Finding the distribution law Throwing a cube twice,
let $X$ be the sum of the two throws

Find the distribution law

My attempt
$$
\begin{array}{c|lcr}
 &2&3&4&5&6&7&8&9&10&11&12 \\
\hline
\text{P}_\text{X}(x) & 1/12 & 1/12 & 1/12&1/12&1/12&1/12&1/12&1/12&1/12&1/12&1/12 \\
\end{array}
$$
But shouldn't the sum be $1$? Currently the sum is $11/12$
 A: The probabilities are not equal. Have a look at the following table.
$$
\left.\begin{array}{c|c|c|c|c} & 1 & 2 & 3 & 4 & 5& 6\\\hline 1 & 2 & 3 & 4 & 5& 6& 7 \\\hline 2 & 3 & 4 & 5 & 6 & 7& 8\\\hline 3 & 4 & 5 & 6 & 7 & 8& 9\\\hline 4 & 5 & 6 & 7 & 8 & 9& 10\\\hline 5 & 6 & 7 & 8 & 9 & 10& 11\\\hline 6 & 7 & 8 & 9 & 10 & 11& 12\end{array}\right.
$$
It is easy to tell various probabilities of the outcomes of this experiment using the table above. For example, the probability that we get $7$ is $6/36=1/6$.
A: \begin{array}{c|lcr}
 &2&3&4&5&6&7&8&9&10&11&12 \\
\hline
\text{P}_\text{X}(x) & 1/36 & 2/36 & 3/36&4/36&5/36&6/36&5/36&4/36&3/36&2/36&1/36 \\
\end{array}
Now the sum is $1$.
A: What you've tried is wrong:
You can get $2$ just getting a one in each die (1+1), so the probability is $\frac{1}{6} \frac{1}{6} = \frac{1}{36}$. Instead, you have more options to get $5$ (1+4,2+3,3+2,4+1), giving you a total probability of $\frac{2}{18}$
Doing this process for all the numbers you get the correct result.
A: Yes, the final distribution should sum to one. I recommend drawing a 6x6 table and fill in the intersections with the sum of the two values. Then count up all the squares that sum to 2 for example and divide that by 36. Do that for each sum. If that is what you did, then you might have made a small mistake.
