In the game of roulette, a player can place a $66 bet on the number 99 and have a startfraction 1 over 38 endfraction 1 38 probability of winning. if the metal ball lands on 99, the player gets to keep the $66 paid to play the game and the player is awarded an additional $210210. otherwise, the player is awarded nothing and the casino takes the player's $66. find the expected value e(x) to the player for one play of the game. if x is the gain to a player in a game of chance, then e(x) is usually negative. this value gives the average amount per game the player can expect to lose.
the expected value is $
the player would expect to lose about $__

The expect value of the game is $ -0.32.

The player is expected to lose about $ 0.32 or 32 cents per game.

Step-by-step explanation:

Note: The digits in the question statement are pasted twice. Correct values are metal ball lands on 9, player gets to keep his $6, player is awarded $210.

Probability of winning =

Probability of losing = 1 - Probability of winning

So,

Probability of losing =

On winning the player gets to keep his $6 and is awarded an additional $210. So, amount of money he will make on winning is $210. On losing the player will lose his $6.

Expected value is calculated as: Sum product of the probabilities with their payouts. The payout on losing will be negative as the money is being lost.

So, expected value of this game would be:

Therefore, the expected value of the game is $ -0.32.

The player is expected to lose about $ 0.32 or 32 cents per game.

c: lkllmn.

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