Answer to Question #224295 in Statistics and Probability for Agenda

Question

Two dice are used, each die loaded so that the probabilities of 1,2,3,4,5,6 are x1-x/6, 1+2x/6, x-1/6, 1+x/6, 1-2x/6, 1+x/6 respectively.

Compute the probability in rolling the two dice sum up to six

Answer 1

Solution:

Using $\Sigma p_i=1$

$(\dfrac{x}{1}-\dfrac{x}{6})+(\dfrac{1+2x}{6})+(x-\dfrac{1}{6})+(\dfrac{1+x}{6})+(\dfrac{1-2x}{6})+(\dfrac{1+x}{6})=1 \\\Rightarrow x=\dfrac3{13}$

Then, the probability distribution:

$X=1,p=\dfrac{5}{26} \\ X=2,p=\dfrac{19}{78} \\ X=3,p=\dfrac{5}{78} \\ X=4,p=\dfrac{8}{39} \\ X=5,p=\dfrac{7}{78} \\ X=6,p=\dfrac{8}{39}$

$P(S\le 6)=P(X=1,X=1)+P(X=1,X=2)+...+P(X=1,X=5) \\+P(X=2,X=1)+P(X=2,X=2)+...+P(X=2,X=4) \\+P(X=3,X=1)+P(X=3,X=2)+...+P(X=3,X=3) \\+P(X=4,X=1)+P(X=4,X=2) \\+P(X=5,X=1)$

$=\dfrac{5}{26}\times \dfrac{5}{26}+\dfrac{5}{26}\times \dfrac{19}{78}+\dfrac{5}{26}\times\dfrac{5}{78}+\dfrac{5}{26}\times\dfrac{8}{39}+\dfrac{5}{26}\times\dfrac{7}{78}$

$+\dfrac{19}{78}\times \dfrac{5}{26}+\dfrac{19}{78}\times \dfrac{19}{78}+\dfrac{19}{78}\times \dfrac{5}{78}+\dfrac{19}{78}\times \dfrac{8}{39}$

$+\dfrac{5}{78}\times\dfrac{5}{26} +\dfrac{5}{78}\times\dfrac{19}{78} +\dfrac{5}{78}\times \dfrac{5}{78}$

$+\dfrac{8}{39}\times \dfrac{5}{26}+\dfrac{8}{39}\times\dfrac{19}{78}$ $+\dfrac{7}{78}\times\dfrac{5}{26}$

$=\dfrac{155}{1014}+\dfrac{1045}{6084}+\dfrac{5}{156}+\dfrac{136}{1521}+\dfrac{35}{2028} \\=\dfrac{2819}{6084}$