Question 982730: Hello. I desperately need help with this expected value question.
In the game of chance called "birdcage" you have three 6-sided dice. In the
game you can pick any number from 1 to 6 and the three dice are then “rolled” in a cage. If $1 is bet and exactly one of the number that you picked is rolled then you win $1, if exactly two of the number that you picked are rolled you win $2, and if all three of the dice are the number that you picked you win $10 (in each of these cases you also get your initial $1 bet back). If none of your number wind up being rolled you lose your $1 bet. Suppose that you play 5 games of Birdcage and pick the same number each time.
a) What is the probability that triples of your number (all three dice come up your numb
b) What is your total expected win or loss? You answer my indicate both the amount and win or loss.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! P(X=0)=(5/6)^3 = 125/216
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P(X=1) = P(first dice six) + P(second dice six) + P(third dice six) = (1/6)(5/6)(5/6) + (5/6)(1/6)(5/6) + (5/6)(5/6)(1/6) = 3⋅(25/216) = (75/216) = (25/72)
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P(X=2) = P(first dice not six) + P(second dice not six) + P(third dice not six) = (5/6)(1/6)(1/6) + (1/6)(5/6)(1/6) + (1/6)(1/6)(5/6) = 3⋅(5/216) = (15/216) = (5/72)
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P(X=3) = (1/6)^3 = (1/216)
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from the above, we have the three probabilities, P(X=1) is probability of getting 1 die with the number you pick, P(X=2) and P(X=3) are the probabilities of getting 2 die with picked number and getting 3 die with picked number
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a) P(X=3) = 1/216 for one roll
for 5 rolls it would be (1/216)^5 = 1/470184984576
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b) expected value for one roll is -1(5/6) + 3(5/72) + 11(1/216) approx −0.57 which means you lose $.57
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expected value for 5 rolls is 5(-0.57) = −2.85 which means you lose $2.85
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