Question

In: Statistics and Probability

In the carnival game​ Under-or-Over-Seven, a pair of fair dice is rolled​ once, and the resulting...

In the carnival game​ Under-or-Over-Seven, a pair of fair dice is rolled​ once, and the resulting sum determines whether the player wins or loses his or her bet. For​ example, using method​ one, the player can bet $2.00 that the sum will be under​ 7, that​ is, 2,​ 3, 4,​ 5, or 6. For this​ bet, the player wins $2.00 if the result is under 7 and loses $2.00 if the outcome equals or is greater than 7.​ Similarly, using method​ two, the player can bet $2.00 that the sum will be over​ 7, that​ is, 8,​ 9, 10,​ 11, or 12.​ Here, the player wins $2.00 if the result is over 7 but loses $2.00 if the result is 7 or under. A third method of play is to bet ​$2.00 on the outcome 7. For this​ bet, the player wins $8.00 if the result of the roll is 7 and loses $2.00 otherwise.

Outcomes of a two dice roll

1

2

3

4

5

6

1

2

3

4

5

6

7

2

3

4

5

6

7

8

3

4

5

6

7

8

9

4

5

6

7

8

9

10

5

6

7

8

9

10

11

6

7

8

9

10

11

12

a. Construct the probability distribution representing the different outcomes that are possible for a $2.00 bet using method one.

X

​P(X)
  

​

​(Type an exact answer in simplified​ form.)

b. Construct the probability distribution representing the different outcomes that are possible for a $2.00 bet using method two.

X

​P(X)
  

​

​(Type an exact answer in simplified​ form.)

c. Construct the probability distribution representing the different outcomes that are possible for a ​$2.00 bet using method three.

X

​P(X)

​

​(Type an exact answer in simplified​ form.)

d. What is the expected​ long-run profit​ (or loss) to the player for each of the three methods of​ play?

Method one expected profit​ (or loss)

muμ

equals=

​

Method two expected profit​ (or loss)

muμ

equals=

Method three expected profit​ (or loss)

muμ

equals=

​

​$​(Round to the nearest cent as​ needed.)

Solutions

Expert Solution

a)

There are 36 outcomes one would get for rolling a pair of fair dice. For sum to be under 7, the possible outcomes are, 2, 3, 4, 5 and 6. The probabilities of the outcomes are 1/36, 2/36, 3/36, 4/36, 5/36. Assume, the r.v denote the sum of the outcome of dice. Therefore,

the P(X<=6)=1/36+2/36+3/36+4/36+5/36 = 5/12

The sum of 7 or more are 7, 8, 9, 10, 11 and 12.

Thus, P(X>=7)=6/36+5/36+4/36+3/36+2/36+1/36 = 7/12

P($2)=0.4167

P(-$2)=0.5833

b)

Out of 36 outcomes 21 has sum less than 7 or less. So the probability of getting a sum 7 or less than 7 is

P(sum equal to or less than 7) = 21/ 36

And probability of getting a sum of over 7 is

P(sum over 7) = 15/36

c)

X = Outcome of the bet of Method 3 in $

There are two possible values of X :

i) $16 if he wins the bet.

ii) - $4, if he loses the bet.

P(X = +16) = P(rolling a 7)

P(X = +16) = 6/36

P(X = -4) = P(rolling any number but 7)

P(X = -4) = 1 - P(rolling a 7) = 1 - 6/36 = 30/36

Answer: Probability distribution of X (Method Three) =

X P(X)
8 6/36
-2 30/36

d)

Expected value of X = E(X) =

For Method one : Expected profit (or loss) = (4 x 15/36) + (-4 x 21/36) = -24/36 = -2/3

Ans: In Method one, there is an expected loss of $(2/3) = $0.67 = 67 cents.

For Method two : Expected profit (or loss) = (4 x 15/36) + (-4 x 21/36) = -24/36 = -2/3

Ans: In Method two, there is an expected loss of $(2/3) = $0.67 = 67 cents.

For Method three : Expected profit (or loss) = (16 x 6/36) + (-4 x 30/36) = -24/36 = -2/3

Ans: In Method three, there is an expected loss of $(2/3) = $0.67 = 67 cents.

So, we see that in the long run, all the three methods have exactly the same expected loss = 67 cents.

orchestra answered 2 years ago
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