Yfor a daily lottery, a person selects a three-digit number. if the person plays for $1, she can win $500. find the expectation. in the same daily lottery, if a person boxes a number, she will win $80. find the expectation if the number 123 is played for $1 and boxed. (when a number is "boxed," it can win when the digits occur in any order.)

The expected value of the first game is -$0.50 and of the second game is -$0.52.

There are 10³ possible numbers for the lottery, and only 1 of them will match in the correct order; this gives a probability of 1/1000.  To find the expected value, we multiply this by the winnings (499 after the $1 cost); we also multiply the probability of losing (999/1000) by the amount lost (-1):
1/1000(499)+999/1000(-1)
499/1000 - 999/1000 = -500/1000 = -0.50

For the second game, since the number is "boxed", there are 3! ways to get the correct digits; this gives a probability of 6/1000.  Multiply this by the winnings, 79 (after the $1 cost); multiply the probability of losing (994/1000) by the loss (-1):
6/1000(79) + 994/1000(-1) = 474/1000 - 994/1000 = -520/1000= -0.52

Aso sorry if i’m wrong

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