We roll a dice until we get six twice ( we stop after the second 6 ). What is the most likely location of the first six? What is the probability that the modus is the actual location of the first six?
My Attempt:
There is this concept of
forever young,
wherever we tossed out first six, the probability that we toss the second six on the $n$-th throw after the first six is
$$
\frac{1}{6} \cdot \left( \frac{5}{6} \right)^{n-1}
$$
This is a monotonically decreasing function. Thus the maximum is attained when $n=1$, which means if we the second six is tossed on the $n$-th toss, we first 6 is most likely tossed on the $n-1$-th toss.
Then, I use Bayes Theorem,
$$
\mathbb{P} \left( F_{n-1}|E \right)
=
\frac{\mathbb{P} \left( E|F_{n-1} \right)\cdot \mathbb{P}\left(F_{n-1}\right)} {\sum_i\mathbb{P} \left( E|F_i \right) \cdot \mathbb{P}\left(F_i\right)}
$$
where E is the event that the second six is tossed on the $n$-th toss, and $F_i$ is the event that the first six is tossed on the $i$-th toss. This is equal to
$$
\frac {(\frac{1}{6}) \cdot (\frac{5}{6}^{n-2} \cdot \frac{1}{6})} {\sum_i (\frac{1}{6} \cdot \frac{5}{6}^{n-i-1}) \cdot (\frac{5}{6}^{i-1} \cdot \frac{1}{6})}
$$
This simplifies to 1 unfortunately, since we can cross out $(\frac{1}{6})^2$, and the sum on the denominator is independent of $i$, being equal to $ (\frac{5}{6})^{n-2} $.
Can someone please point out my mistake here?
First of all, before I critique your method, the answer to :
I am unfamilar with the term modus. However, the most likely location of the first 6 is independent of your intent to keep rolling the die after the first 6 comes up.
Therefore, by the analysis that you already gave, the most likely location of the first 6 is the first roll. I wikipedia-googled on modus; it seems to be a synonym for mode, which is the same as how I interpreted the question.
As for critiquing your math, I found it perfect, except that you overlooked that the denominator is the sum of $(n-1)$ terms, each of whom compute to the same value.
This means that $p[(F_{(n-1)}|E] = \frac{1}{n-1}.$