Answer to Question #225058 in Statistics and Probability for Atubiga

Question

Two loaded dice were tossed so that the probabilities of 1,2,3,4,5,6 are 1-x/6, 1+2x/6, 1-x/6, 1+x/6, 1-2x/6, x+x/6 respectively. Compute the probability in rolling such dice sum up to six

Answer 1

Let A and B be the number of points on the first and second dice, respectively.

Find the corresponding probabilities for the sum A + B up to six:

$p(A + B = 2) = p(A = 1) \cdot p(B = 1) = {\left( {1 - \frac{x}{6}} \right)^2}$

$p(A + B = 3) = p(A = 1) \cdot p(B = 2) + p(A = 2) \cdot p(B = 1) = \left( {1 - \frac{x}{6}} \right)\left( {1 + \frac{{2x}}{6}} \right) + \left( {1 + \frac{{2x}}{6}} \right)\left( {1 - \frac{x}{6}} \right) = 2\left( {1 - \frac{x}{6}} \right)\left( {1 + \frac{{2x}}{6}} \right)$

$p(A + B = 4) = p(A = 1) \cdot p(B = 3) + p(A = 3) \cdot p(B = 1) + p(A = 2) \cdot p(B = 2) = {\left( {1 - \frac{x}{6}} \right)^2} + {\left( {1 - \frac{x}{6}} \right)^2} + {\left( {1 + \frac{{2x}}{6}} \right)^2} = 2{\left( {1 - \frac{x}{6}} \right)^2} + {\left( {1 + \frac{{2x}}{6}} \right)^2}$

$p(A + B = 5) = p(A = 1) \cdot p(B = 4) + p(A = 2) \cdot p(B = 3) + p(A = 3) \cdot p(B = 2) + p(A = 4) \cdot p(B = 1) = \left( {1 - \frac{x}{6}} \right)\left( {1 + \frac{x}{6}} \right) + \left( {1 + \frac{{2x}}{6}} \right)\left( {1 - \frac{x}{6}} \right) + \left( {1 - \frac{x}{6}} \right)\left( {1 + \frac{{2x}}{6}} \right) + \left( {1 - \frac{x}{6}} \right)\left( {1 + \frac{x}{6}} \right) = 2\left( {1 - \frac{{{x^2}}}{{36}}} \right) + 2\left( {1 - \frac{x}{6}} \right)\left( {1 + \frac{{2x}}{6}} \right)$

$p(A + B = 6) = p(A = 1) \cdot p(B = 5) + p(A = 2) \cdot p(B = 4) + p(A = 3) \cdot p(B = 3) + p(A = 4) \cdot p(B = 2) + p(A = 5) \cdot p(B = 1) = \left( {1 - \frac{x}{6}} \right)\left( {1 - \frac{{2x}}{6}} \right) + \left( {1 + \frac{x}{6}} \right)\left( {1 + \frac{{2x}}{6}} \right) + {\left( {1 - \frac{x}{6}} \right)^2} + \left( {1 + \frac{x}{6}} \right)\left( {1 + \frac{{2x}}{6}} \right) + \left( {1 - \frac{x}{6}} \right)\left( {1 - \frac{{2x}}{6}} \right) = 2\left( {1 - \frac{x}{6}} \right)\left( {1 - \frac{{2x}}{6}} \right) + 2\left( {1 + \frac{x}{6}} \right)\left( {1 + \frac{{2x}}{6}} \right) + {\left( {1 - \frac{x}{6}} \right)^2}$

We have a probability distribution:

$\begin{matrix} {A + B}&P\\ 2&{{{\left( {1 - \frac{x}{6}} \right)}^2}}\\ 3&{2\left( {1 - \frac{x}{6}} \right)\left( {1 + \frac{{2x}}{6}} \right)}\\ 4&{2{{\left( {1 - \frac{x}{6}} \right)}^2} + {{\left( {1 + \frac{{2x}}{6}} \right)}^2}}\\ 5&{2\left( {1 - \frac{{{x^2}}}{{36}}} \right) + 2\left( {1 - \frac{x}{6}} \right)\left( {1 + \frac{{2x}}{6}} \right)}\\ 6&{2\left( {1 - \frac{x}{6}} \right)\left( {1 - \frac{{2x}}{6}} \right) + 2\left( {1 + \frac{x}{6}} \right)\left( {1 + \frac{{2x}}{6}} \right) + {{\left( {1 - \frac{x}{6}} \right)}^2}} \end{matrix}$