# Explicit Probability Formula of a Die Roll

The Question:

Suppose a pair of balanced dice is tossed.Let $E_x$ be the event that the sum of two numbers obtained is equal to $x$, $x = 2, 3, ..., 12$.

Develop an explicit expression for $P(E_x)$.

$P(E_x) = {1 \over 36} \min \{ x-1, 13-x\}$ where $x = i + j$, and $i,j = 1,..6$.

My Question:

How was the minimum expression found? Can anyone explain to me the reasoning as I cannot arrive at it?

Thank you.

Think of the two die outcomes as independent events and calculate the possible sums $$\text{Sum of dice}$$ \begin{array}{|c||c|c|c|} \hline \text{Die 1 $\rightarrow$}\\\text{Die 2 $\downarrow$}&1& 2 & 3 & 4 & 5 & 6 \\ \hline 1 &2 &3 &4&5&6&7\\ \hline 2 &3 &4&5&6&7&8\\ \hline 3 &4 &5&6&7&8&9\\ \hline 4 &5 &6&7&8&9&10\\ \hline 5 &6 &7&8&9&10&11\\ \hline 6 &7 &8&9&10&11&12\\ \hline \end{array} Each outcome will occur with equal probability $1/36$. Look at the diagonals of the table to get that \begin{array}{|c|c|} \hline x & P(E_x=x) \\ \hline 2 &1/36 \\ \hline 3 &2/36 \\ \hline 4 &3/36 \\ \hline 5 &4/36 \\ \hline 6 &5/36 \\ \hline 7 &6/36 \\ \hline 8 &5/36 \\ \hline 9 &4/36 \\ \hline 10 &3/36 \\ \hline 11 &2/36 \\ \hline 12 &1/36 \\ \hline \end{array} and so $P(E_x=x)=(x-1)/36$ for $x \leq 7$ and $P(E_x=x)=(13-x)/36$ for $x \geq7$. Finally note that $(x-1) \leq (13-x)$ for $x \leq 7$ and $(13-x) \leq (x-1)$ for $x \leq 7$, thus $P(E_x=x)= \min\{(x-1)/36,(13-x)/36\}$.

• Did you just come up with x-1 by looking at the table or did you derive it mathematically? If you did can you show me how? – ATP Aug 26 '17 at 0:53
• What do you mean by deriving mathematically? Calculating each value explicitly is a valid derivation. Writing out tables of outcomes helps to see patterns in the probabilities and thus accurately describe a probability density function. – Alex Aug 26 '17 at 1:12

Think of the ways you can get $x=2$:

$A =1$ and $B =1$.

Think of the ways you can get $x = 3$:

$A = 1$ and $B =2$;

$A = 2$ and $B = 1$.

Think of the ways you can get $x = 4$:

$A = 1$ and $B = 3$;

$A = 2$ and $B = 2$;

$A = 3$ and $B = 1$.

...

Think of the ways you can get $x = 12$:

$A = 6$ and $B = 6$.

Each sub-event has a probability of ${1 \over 6}$.

Now generalize for every condition...