Page 96 - Probability Demystified

CHAPTER 5 Odds and Expectation 85

Since the person who buys a ticket does not get his or her $1 back, the net
gain if he or she wins is $750 $1 ¼ $749. The probability of winning is one
chance in 1000 since 1000 tickets are sold. The net loss is $1 denoted as
999
negative and the chances of not winning are 1000 1 or 1000 : Now EðXÞ¼
1000
$749 1 þð $1Þ 999 ¼ $0:25:
1000 1000
Here again it is necessary to realize that one cannot lose $0.25 but what
this means is that the house makes $0.25 on every ticket sold. If a person
purchased one ticket for raffles like this one over a long period of time, the
person would lose on average $0.25 each time since he or she would win on
average one time in 1000.
There is an alternative method that can be used to solve problems when
tickets are sold or when people pay to play a game. In this case, multiply the
prize value by the probability of winning and subtract the cost of the ticket or
the cost of playing the game. Using the information in the previous example,
the solution looks like this:
1
EðXÞ¼ $750 $1 ¼ $0:75 $1 ¼ $0:25:
1000
When the expected value is zero, the game is said to be fair. That is, there is a
fifty–fifty chance of winning. When the expected value of a game is negative, it
is in favor of the house (i.e., the person or organization running the game).
When the expected value of a game is positive, it is in favor of the player. The
last situation rarely ever happens unless the con man is not knowledgeable of
probability theory.

EXAMPLE: One thousand tickets are sold for $2 each and there are four
prizes. They are $500, $250, $100, and $50. Find the expected value if a
person purchases 2 tickets.

SOLUTION:
Find the expected value if a person purchases one ticket.

Gain, X $499 $249 $99 $49 $1
1 1 1 1 996
Probability P(X)
1000 1000 1000 1000 1000

1 1 1 1 996
EðXÞ¼ $499 þ $249 þ $99 þ $49 $1
1000 1000 1000 1000 1000
¼ $0:10