Question 1076597: For many years, organized crime ran a numbers game that is now run legally by many state governments. The player selects a three-digit number from 000 to 999. There are 1000 such numbers. A bet of $2 is placed on a number, say number 115. If the number is selected, the player wins $1000. If any other number is selected, the player wins nothing. Find the expected value for this game and describe what this means.
The expected value of the numbers game is $
Found 2 solutions by jorel1380, Boreal:
Answer by jorel1380(3719)   (Show Source):
You can put this solution on YOUR website! The odds of winning are 1/1000, with a payoff of $1000 if your numbers are picked. So 1000 x 1/1000=$1. Since tickets are $2, your expected value is -$1. Or, in other words, a loss of at least $1. ☺☺☺☺
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! E(x)=1000*(1/1000)-2(999/1000)=$1-$1.998=-$0.998, assuming that if one wins, he gets the money back that was originally spent on the ticket. If not, then the cost of the ticket makes the expected value-$1.
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