- #1

nicholasch

- 8

- 0

## Homework Statement

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 don't 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