In the game of roulette, a player can place a $ bet on the number and have a probability of winning. If the metal ball lands on , the player gets to keep the $ paid to play the game and the player is awarded an additional $. Otherwise, the player is awarded nothing and the casino takes the player's $. What is the expected value of the game to the player? If you played the game 1000 times, how much would you expect to lose?

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okpalawalter8 okpalawalter8 · Answer 1 · 2020-10-21T18:37:41+02:00

Complete Question

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Answer:

a

The expected amount to be lost is 0.5263 cents per game

b

The expected amount to be lost is

[tex]P = \$ 526 .3 [/tex]

Step-by-step explanation:

From the question we are told that

The probability of winning is [tex]p = \frac{1}{38} = 0.0263[/tex]

Generally the probability of losing is mathematically evaluated as

[tex]q = 1-p [/tex]

=> [tex]q = 1-0.0263 [/tex]

=> [tex]q = 0.9737[/tex]

Generally the expected value is mathematically represented as

[tex]E(X) = \sum x_i * P(x_i)[/tex]

Here [tex]x_i[/tex] is a discrete variable (i.e it is a countable ) are outcomes of the player winning and the player losing

So

[tex]E(X) = x_1 * P(x_1) + x_2 * P(x_2) [/tex]

Now the outcome of winning is making $350 so

[tex]x_1 = \$ 350[/tex]

the outcome of losing is losing $10 so

[tex]x_2 = - \$10[/tex]

So

[tex]E(X) = 350* 0.0263 -10 * 0.9737 [/tex]

[tex]E(X) = -0.5263 [/tex]

hence the expected amount to be lost is 0.5263 cents per game

If you played the game 1000 times, the amount that is expected to be lost is

[tex]P = 1000 * -0.5263[/tex]

=> [tex]P = -\$ 526 .3 [/tex]

Hence the expected amount to be lost is

[tex]P = \$ 526 .3 [/tex]