Question 1083976: Suppose you buy a ticket for $1.00 out of a lottery of 1,000 tickets in which the prize for the one winning ticket is to be $500. What is your expected value?
a. $0.00
b. –$1.00
c. –$0.50
d. –$0.40
Answer by jim_thompson5910(35256)   (Show Source):
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Answer: Choice C) -$0.50
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Explanation:
Let's define the following events
A = event of winning the lottery
B = event of losing the lottery
The events are complementary. One or the other (but not both at the same time) must happen. This means
P(A) + P(B) = 1
where,
P(A) = probability of event A happening
P(A) = probability of winning the lottery
P(B) = probability of event B happening
P(B) = probability of losing the lottery
There are 1000 tickets and only 1 winning ticket. The probability of winning is
P(A) = (number of winning tickets)/(number of tickets total)
P(A) = 1/1000
P(A) = 0.001
The probability of losing is
P(B) = (number of losing tickets)/(number of tickets total)
P(B) = 999/1000
P(B) = 0.999
Take note how
P(A) + P(B) = 0.001+0.999 = 1
So an alternative to finding P(B) is to do
P(B) = 1-P(A)
P(B) = 1-0.001
P(B) = 0.999
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Now let's introduce the idea of net value. I'm going to use the function notation N(A) to indicate the net value of event A happening. It's the net result the person gets if the event happens. In this case,
N(A) = 500-1 = 499
meaning that the person's net winnings is $499 if they win.
Don't forget to subtract off the cost of the ticket.
Likewise
N(B) = -1
is the net value for the ticket holder if the person doesn't win the lottery.
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In summary so far, we have
P(A) = 0.001
P(B) = 0.999
N(A) = 499
N(B) = -1
Multiply the probabilities with their corresponding net values. Then add up the results to get the expected value E
E = P(A)*N(A) + P(B)*N(B)
E = 0.001*499 + 0.999*(-1)
E = 0.499 - 0.999
E = -0.50
On average, the expected loss is 0.50 dollars, which is a 50 cent loss.
This may not seem like much, but the losses add up.
Note: this is not a fair game. Mathematically a fair game is when the expected value is 0.
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