Question

What is the cdf of tossing a pair of dice several times, let take X RV is the sum of the dice possiple outcomes.

Answer 1

Favorite outcome for X=2 is {1,1}.

Favorite outcomes for X=3 are {1,2},{2,1}.

Favorite outcomes for X=4 are {1,3},{2,2},{3,1}.

Favorite outcomes for X=5 are {1,4},{2,3},{3,2},{4,1}.

And so on.

Thus, theoretical probability of the 11 outcomes from tossing two dice:

$\begin{matrix} X\;\;\;\;\;| & 2&3&4&5&6&7&8&9&10&11&12 \\ P(X)|& \frac{1}{36}&\frac{2}{36}&\frac{3}{36}&\frac{4}{36}&\frac{5}{36}&\frac{6}{36}&\frac{5}{36}&\frac{4}{36}&\frac{3}{36}&\frac{2}{36}&\frac{1}{36} \end{matrix}$

Cumulative distribution function F(X):

$F(2)=Pr(X\le2)=\frac{1}{36}.$

$F(3)=Pr(X\le3)=F(2)+Pr(X=3)=\frac{1}{36}+\frac{2}{36}=\frac{3}{36}.$

$F(4)=Pr(X\le4)=F(3)+Pr(X=4)=\frac{3}{36}+\frac{3}{36}=\frac{6}{36}.$

$F(5)=Pr(X\le5)=F(4)+Pr(X=5)=\frac{6}{36}+\frac{4}{36}=\frac{10}{36}.$

And so on.

Finally,

$\begin{matrix} X\;\;\;\;\;| & 2&3&4&5&6&7&8&9&10&11&12 \\ F(X)|& \frac{1}{36}&\frac{3}{36}&\frac{6}{36}&\frac{10}{36}&\frac{15}{36}&\frac{21}{36}&\frac{26}{36}&\frac{30}{36}&\frac{33}{36}&\frac{35}{36}&1 \end{matrix}$