PDF and CDF of Sum of 2 dice

Question

i'm trying to obtain the PDF and CDF of the sum of 2 dice toss.

There are tons of elementary exercise where is asked to find the exact probability,

but what about the PDF and CDF?

i thought this is a convolution of discrete uniform PDF. but i don't know where to start to find it.

it's like $P(X_1+X_2<y)=$ but from here i don't know how to continue... i can't substitute $1/n=X_1$ and $1/n=X_1$

this would make nosense.

Thank You

Answer 1

Clearly the "exact probability of the sum" is the PDF.

$\mathsf P(X_1{+}X_2{=}y)=\dfrac 1{36}\times\begin{cases}(y-1) &:& y\in\{2,3,4,5,6,7\}\\[1ex] (13-y) &:& y\in\{8,9,10,11,12\}\\0 &:& \textsf{else}\end{cases}$

And the accumulation of this is the CDF.

$\mathsf P(X_1{+}X_2{\leq}y)=\dfrac 1{72}\times\begin{cases}0&:& y\leq 1\\[1ex]\lfloor y\rfloor(\lfloor y\rfloor-1) &:& \lfloor y\rfloor\in\{2,3,4,5,6,7\}\\[1ex] 25\lfloor y\rfloor-\lfloor y\rfloor^2-84 &:& \lfloor y\rfloor\in\{8,9,10,11\}\\[1ex] 72 &:& y\geq 12 \end{cases}$

Answer 2

How many ways are there to get a sum of $2$? We must get $1$ and then $1$ again with probability $\frac{1}{36}$.

How many ways are there to get a sum of $3$? We must either get a $1$ and then $2$ or vice versa with probability $\frac{2}{36}$.

How many ways are there to get a sum of $4$? We must either get a $1$ and then $3$, a $3$ and then $1$, or a $2$ and a $2$, giving a probability of $\frac{3}{36}$

Carrying on in this manner, we find that the probabilities appear to follow triangular shape. As commented, letting $X$ be the sum of the the two die, $X$ is a discrete random variable so we need a pmf.

$$ p_{X}(x)= \begin{cases} \frac{1}{36} & x = 12 \\ \frac{2}{36} & x=11 \\ \frac{3}{36} & x=10 \\ \frac{4}{36} & x=9 \\ \frac{5}{36} & x=8 \\ \frac{6}{36} & x=7 \\ \frac{5}{36} & x=6 \\ \frac{4}{36} & x=5 \\ \frac{3}{36} & x=4 \\ \frac{2}{36} & x=3 \\ \frac{1}{36} & x=2 \\ 0 & \text{otherwise} \\ \end{cases} $$

If you wish to construct a probability density function, despite it not making much sense in this situation, you could so something like

$$ f_{X}(x)= \begin{cases} \frac{1}{36} & x \geq 12 \\ \frac{2}{36} & 11 \leq x \lt 12 \\ \frac{3}{36} & 10 \leq x \lt 11 \\ \frac{4}{36} & 9 \leq x \lt 10 \\ \frac{5}{36} & 8 \leq x \lt 9 \\ \frac{6}{36} & 7 \leq x \lt 8 \\ \frac{5}{36} & 6 \leq x \lt 7 \\ \frac{4}{36} & 5 \leq x \lt 6 \\ \frac{3}{36} & 4 \leq x \lt 5 \\ \frac{2}{36} & 3 \leq x \lt 4 \\ \frac{1}{36} & 2 \leq x \lt 3 \\ 0 & x \lt 2 \\ \end{cases} $$

And the cumulative density function would be something like

$$ F_{X}(x)= \begin{cases} 1 & x \geq 12 \\ \frac{35}{36} & 11 \leq x \lt 12 \\ \frac{33}{36} & 10 \leq x \lt 11 \\ \frac{30}{36} & 9 \leq x \lt 10 \\ \frac{26}{36} & 8 \leq x \lt 9 \\ \frac{21}{36} & 7 \leq x \lt 8 \\ \frac{15}{36} & 6 \leq x \lt 7 \\ \frac{10}{36} & 5 \leq x \lt 6 \\ \frac{6}{36} & 4 \leq x \lt 5 \\ \frac{3}{36} & 3 \leq x \lt 4 \\ \frac{1}{36} & 2 \leq x \lt 3 \\ 0 & x \lt 2 \\ \end{cases} $$

Answer 3

Let $X_1$ and $X_2$ be the outcome of the first and second dice. We are interested in their sum $X = X_1 + X_2$. It is important to note that since $X$ can only take on a finite number of outcome, namely 2 to 12, it is still discrete.

You could try convolution. Note the outcomes here are so simple that you could just enumerate them. Look at look at the probability table here. Essentially you will just count the number of outcome associated to each sum. Then you get $P(X=k) =$ the number of outcome associated to the sum k / 36.

CDF is defined for both discrete and continuous setting. Here you could just look at the table and for each k, come up with a formula to include the "triangle" that include all the outcomes lower than k.

Hope this helps.