In the game of roulette, a player can place a $5 bet on the number 11 and have a startfraction 1 over 38 endfraction probability of winning. if the metal ball lands on 11, the player gets to keep the $5 paid to play the game and the player is awarded an additional $175. otherwise, the player is awarded nothing and the casino takes the player's $5. 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?

we know that
P(winning) = 1/38
P(losing) = 1-1/38 = 37/38
Expected value = 175*1/38 - 5*37/38
175/38 - 185/38 = - 10/38 =-$0.2632

the answer part A) is
The player's expected value is -$0.2632 ( is losing)

b) if you played the game 1000 times, how much would you expect to lose?
-$0.2632*1000=-$263.20

the answer Part B) is $263.20

Answer Link

shubheetutor shubheetutor · Answer 2 · 2021-10-19T20:58:47+02:00

if you played the game 1000 times, [tex]\$ 263.20[/tex] are expected to lose.

[tex]Probability=\dfrac{Favourable\; outcomes}{Total\; number\; of\; outcomes}[/tex]

Probability of winning[tex]=\dfrac{1}{38}[/tex]

if out of total [tex]38[/tex], [tex]1[/tex] game has been won, so rest have been lost,so

Probability of losing [tex]=\dfrac{37}{38}[/tex]

Hence Expected of the game to the player [tex]=175 \times \dfrac{1}{38}-5 \times \dfrac{37}{38}[/tex]

[tex]=\dfrac{175}{38}- \dfrac{185}{38}\\\\=\$ -0.2632[/tex]

A)The expected value is [tex]\$ -0..2632[/tex] (is losing)

B)if you played the game [tex]1000[/tex] times,

so it will be [tex]1000 \times \$ -0.2632=\$263.20[/tex]

Learn more about Probability here;

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