Question

Suppose that you are offered the following "deal." You roll a six sided die. If you roll a 6, you win $15. If you roll a 3, 4 or 5, you win $1. Otherwise, you pay $4.

a. Complete the PDF Table. List the X values, where X is the profit, from smallest to largest. Round to 4 decimal places where appropriate.

Probability Distribution Table

X	P(X)

b. Find the expected profit. $ ____ (Round to the nearest cent)

c. Interpret the expected value.

___ If you play many games you will likely win on average very close to $1.67 per game.

____ You will win this much if you play a game.

____ This is the most likely amount of money you will win.

d. Based on the expected value, should you play this game?

__ Yes, since the expected value is 0, you would be very likely to come very close to breaking even if you played many games, so you might as well have fun at no cost.

__ Yes, since the expected value is positive, you would be very likely to come home with more money if you played many games.

__ No, since the expected value is negative, you would be very likely to come home with less money if you played many games.

__ Yes, because you can win $15.00 which is greater than the $4.00 that you can lose.

__ No, this is a gambling game and it is always a bad idea to gamble.

Answer 1

Event : Rolling a six sided die.

When a die is rolled , it will result in an equally likely event of getting any one of the six numbers (1 through 6).

Thus , P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6

Outcome	Probability	Payout or Profit
1	1/6	-4
2	1/6	-4
3	1/6	1
4	1/6	1
5	1/6	1
6	1/6	15

Thus,

Profit is $15 : 6 is rolled

P(6) = 1/6 = P($15)

Profit is $1: 3,4 or 5 is rolled

P ( 3 or 4 or 5) = P(3) + P(4) + P(5)

=1/3 + 1/3 + 1/3

P(3 or 4 or 5) = 3/6 = P($1)

Profit is -4 : 1 or 2 is rolled

P(1 or 2) = P(1) + P(2)

= 1/6 + 1/6

P(1 or 2) = 2/6 = P(-4)

Thus , Probability distribution table is:

X : Profit	Probability
-4	0.3333
1	0.5
15	0.1667

b) Expected Value is given by :

E(x) = X*p(x)

X	P(x)	X * P(x)
-4	0.3333	-1.3332
1	0.5000	0.5000
15	0.1667	2.5005

E(x) = -1.3332 + 0.5000 + 2.5005

= 1.6673

Thus , Expected Profit is $1.6673

c) Interpretation: If you play many games you will likely win on average very close to $1.67 per game.

d) Yes , since the expected value is positive, you would be very likely to come home with more money if you played many games.

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