in a game of roulette a player can place a 10 dollar bet on

in a game of roulette a player can place a 10 dollar bet on number 16 and have 1/38 probability of winning.if the metal ball lands on 16 the players get to keep 10 dollars paid to play the game and the player is awarded an additional 350 dollars otherwise the player is awarded nothing and the casino takes the players 10 dollars . what is the expectd 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 $ (round to the nearest cent as needed)

the player would expect to losse about $ (round to the nearest cent as needed)

Comment

Solution

expected value is  x1P1 + x2P2 + x3P3 + . . . + xnPn

if we win we will get 350 and probability of winning id 1/38 therefore expected will be 1/38*350 = 9.21

if we lose we will lose 10 and proability of losing is 37/38 then expected value is = -10*37/38 = - 9.7

therefore total expected value will be 9.21 - 9.7 = -0.52

now if we play the game 1000 times we are expected to loose 1000 * 0.52 = 520 dollar we will lose.