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You place a 1-dollar bet on the number 17 at Las Vegas, and your friend places a 1-dollar bet on black (see Exercises 1.1.6 and 1.1.7). Let X be your winnings and Y be her winnings. Compare E(X), E(Y ), and V(X), V(Y). What do these computations tell you about the nature of your winnings if you and your friend make a sequence of bets, with you betting each time on a number and your friend betting on a color

Sagot :

nuhulawal20

Answer:

E(x)  = -$0.05263

E(y) = -$0.05263

V(x) = 0.9972

V(y) = 32.23

the expectations are the same, both will lose the same amount of money if they play for 38 times.

Though Variance is different. The girl betting on colors has little chance to gain  and lose but the boy betting on number has huge deviation

Step-by-step explanation:

Given that; you place a 1-dollar bet on the number 17 at Las Vegas and your friend places a 1-dollar bet on black.

Let x represent your winnings and y represent your friends winnings

we know that, There are 18 black slots, 18 red slots and two green slots (0 and 00) on a US roulette wheel.

so, the chances of getting black(red) = 18/38

now, E(y) = (18/38)$1 + (20/38)(-$1)

E(y) = -$0.05263

Now, bets made on the chart of the various numbers 1 to 36 and zero, Just put a chip down in the middle of the square with the number you select. If it comes up, it pays 35-to-1. It is called a straight up bet.

hence, the chance of getting selected number will be 1/38

so E(x) will be;

E(x) = (1/38)$35 + (37/38)(-$1)

E(x)  = -$0.05263

So, V(x) will be;

V(x) = [ 1 - (1/19)² ]

= 1 - 0.00277

V(x) = 0.9972

V(y) = { [ (35)² × (1/38) ] - (1/19)² }

= 32.2368 - 0.00277

V(y) = 32.23

Therefore, the expectations are the same, both will lose the same amount of money if they play for 38 times.

Though Variance is different. The girl betting on colors has little chance to gain  and lose but the boy betting on number has huge deviation.

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