Question 1173901: A lottery system has balls numbered 1 to 60 and randomly selects 7 of the lottery balls. There is only one prize of $5,600,000.00 which is awarded only if a lottery player selects the correct set of 7 lottery balls.
a) If a lottery ticket costs $1.00, what is a lottery player's expected value?
b) How much would the lottery prize have to be worth if it was to be a fair game?
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! The number of combinations of 7 numbers is N = 60*59*58*57*56*55*54/7! = 386206920
P(winning) = 1/N. Hence the probability of NOT winning = (N-1)/N
The prize amount is $5.6M, and the ticket cost is $1.
The expected value is therefore P(winning)*prize - P(losing)*ticket cost =
(1/N)*$5.6M - ((N-1)/N)*$1 = -0.9855 ~-98.6.
So you will lose almost 99 cents for every dollar bet.
For a fair game, i.e. break even, P/N = (1 - 1/N) -> P = N - 1 = 386206919 = $386.2M
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