Two distinct dice, expected value

Question

We have two distinct dice. $X$ will be the number of pips showing on the top face of the die after the toss. $Y$ is the same for the second die.

What is the expected value $$Z_2 = greatest\; common\; divisor(X,Y) $$

Can anyone help me how to solve that please? I just know the expected value of $X$, which is $3.5$ and $Y$ is also $3.5$ But I don't know what to do in case of the greatest common divisor.

Answer 1

You can make a table. This is mostly a good idea in cases of two dice. The first column and row are the possible outcomes of the two dice each. For each combination the $gcd$ has been inserted. Please check the entries.

$$ \begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} \hline \text{ d1/d2 } 1 & 1 &2 &3 &4 &5 &6 \\ \hline \hline \hline 1 &1 &1 &1 &1 &1 &1 \\ \hline 2& 1 & 2 &1 &2 &1&2 \\ \hline 3& 1 &1 &3 &1 &1&3\\ \hline 4 &1 &2 &1&4&1&2 \\ \hline 5 &1 &1&1&1&5&1 \\ \hline 6&1&2&3&2&1&6 \\ \hline \end{array}$$

Each outcome has a probability of $\frac1{36}$. Now it is not difficult to calculate $E(Z_2)$.

Answer 2

Make a $6 \times 6$ table, in which each of the 36 cells shows the GCD. I've filled in the first three rows. (This is before morning coffee for me, so better check!) Can you finish?

X\Y:   1  2  3  4  5  6 
1      1  1  1  1  1  1
2      1  2  1  2  1  2
3      1  1  3  1  1  3
...

Then count to find numerators of fractions of which denominators are 36.