Question

A player pays $ 13 to roll three six-sided balanced dice. If the sum of the 3 dice is less than 13, then the player will receive a prize of $ 70. Otherwise, you lose the $13.
a. Find the expected value of profit.

Answer 1

Let X denotes the profit.

X is either (70-13) = 57 (in case of winnin)

or -13 (in case of losing)

P(win) = P(sum of 3 dice is < 13)

p(lose) = 1 - p(win)

If 3 dice are rolled, there are 216 possible outcomes. We have to count those outcomes for which, sum is less than 13.

• There is one possible way three dice can total 3
• 3 ways for 4
• 6 for 5
• 10 for 6
• 15 for 7
• 21 for 8
• 25 for 9
• 27 for 10
• 27 for 11
• 25 for 12

Total = 1 + 3 + 6 + 10 + 15 + 21 + 25 + 27 + 27 + 25 = 160

So, P(win) = 160/216 = 0.74

Expected value of profit = 57*0.74 + (-13)*(1-0.74) = 38.8

Counting note:

I am providing some of the possibilities (not whole, it will take lot of space)

• 3 = 1 + 1 + 1
• 4 = 1 + 1 + 2
• 5 = 1 + 1 + 3 = 2 + 2 + 1
• 6 = 1 + 1 + 4 = 1 + 2 + 3 = 2 + 2 + 2
• 7 = 1 + 1 + 5 = 2 + 2 + 3 = 3 + 3 + 1 = 1 + 2 + 4
• 8 = 1 + 1 + 6 = 2 + 3 + 3 = 4 + 3 + 1 = 1 + 2 + 5 = 2 + 2 + 4
• 9 = 6 + 2 + 1 = 4 + 3 + 2 = 3 + 3 + 3 = 2 + 2 + 5 = 1 + 3 + 5 = 1 + 4 + 4
• 10 = 6 + 3 + 1 = 6 + 2 + 2 = 5 + 3 + 2 = 4 + 4 + 2 = 4 + 3 + 3 = 1 + 4 + 5
• 11 = 6 + 4 + 1 = 1 + 5 + 5 = 5 + 4 + 2 = 3 + 3 + 5 = 4 + 3 + 4 = 6 + 3 + 2
• 12 = 6 + 5 + 1 = 4 + 3 + 5 = 4 + 4 + 4 = 5 + 2 + 5 = 6 + 4 + 2 = 6 + 3 + 3

Note: 12 can be obtained as 5 + 4 + 3 , which is a permutation of second sum in row corresponding to 12 (in bold) .

As I wrote earlier, i am providing only some of possibilities.