Question

Problem 2.

You pay 5$ /round (nonrefundable) to play the game of rolling a pair of fair dice.

If you roll an even sum, you lose, no pay off. If you roll an odd sum, that's your win (say, roll of 7 pays you 7$).

Discrete random variable X represents the winnings.

• For example, the lowest value of X is x=0 when you roll an even sum.
• For example, x=3 only when you roll {1,2} or {2,1}. You win 1+2=3$ (odd sum). AndP ( x = 3 ) = P ( { 1 , 2 } ) + P ( { 2 , 1 } ) = 1 / 36 + 1 / 36 = 1 / 18.

a) Find all possible values of X with their probabilities. Make the table as in Problem 1, a) for the probability distribution of X. Above, we calculated just one row of the table:

x	Add favorable dice Probabilities	P(x)
0	…	…
3	1/36 + 1/36	1/18
…	…	…

Dice related probabilities are discussed in 5.1, page 252, Example 5.

b) Find the expected value of X and interpret it.

Expected value is discussed in 6.1, page 320 (as mean), 321 and Examples 5,6,7.

c) Does it make mathematical sense to play the game? Remember, you have to pay 5$/game to play, what is your net gain/loss per game in the long run?

d) What price a (instead of 5$) would make the game fair? It is called the fair price as you break even in the long run: μ ( X ) − a = 0.

Answer 1

Answer:

Given data

2.

a)Find all possible values of X with their probabilities. Make the table as in Problem 1, a) for the probability distribution of X.

The probability distribution of X is as follows.

X=x	Favourable events	Sum of favourable dice probabilities	P(X=x)
0	(1,1),(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5),(2,2),(2,4),(2,6),(4,2),(4,4),(4,6),(6,2),(6,4),(6,6)	1/36 + 1/36 +1/36 +1/36 +1/36 + 1/36 + 1/36+ 1/36 +1/36+ 1/36 + 1/36+ 1/36+ 1/36+ 1/36+ 1/36 +1/36+ 1/36+ 1/36	18/36 = 1/2
3	(1,2),(2,1)	1/36 +1/36	2/36 = 1/18
5	(1,4),(4,1),(2,3),(3,2)	1/36+ 1/36 +1/36+ 1/36+	4/36 = 1/18
7	(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)	1/36+ 1/36+ 1/36 +1/36+ 1/36+ 1/36	6/36 = 1/6
9	(3,6),(6,3),(4,5),(5,4)	1/36 +1/36+ 1/36+ 1/36	4/36 = 1/9
11	(5,6),(6,5)	1/36+ 1/36	2/36 = 1/18

b) Find the expected value of X and interpret it.

Expected value is discussed in 6.1, page 320 (as mean), 321

E(X) = * P( X = x ) ==3.5

In long run, it is expected to win $ 3.5 per game.

c)  Does it make mathematical sense to play the game? Remember, you have to pay 5$/game to play, what is your net gain/loss per game in the long run

Expected win from a game is $ 3.5

Price to play a game is $ 5

So, expected profit to play a game = $ (3.5 - 5 ) = $ -1.5 i.e. expected loss to play a game is $ 1.5

Hence, there is no mathematical sense to play the game.

d) What price a (instead of 5$) would make the game fair? It is called the fair price as you break even in the long run: μ ( X ) − a = 0.

A game is fair if expected profit to play a game is 0.

Expected win from a game is $ 3.5

To be a fair game, price to play a game should be $ 5.