Asked by Anonymous
In a particular game, a fair die is tossed. If the number of spots showing is either 4 or 5 you win $1, if the number of spots showing is 6 you win $4, and if the number of spots showing is 1, 2, or 3 you win nothing. Let X be the amount that you win.
What is the probability distribution for the situation?
What is the probability distribution for the situation?
Answers
Answered by
Reiny
I consider this to be an "expected value" question.
It usually asks for the expected profit or loss if the game is played repeatedly.
Prob(4 or 5) = 2/6 = 1/3
prob(6) = 1/6
prob(1,2, or 3) = 3/6 = 1/2
(note 1/3 + 1/6 + 1/2 = 1)
Expected value = (1/3)(1) + (1/6)(4) + (1/2)(0) = 1
This is a totally fair game, that is, if you were to pay $1 to play this game, you would expect to get back $1 each time you play , that is , you would break even.
It usually asks for the expected profit or loss if the game is played repeatedly.
Prob(4 or 5) = 2/6 = 1/3
prob(6) = 1/6
prob(1,2, or 3) = 3/6 = 1/2
(note 1/3 + 1/6 + 1/2 = 1)
Expected value = (1/3)(1) + (1/6)(4) + (1/2)(0) = 1
This is a totally fair game, that is, if you were to pay $1 to play this game, you would expect to get back $1 each time you play , that is , you would break even.
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