Question

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, the player can bet $1 that the sum will be under 7—that is, 2, 3, 4, 5, or 6. For this bet, the player wins $1 if the result is under 7 and loses $1 if the outcome equals or is greater than 7. Similarly, the player can bet $1 that the sum will be over 7—that is, 8, 9, 10, 11, or 12. Here, the player wins $1 if the result is over 7 but loses $1 if the result is 7 or under. A third method of play is to bet $1 on the outcome 7. For this bet, the player wins $4 if the result of the roll is 7 and loses $1 otherwise.

(a) Construct the probability distribution representing the different outcomes that are possible for a $1 bet on being under 7.

(b) Construct the probability distribution representing the different outcomes that are possible for a $1 bet on being over 7.

(c) Construct the probability distribution representing the different outcomes that are possible for a $1 bet on 7.

(d) Show that the expected long-run profit (or loss) to the player is the same, no matter which method of play is used.

Answer 1

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Since there are two dice which have been rolled. Therefore Total # of observations possible are 6^2 = 36.

Sample space is drawn below with random variable as sum of both the rolls specified.

Dice_1	Dice_2	sum
1	1	2
2	1	3
3	1	4
4	1	5
5	1	6
6	1	7
1	2	3
2	2	4
3	2	5
4	2	6
5	2	7
6	2	8
1	3	4
2	3	5
3	3	6
4	3	7
5	3	8
6	3	9
1	4	5
2	4	6
3	4	7
4	4	8
5	4	9
6	4	10
1	5	6
2	5	7
3	5	8
4	5	9
5	5	10
6	5	11
1	6	7
2	6	8
3	6	9
4	6	10
5	6	11
6	6	12

a). Let X1 be the random variable representing outcomes of sum of both rolls of dice under 7.

therefore probability distribution is as under :

X1	Freq.	P (X1)
2	1	0.027778
3	2	0.055556
4	3	0.083333
5	4	0.111111
6	5	0.138889

b). Let X2 be the random variable representing outcomes of sum of both rolls of dice above 7.

therefore probability distribution is as under

X2	Freq.	P (X2)
8	5	0.138889
9	4	0.111111
10	3	0.083333
11	2	0.055556
12	1	0.027778

c).   Let X3 be the random variable representing outcomes of sum of both rolls of dice equals to  7.

therefore probability distribution is as under :

X3	Freq.	P (X3)
7	6	0.166667

When we add all P(X1) + P(X2) + P(X3) = 1, therefore our distribution is correct.

d).

Therefore since E(Y1), E(Y2) & E(Y3) are all equal to -0.1667 which is same. Hence proved.