probability math

Question

Tickets for a raffle cost 9. There were 616 tickets sold. One ticket will be randomly selected as the winner, and that person wins 1800 and also the person is given back the cost of the ticket. For someone who buys a ticket, what is the Expected Value (the mean of the distribution)?

Answer 1

P(winning) = 1/616 1800 x P(winning) = 1800(1/616) = $2.92

P(losing = 615/616 = .99838 P(losing)(-1) = about -$1

$2.92-$1 = expected value of the raffle ticket of $1.92.

It's the mean, highly skewed. The mode is -$1, the median = -$1. which are better estimates of the likely income when there is an outlier, especially this far of an outlier.

If you played this raffle game 616 times, on average you'd win $1800 one time and lose $615 for a net

gain of 1800-615 = $1185. divide by 616 = gain per raffle ticket = $1.92

Buy up all 616 tickets and you win $1185