1. ## Probability

Help i've been trying to figure his out all week?

In the game of roulet a player can place a $5 bet on the number 17 and have a 1/38 probability of winning ifthe metal ball lands on 17 the player wins$175, otherwise the casino takes the players $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? 2. Hello, kandela573! Are you familiar with the Expected Value formula? In the game of roulette, a player can place a$5 bet on the number 17.
He has a 1/38 probability of winning $175. Otherwise, the casino takes the player's$5.

(a) What is the expected value of the game to the player?
(b) If you played the game 1000 times how much would you expect to lose?
It's really straight-forward . . .

. . $\displaystyle \begin{array}{ccc}P(\text{win \$175}) &=& \dfrac{1}{38} \\ \\[-3mm]
P(\text{lose \$5}) &=& \dfrac{37}{38} \end{array}$

$\displaystyle E \;=\;\left(\frac{1}{38}\right)(+175) + \left(\frac{37}{38}\right)(-5) \;=\;-\frac{10}{38} \;=\;-0.263157895$

He can expect to lose an average of about $\displaystyle 26.3$ cents per game. (a)

In 1000 games: . $\displaystyle E \;=\;1000 \times (-0.263157895) \;= \;-263.157895$

. . We can expect to lose about $263.16 (b) 3. ## actuallly You got it wrong you forgot to take out the$5 dollars that he placed as his original bet so the answer actually was

170*1/38+(-5)*37/38=-.395 for his expected loses for each role
for 1000 games youthen expected to lose $395 4. Hello, kandela573! Sorry, you're way off-base . . . You got it wrong. You forgot to take out the$5 dollars that he placed as his original bet.
No, I didn't . . . Did you look at my reply?

170(1/38) + (-5)(37/38) = -0.395 for his expected losses for each roll.
. ?

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# 1:38 probability

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