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The following describes a dice game being offered to you. You will roll a die once, and if you get a multiple of three, you win 33.00 dollors. If you roll a 1, you get a second chance to roll the die. On your second roll, if you get a multiple of three, you win 66.00 dollars. If you get a 1 on your second roll, you win 27.00 dollars. Otherwise, you lose.

Here I want to create a probability distribution for X, where my X=amount that you win from the game

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X= 33 when you get multiple of 3 in first throw . Probablity= $\frac{1}{3}$
X= 27 when you get 1 in first throw and second throw. Probablity= $\frac{1}{6}.\frac{1}{6}=\frac{1}{36}$
X= 66 when you get 1 in first throw and multiple of 3 in second . Probablity= $\frac{1}{6}.\frac{1}{3}=\frac{1}{18}$

X=0 in all other cases=$1-\frac{5}{12}=\frac{7}{12}$

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  • $\begingroup$ So, it can't be called a fair game? $\endgroup$ – JimmyFails Apr 22 '17 at 8:25
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If you want to create a probability distribution for X, you first need to determine the (discrete) values that X can be, which are 0, 27, 33, and 66. Next you need to assign to each the probability of that outcome:

P(0) = prob of losing on first roll (3/6) + prob of getting to second roll (1/6) and then losing (3/6) = (3/6) + (1/6)*(3/6) = 7/12

P(27) = prob of getting to second roll (1/6) and then rolling a 1 (1/6) = (1/6)*(1/6) = 1/36

P(33) = prob of getting a 3 or 6 on first roll = 2/6 = 1/3

P(66) = prob of getting to second roll (1/6) and then rolling a 3 or 6 (2/6) = (1/6)*(2/6) = 1/18

Check: 7/12 + 1/36 + 1/3 + 1/18 = 1.

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