In the game of roulette, a player can place a $10 bet on the number 6 and have a 1/38 probability of winning. If the metal ball lands on 6, 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?The expected value is $___The player would expect to lose about $__

Answer:The expected value is $ - 0.53 (negative means losing)If you the player played the game 1,000 times, the player would expect to lose $530.Explanation:The expected value is calculated as the net result of the sum of the products of every probability times each value, less the cost.That is: expected value = [ ∑ (probability × value)] - costFor the game of roulette you have:cost of the bet: $10probability of wining: 1/38value of wining: $ 350 + $ 10 = $ 360probability of losing: 1 - 1/38 = 37/38value of losing: 0Expected value = $ 360 × 1/38 + $ 0 × 37/38 - $10 = $9.47 - $10 = - $ 0.53Since, each time you play is independent of the others plays,  if you played 1,000 times, you would expect to lose 1,000 times 0.53, i.e 1,000 × 0.53 = $ 530.