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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 ?

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  • $\begingroup$ Just to be clear: order matters? That is to say, you are regarding the roll $(1,6)$ as different from $(6,1)$? $\endgroup$ – lulu Dec 3 '16 at 14:23
  • $\begingroup$ @lulu yes exactly. $\endgroup$ – SpiderRico Dec 3 '16 at 14:24
  • $\begingroup$ In that case the posted solution is complete (N.B. the calculation gets a lot messier if you want to ignore the order). $\endgroup$ – lulu Dec 3 '16 at 14:25
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You want the product, not the sum, of the probabilities.

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

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

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  • $\begingroup$ Yes ! :) The rest seems ok though, right ? $\endgroup$ – SpiderRico Dec 3 '16 at 14:22

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