Question 446751: the carnival has come to town, and lucky Louie has once again set up his pavilion.he offers a simple game consisting of rolling two normal six sided dice at the same time for the extraordinary law price of 3$ per play.if the player rolls a sum of 5, the player win 5$,if the sum is a 7, the player wins 7$;and if the sum is a 10 then the player win 10$.otherwise the player does not win anything. how much should you win or lose per game (after paying 3$)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! the carnival has come to town, and lucky Louie has once again set up his pavilion.he offers a simple game consisting of rolling two normal six sided dice at the same time for the extraordinary
low price of 3$ per play.
if the player rolls a sum of 5, the player win 5$,
if the sum is a 7, the player wins 7$;
and if the sum is a 10 then the player win 10$.
otherwise the player does not win anything.
how much should you win or lose per game (after paying 3$)
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Roll result::::::::::::::::::::::::5::::::7::::::10
Random "winning" values:::::::: ...2......4......7.....-3
Corresponding probabilities:::: ..4/36...6/36...3/36..23/36
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Multiply the probabilities times the "winning" values.
Add the products to get---
Expected winnings: [8 + 24 + 21 - 69]/36 = -44.4 cents
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Cheers,
Stan H.
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