Thread: Probability - Roulette View Single Post
 P: 8 1. The problem statement, all variables and given/known data This particular roulette wheel has 37 slots - 0,1,2...36. A gambler can bet on different combinations of numbers. Louise loves to bet on a block of 4 numbers, called a corner. The payout on a corner is 8 to 1. (a) Let X be a gambler's winning from a $1 bet on a corner. What is the probablity distribution for X? (Hint X can be negative) (b) What is the expected value of a$1 bet on a corner. 2. The attempt at a solution (a) Okay i figured that the chance of a corner would be. Pr(corner) = (1/37)X4 = 4/37. However, i dont think this counts as the probablity distribution? Not quite sure how to approach this. (b) Excepted value for a $1 bet. Need to take account both winning and losing? Pr (corner) = 4/37 Pay out is 8 to 1. therefore is excepted value is 4/37 X$8 = 32/37 Pr (no corner) = 33/37 ((1-(4/37)) Expected value is 33/37 X -\$8 = -264/37 Excepted value of 1 dollar bet = (-264/37) + (33/37) = -232/37 = -6.27 You would expect to lose -6.27?? Thanks