Question

In the American version of the game of roulette, a wheel consists of 38 slots numbered 0, 00, 1, 2, … , 36. To play the game, a metal ball is spun around the wheel and is allowed to fall into one of the numbered slots. A “street” bet is a bet on three consecutive numbers (e.g. 11, 12, and 13 – if the ball lands on any of these, the player wins). If the payout for a “street” bet is 11 to 1 (the original bet is returned, along with 11 times its value, so a $1 bet nets a profit of $11), what is the expected value of $1 “street” bet?

Round your answer to the nearest cent.

Answer 1

P(win) = P(getting 3 consecutive numbers out of 38) = 3/38

P(lose) = P(getting numbers other than 3 consecutive numbers) = 1 - 3/38

= 35/38

Expected value = P(win)*(winning amount) +P(lose)*(losing amount)

= (3/38)*(11) + (35/38)*(-1)

=33/38 - 35/38

=-2/38

= -0.05