 6 combinations out of 72

the probability of me paying you $31 is 6/72 or 1/12
the probability of you paying me is 11/12

the max you could pay me is $12

31(1/12) vs 12(1/72) + 11(1/36) + 10(1/24) + 9(1/18) + 8(5/72) + 6(5/72) + 5(1/18) + 4(1/24) + 3(1/36) + 2(1/72)

31/12 vs 35/12
five dollar difference
we could divide that in half "to be fair" and get

A) $2.24

Answer 4

prob(7) = 6/36
prob(6) = p(8) = 5/36
p(5) = p(9) = 4/36
p(4) = p(10) = 3/36
p(3) = p(11) = 2/36
p(2) = p(12) = 1/36, so expected value is
-6/36(31) + 5/36(6) + 5/36(8) + 4/36(5 + 9) + 3/36(4 + 10) + 2/36(3+11) + 1/36(2+12) =
-31/6 + 14(5+4+3+2+1)/36 =
-31/6 + 14(15)/36 =
-31/6 + 35/6 =
4/6 =
2/3 ≈ $0.67

Answer 5

a.$2.24- If you roll each number once it totals 28, just $3 dollars less than rolling the jackpot.

Answer 6

$4.11 ??