20. In the game of roulette, there are 18 red spaces, 18 black spaces and 2 gree

Question

20. In the game of roulette, there are 18 red spaces, 18 black spaces and 2 green spaces. Each of the spaces also has a unique number on it. A ball is placed on the wheel and it lands in one of the spaces. a) One of the bets that can be made is to bet on black or red. If you bet $1 on red and the ball lands on one of the red spaces, you win $1 (not get - win). If the ball does not land on red, you lose your dollar. Determine the expected value of this bet. b) Another bet you can make is to bet on a specific number. If the ball lands on that number, you win $35. If not, you lose $1. Determine the expected value of this bet.

Explanation / Answer

a)

Out of 38 slots, 18 are red so

P(red) = 18/38

P(not red) = 1- 18/38 = 20/38

Let X is a random variable shows the winning amount. When ball is on red then you win X= $1-$1 = 0. When ball is not on red slot then X= -$1.

So expected value is

E(X) = (18/38) * ($0) + (20/38) * (-$1) = -20/38 = -$0.53

b)

Since each number of unique so probability of getting a number of which you bet is

P(win) = 1/38

P(not win) = 1- 1/38 = 37/38

Let X is a random variable shows the winning amount. When ball is on requried slot then you win X= $54-$53. When ball is not on required slot then X= -$1.

So expected value is

E(X) = (1/38) * ($34) + (37/38) * (-$1) = -3/38 = -$0.08

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