Page 97 - Probability Demystified

86 CHAPTER 5 Odds and Expectation

The expected value is $0.10 for one ticket. It is 2( $0.10) ¼ $0.20 for
two tickets.

Alternate solution

1 1 1 1
EðXÞ¼ $500 þ $250 þ $100 þ $50 $1
1000 1000 1000 1000
¼ $0:10

2ð $0:10Þ¼ $0:20

Expectation can be used to determine the average amount of money the
house can make on each play of a gambling game. Consider the game called
Chuck-a-luck. A player pays $1 and chooses a number from 1 to 6.
Then three dice are tossed (usually in a cage). If the player’s number comes
up once, the player gets $2. If it comes up twice, the player gets $3, and if it
comes up on all three dice, the player wins $4. Con men like to say that the
1
probability of any number coming up is on each die; therefore, each number
6
1
3
has a probability of or of occurring, and if it occurs more than once, the
6 2
player wins more money. Hence, the game is in favor of the player. This is
not true. The next example shows how to compute the expected value for the
game of Chuck-a-luck.
EXAMPLE: Find the expected value for the game Chuck-a-luck.

SOLUTION:

There are 6 6 6 ¼ 216 outcomes in the sample space for three dice. The
5
probability of winning on each die is 1 and the probability of losing is :
6 6
1
1
1
The probability that you win on all three dice is ¼ 1 :
6 6 6 216
5
5
5
The probability that you lose on all three dice is ¼ 125 :
6 6 6 216
1
5
1
The probability that you win on two dice is ¼ 5 , but this can occur
6 6 6 216
in three different ways: (i) win on the first and the second dice, and lose on the
third die, (ii) win on the first die, lose on the second die, and win on the third
die, (iii) lose on the first die, and win on the second and third dice. Therefore,
the probability of winning on two out of three dice is 3 5 ¼ 15 :
216 216
5
5
1
The probability of winning on one die is ¼ 25 , and there are three
6 6 6 216
different ways to win. Hence, the probability of winning on one die is
3 25 ¼ 75 :
216 216