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Ticket probability

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2 Answers

Look at the possible outcomes:
Holder of first number: probability 1/1000, winning $500-$1=$499
Holder of remaining three numbers: probability 3/1000, winnings $100-$1=$99
Holder of remaining 996 numbers: probability 996/1000, loss -$1
The expected winnings are:
(1/1000)($499)+(3/1000)($99)+(996/1000)(-$1) = -$0.20
On average, you lose 20 cents.
This is a confusing problem because it can be read in a number of ways.  Depending on how you read it, you don't  know a lot of things.  For example,
(1) The problem says that "you" paid $1 per ticket for 1000 tickets.  It does not say that "you" sold any of those.  If you kept all the tickets, then you spent $1000 and won back only $800 so "you" actually lost $200 or an average of $.20 per ticket.  This is an actual loss not an expected one. 
(2) How many tickets total were actually sold at the fair.  Without this information you don't know what the expected loss per ticket was.  If N tickets were sold at the fair, then the actual expected winnings per ticket for a random ticket would be 
(1/N)($499)+(3/N)($99) + ((N-4)/N)-$1) = -($1) + ($800)/N
So as the total number of tickets sold, N, gets large, the average ticket holder's expected winnings approach -$1.