<s>[INST] Basic Probability Question about dice with 112233 sides instead of 123456 We have two special $6$-sided dices. each of them has sides $112233$ instead of $123456$.
We roll them. what is the probability that sum of them is $5$? What is the probability that sum of them is odd?
My problem is that I don't know if $n(S) = 6 \times 6 = 36$ or $n(S) = 3 \times 3 = 9$. I want to know should I consider the same sides just one side or not? Is it right to say it is like we have a $3$-sided dice with $123$ on it or not? 
 [/INST] First consider it intuitively:
We can draw up a table showing all possible sums:
$$\begin{array}{|c|cccccc|}\hline&1&1&2&2&3&3\\
\hline
1&2&2&3&3&4&4\\
1&2&2&3&3&4&4\\
2&3&3&4&4&5&5\\
2&3&3&4&4&5&5\\
3&4&4&5&5&6&6\\
3&4&4&5&5&6&6\\
\hline
\end{array}$$
Therefore we can see that $P(S=5) = \frac{8}{36}=\frac 29$
And $P(S \text{ is odd}) = \frac{16}{36}=\frac 49$

Now consider the maths behind this:
We can say that $P(1)=P(2)=P(3)=\frac 26 =\frac 13$
For $S=5$, we must have one dice being $2$ and the other $3$, so we can say that 
\begin{align}P(S=5)&=P((D_1=2\text{ and }D_2=3)\text{ or }(D_1=3\text{ and }D_2 =2))\\
&=(P(D_1=2)\times P(D_2=3)) + (P(D_1=3)\times P(D_2=2))\\
&=\left(\frac 13 \times \frac 13\right) + \left(\frac 13 \times \frac 13\right)\\
&=\frac 19 + \frac 19\\
&= \frac 29
\end{align}
Can you work out how to do it for $P(S\text{ is odd})$?
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