(Round to two decimal places as needed.) In the game of ave a of winning. If the
ID: 2934804 • Letter: #
Question
(Round to two decimal places as needed.) In the game of ave a of winning. If the metal 38 ball lands on 30, the player gets to keep the $10 paid to play the game and the player is awarded an additional $350 Otherwise, the player is awarded nothing and the casino takes the player's $10. What is the expected value of the game to the player? If you played the game 1000 times, how much would you expect to lose? Note that the expected value is the amount, on average, one would expect to gain or lose each game. The expected value is$ (Round to the nearest cent as needed.) The player would expect to lose about $ (Round to the nearest cent as needed.)Explanation / Answer
a) The expected value per game to the player here is computed as:
= ( 350 ) * Probability of winning bet + (-10)*Probability of losing the bet
= (350)*(1/38) - 10*(37/38 )
= 0.53
Therefore - $0.53 that is - 53 cents is the value of the game to the player here.
b) Now for 1000 times if the game is played, the expected amount that a player is to lose is computed as:
= 1000*0.53
= $526.32
Therefore the player would be expected to lose about $526.32
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