# Pairwise sampling with dice rolls

I'm attempting to solve the following question.

An unbiased dice is throwed 12 times to obtain the samples $x_1, x_2, \dots ,x_{12} \in \{1, \dots, 6\}$.

I'm asked the probability of $$\forall i,j \in \{1, \dots, 6\}, i \neq j, y_i \neq y_j$$ where $y_k$ is defined as $y_k = (x_{2k-1}, x_{2k})$, i.e $y_1 = (x_1, x_2), y_2 = (x_3, x_4), \dots, y_6 = (x_5, x_6).$

My attempt

We want $y_i \neq y_j$ for all i and j. So, we can write the desired probability as follows $$\sum_{i=1}^6 \sum_{j=i+1}^6 P_{r}(y_i \neq y_j)$$

where $$P_{r}(y_i \neq y_j) = 1 - P_{r}(y_i = y_j) = \frac{35}{36}$$

So, then my solution becomes

$$15\times \frac{35}{36}$$ which is clearly wrong.

What is wrong with my reasoning ?

• Just to be clear: order matters? That is to say, you are regarding the roll $(1,6)$ as different from $(6,1)$? – lulu Dec 3 '16 at 14:23
• @lulu yes exactly. – SpiderRico Dec 3 '16 at 14:24
• In that case the posted solution is complete (N.B. the calculation gets a lot messier if you want to ignore the order). – lulu Dec 3 '16 at 14:25

\begin{align}\Pr(\bigcap_{i=1}^5\bigcap_{j=i+1}^6 y_i\neq y_j) & = \prod_{i=1}^5\prod_{j=i+1}^6 \Pr(y_i\neq y_j) \\[1ex] & = {\Bigl(\frac{35}{36}\Bigr)}^\binom{6}{2} \\[1ex] & = {\Bigl(\frac{35}{36}\Bigr)}^{15}\end{align}