Prove that you cannot load two dice in such a way that the probabilities for any sum from 2 to 12 are the same. (Be sure to consider the case where one or more sides turn up with probability zero.)
For each dice, there are 6 probabilities - P1, P2, P3, P4, P5 and P6. Let us assume that we can influence these values in any way we want (assuming the laws of probability are not borken). This means that the only restrictions on our assignments are that:
\(\sum_{i=1}^{i=6} P_{i} = 1\)
\(P_{i} \ge 0\)
Since they are all equal, \(P_{i}=\frac{1}{11}\)
For clarity, let us use the following 3 designations:
\(P_{i}\) as the probability of the sum of the two die being i
\(A_{i}\) as the probability of the first dice being i
\(B_{i}\) as the probability of the second dice being i
The probability of each sum is then:
\[P_{12} = A_{6}*B_{6}=\frac{1}{11}\]
\[P_{11} = A_{6}*B_{5} + A_{5}*B_{6}=\frac{1}{11}\]
\[P_{10} = A_{6}*B_{4} + A_{5}*B_{5} + A_{4}*B_{6}=\frac{1}{11}\]
\[P_{9} = A_{6}*B_{3} + A_{5}*B_{4} + A_{4}*B_{5} + A_{3}*B_{6}=\frac{1}{11}\]
\[P_{8} = A_{6}*B_{2} + A_{5}*B_{3} + A_{4}*B_{4} + A_{3}*B_{5} + A_{2}*B_{6}=\frac{1}{11}\]
\[P_{7} = A_{6}*B_{1} + A_{5}*B_{2} + A_{4}*B_{3} + A_{3}*B_{4} + A_{2}*B_{5} + A_{1}*B_{6}=\frac{1}{11}\]
\[P_{6} = A_{5}*B_{1} + A_{4}*B_{2} + A_{3}*B_{3} + A_{2}*B_{4} + A_{1}*B_{5}=\frac{1}{11}\]
\[P_{5} = A_{4}*B_{1} + A_{3}*B_{2} + A_{2}*B_{3} + A_{1}*B_{4}=\frac{1}{11}\]
\[P_{4} = A_{3}*B_{1} + A_{2}*B_{2} + A_{1}*B_{3}=\frac{1}{11}\]
\[P_{3} = A_{2}*B_{1} + A_{1}*B_{2}=\frac{1}{11}\]
\[P_{2} = A_{1}*B_{1}=\frac{1}{11}\]
Assuming the two die are weighted the same (AKA \(A_{i} = B_{i}\)):
\[P_{12} = (A_{6})^{2}=\frac{1}{11}\]
\[P_{11} = 2(A_{6}*A_{5})=\frac{1}{11}\]
\[P_{10} = 2(A_{6}*A_{4}) + (A_{5})^{2}=\frac{1}{11}\]
\[P_{9} = 2(A_{6}*A_{3}) + 2(A_{5}*A_{4})=\frac{1}{11}\]
\[P_{8} = 2(A_{6}*A_{2}) + 2(A_{5}*A_{3}) + (A_{4})^{2}=\frac{1}{11}\]
\[P_{7} = 2(A_{6}*A_{1}) + 2(A_{5}*A_{2}) + 2(A_{4}*A_{3})=\frac{1}{11}\]
\[P_{6} = 2(A_{5}*A_{1}) + 2(A_{4}*A_{2}) + (A_{3})^{2}=\frac{1}{11}\]
\[P_{5} = 2(A_{4}*A_{1}) + 2(A_{3}*A_{2})=\frac{1}{11}\]
\[P_{4} = 2(A_{3}*A_{1}) + (A_{2})^{2}=\frac{1}{11}\]
\[P_{3} = 2(A_{2}*A_{1})=\frac{1}{11}\]
\[P_{2} = (A_{1})^{2}=\frac{1}{11}\]
Based on this in order for all the sums to have the same probability, we can put together many equations:
\[A_{6} = \sqrt{\frac{1}{11}}=A_{1}\]
A_6 <- sqrt(1/11)
A_1 <- A_6\(P_{12}=P_{11}\)
\[(A_{6})^{2} = 2(A_{6}*A_{5})\]
\[A_{5} = \frac{A_{6}}{2}=A_{2}\]
A_5 <- A_6/2
A_2 <- A_5\(P_{12}=P_{10}\)
\[(A_{6})^{2} = 2(A_{6}*A_{4}) + (A_{5})^{2}\]
\[(A_{6})^{2} = 2(A_{6}*A_{4}) + (\frac{A_{6}}{2})^{2}\]
\[A_{6} = 2(A_{4}) + \frac{A_{6}}{4}\]
\[\frac{3A_{6}}{4} = 2(A_{4})\]
\[A_{4} = \frac{3A_{6}}{8}\]
A_4 <- 3*A_6/8\(P_{7}=\frac{1}{11}\)
\[2(A_{6}*A_{1})+2(A_{5}*A_{2})+2(A_{4}*A_{3})=\frac{1}{11}\]
\[A_{6}*A_{1}+A_{5}*A_{2}+A_{4}*A_{3}=\frac{1}{22}\]
\[A_{4}*A_{3}=\frac{1}{22}- A_{6}*A_{1}-A_{5}*A_{2}\]
\[A_{3}=\frac{\frac{1}{22}- A_{6}*A_{1}-A_{5}*A_{2}}{A_{4}}\]
A_3 <- (1/22-A_6*A_1-A_5*A_2)/A_4This calculation leads to a negative number for \(A_{3}=\) -0.6, which means such a situation is impossible!