On the average, how many times must a die be thrown until on gets a 6 -- where am I doing wrong

Question

I've already known one of the right solutions which is $$\sum_{n=0}^\infty n \cdot p \cdot q ^{n−1}$$ However I don't know why this logic doesn't work: We are calculating the quantity -- $$E(N|Last=6)$$ where N is the number of rolls and Last=6 means only the last roll results in 6. So we have $$E(N|Last=6) = \sum_{n=0}^\infty n \cdot P(n|Last=6)$$ where according to Bayes rule $$P(n|Last=6) = P(Last=6|n) \cdot P(n)/P(Last=6)$$ where $P(n)$ is uniform and $P(Last=6) = 1/6$ and $P(Last=6|n) = p \cdot q^{n-1}$ $$$$ If you plug these back in $E(N|Last=6) = \sum_{n=0}^\infty n \cdot P(n|Last=6)$, the result is actually different. $$$$ Where am I doing wrong here and what's the right way of applying Bayes rule here?

Answer 1

Think of X- number of times until getting a 6.

You need to get different numbers from 6 in the (x-1)-th tries and at the x-th try you get a 6

X has geometric distribution, $p=1/6$ (probability of success) and $q=5/6$.

Now just compute $E(X)$