# Trust me, its fair

Two friends, Alice and Bob, were bored so Bob came up with a game for them to play. The game is fairly simple. They roll and six sided die twice and concatenate the results to get a single two digit number. If this number is divisible by 3 Alice wins and if it is not, Bob wins. (For example, if the first roll is a 4 and the second role is a 2 then the number is $$42=3\times14$$ and Alice wins). Alice is skeptical about this game because she feels she is at a disadvantage, but Bob assures her that he is using a special weighted die that makes the game fair.

Assume that Bob's die is weighted such that $$P(2)=P(3)=P(4)=P(5)=\frac{1}{6}$$ and $$P(1)=p$$.

Find the value of $$p$$ that makes this game fair (i.e. Alice wins with probabilty $$\frac{1}{2}$$). Enter -1 as your answer if you think that the game cannot be made fair.

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