Question

In the game of​ roulette, a player can place a ​$5 bet on the number 3 and have a startfraction 1 over 38 endfraction probability of winning. if the metal ball lands on 3​, 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 ​$5. what is the expected value of the game to the​ player? if you played the game 1000​ times, how much would you expect to​ lose? note that the expected value is the​ amount, on​ average, one would expect to gain or lose each game. the expected value is ​$ nothing. ​(round to the nearest cent as​ needed.) the player would expect to lose about ​$ nothing. ​(round to the nearest cent as​ needed.)

Answer 1

The expected value per game is -0.26.  Over 1000 games, you can expect to lose $263.16.

To find the expected value, we multiply the probability of winning by the amount of winnings, the probability of losing by the amount of loss, and adding those together.

We have a 1/38 chance of winning; 1/38(175) = $4.61.  We also have a 37/38 chance of losing; 37/38(5) = $4.87.

$4.61-$4.87 = -$0.26 (rounded)

To five decimal places, our answer is -0.26136; multiplied by 1000 games, this is $261.36 lost.