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 × 14 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 ) = 1 6 P(2)=P(3)=P(4)=P(5)=\frac{1}{6} and P ( 1 ) = p P(1)=p .

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


The answer is -1.

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2 solutions

汶良 林
Jun 21, 2015

P(2) = P(3) = P(4) = P(5) = 1/6

P(1) = k, P(6) = 1/3 - k

12, 15, 21, 24, 33, 36, 42, 45, 51, 54, 63, 66

k(1/6) + k(1/6) + (1/6)k + (1/6)(1/6) + (1/6)(1/6) + (1/6)(1/3 - k) + (1/6)(1/6) + (1/6)(1/6) + (1/6)k + (1/6)(1/6) + (1/3 - k)(1/6) + (1/3 - k)(1/3 - k) = 1/2

6k + 6k + 6k + 1 + 1 + (2 - 6k) + 1 + 1 + 6k + 1 + (2 - 6k) + (2 - 6k)² = 18

36k² - 12k - 3 = 0

12k² - 4k - 1 = 0

(6k + 1)(2k - 1) = 0

k = -1/6 (invalid),

k = 1/2 → 1/3 - k = -1/6 (invalid).

so the answer is -1.

Niven Achenjang
Jun 7, 2015

To begin, let's define some variables. Let n n be the first number rolled, A \textbf{A} be the event that Alice wins, 0 \textbf{0} be the event that n 0 ( m o d 3 ) n\equiv0 (mod\quad3) , 1 \textbf{1} be the event that n 1 ( m o d 3 ) n\equiv1 (mod\quad3) , and 2 \textbf{2} be the event that n 2 ( m o d 3 ) n\equiv2 (mod\quad3) . We then have the following:

P ( 0 ) = P ( 3 ) + P ( 6 ) = 1 2 p P(\textbf{0}) = P(3)+P(6) = \frac{1}{2}-p

P ( 1 ) = P ( 1 ) + P ( 4 ) = 1 6 + p P(\textbf{1}) = P(1)+P(4) = \frac{1}{6}+p

P ( 2 ) = P ( 2 ) + P ( 5 ) = 1 3 P(\textbf{2}) = P(2)+P(5) = \frac{1}{3}

P ( A 0 ) = P ( 3 ) + P ( 6 ) = 1 2 p P(\textbf{A}|\textbf{0}) = P(3) + P(6) = \frac{1}{2}-p

P ( A 1 ) = P ( 2 ) + P ( 5 ) = 1 3 P(\textbf{A}|\textbf{1}) = P(2) + P(5) = \frac{1}{3}

P ( A 2 ) = P ( 1 ) + P ( 4 ) = 1 6 + p P(\textbf{A}|\textbf{2}) = P(1) + P(4) = \frac{1}{6}+p

Putting these six equations together, we have P ( A ) = P ( 0 ) P ( A 0 ) + P ( 1 ) P ( A 1 ) + P ( 2 ) P ( A 2 ) = p 2 1 3 p + 13 36 P(\textbf{A}) = P(\textbf{0})P(\textbf{A}|\textbf{0})+P(\textbf{1})P(\textbf{A}|\textbf{1})+P(\textbf{2})P(\textbf{A}|\textbf{2}) = p^2-\frac{1}{3}p+\frac{13}{36} for 0 p 1 3 0\le p\le\frac{1}{3}

Solving for P ( A ) = 1 2 P(\textbf{A})=\frac{1}{2} simply requires using the quadratic formula and yields p = 1 ± 6 6 p=\frac{1\pm\sqrt{6}}{6} . You can verify with a calculator that neither or these values are in the range 0 p 1 3 0\le p\le\frac{1}{3} so the answer is -1.

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