Basic Probability Question about dice with 112233 sides instead of 123456

Question

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?

Answer 1

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 $P(S \text{ is odd}) = \frac{16}{36}=\frac 49$

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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 $P(S\text{ is odd})$?

Answer 2

You can take $n(S)=6\times6=36$, as you consider each side distinct regardless of the sides having same numbers or not.

Favorable outcomes are $(2,3),(2,3),(2,3),(2,3),(3,2),(3,2),(3,2),(3,2)$

Probability is therefore $\dfrac{8}{36}=\dfrac{2}{9}$

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Alternatively, as @lulu mentioned, consider it as a 3 sided die with sides $(1,2,3)$. Since each number is twice, such a consideration does not affect the probability. In this case, $n(s)=3\times 3=9$ and favorable outcomes are $(2,3),(3,2)$. Thus we get same probability $\dfrac{2}{9}$