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Let $X_n$ be the event that the dice takes $n$ rolls to get the first 6, given all the rolls are even. Let $A_n$ be the event that it takes $n$ rolls to get the first 6, and let $B_n$ be the event that all the n rolls are even.

$P(X_n)=P(A_n|B_n)=\dfrac{P(B_n|A_n)P(A_n)}{P(B_n)}$ (using Bayes' theorem)

Now:

$P(A_n)=\dfrac{1}{6}\cdot\left(\dfrac{5}{6}\right)^{n-1}$ (since the first $n-1$ rolls are not 6 and the last roll is)

$P(B_n)=\left(\dfrac{1}{2}\right)^n$ (since there are three even numbers and three odd numbers)

$P(B_n|A_n)=\left(\dfrac{2}{5}\right)^{n-1}$ (since the last roll is even but for all the other rolls there are 2 even outcomes and 3 odd outcomes)

So $P(A_n|B_n)=\dfrac{\dfrac{1}{6}\cdot\left(\dfrac{2}{5}\right)^{n-1}\cdot\left(\dfrac{5}{6}\right)^{n-1}}{\left(\dfrac{1}{2}\right)^n}=\dfrac{1}{3}\cdot\dfrac{2^{n-1}\cdot5^{n-1}\cdot2^{n-1}}{5^{n-1}\cdot6^{n-1}}=\dfrac{1}{3}\cdot\left(\dfrac{2}{3}\right)^{n-1}$

Now, we wish to find $\displaystyle\sum^{\infty}_{i=1}i\cdot P(X_i)=\displaystyle\sum^{\infty}_{i=1}i\cdot\dfrac{1}{3}\cdot\left(\dfrac{2}{3}\right)^{i-1}$$=\dfrac{1}{2}\cdot\displaystyle\sum^{\infty}_{i=1}i\left(\dfrac{2}{3}\right)^{i}$ $=\dfrac{1}{2}\cdot\displaystyle\sum^{\infty}_{i=0}\left(\left(\dfrac{2}{3}\right)^{i}\cdot\displaystyle\sum^{\infty}_{j=1}\left(\dfrac{2}{3}\right)^{j}\right)$$=\dfrac{1}{2}\cdot\displaystyle\sum^{\infty}_{i=0}\left(\dfrac{2}{3}\right)^{i}\cdot2$$=\displaystyle\sum^{\infty}_{i=0}\left(\dfrac{2}{3}\right)^{i}=3$

So it takes on average 3 rolls for the dice to get a 6, given that all the rolls are even.