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 ?

You want the product, not the sum, of the probabilities.

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

(PS: Also, there are fifteen ways to select two items from six, not thirty.)