Expected Dice Rolling (EDITED) Q- What is the expected no of rolls of a Dice to get a 6 CONDITIONED  that all previous rolls ( if any) were even numbers ?
How I attempted this question - We would have a series of 2,4 which would terminate to 6. Sample space - (2,4,4,6) ,(6), (2,6),(4,4,4,4,2,6) etc. Probability that length of chain is 1 = 1/3( we get a six in first roll). If expected length of chain is x then Probability that length of chain is (x+1)= 2/3 ( as we missed out the first roll expected length is x+1.)
x= 1/3×1 + 2/3×(x+1) 
x=3.
But the ans is 1.5 as given in YouTube video - "Most mathematicians miss this probability question" by the famous Presh talwalkar. Explain the solution as in that video. Thank you!
 A: This is a subtle question, and the solution given in the video is pretty clever.
I think a good way to imagine it is this: Take an auditorium of a bunch people. You have them all roll a die until they get a 6, at which point they stop. Then, you select just the ones who only rolled an even number prior to the 6, and ask them how many times they had to roll a die to get the 6, and take the average.
So you're excluding all the people who rolled an odd number. They know this, so once they roll an odd number, instead of rolling until they get a six, they might as well just stop.
Then you've divided the auditorium into four groups: Those who only rolled 2's and 4's until they rolled a 6, then stopped (these are the ones you care about). Those who rolled 2's and 4's until they stopped on a 5. Those who stopped on a 3. Those who stopped on a 1.
But if you think about it- there's no reason why the people in the "6" group (the ones you care about) would have a different average length than the "1", or "3", or "5" group- so you might as well combine all four groups together!
So the question is equivalent to "Roll a die until you roll a number which is not a 2 or 4, what's the expected number of die you roll".
And the answer to this is $3/2$, using a similar logic to in your original post.
A: The specified condition constrains only the prior rolls.

Thus, the sample points of length $n$ are $n$-tuples with each of the first $(n-1)$ components equal to $2$ or $4$, but with no constraint on the $n$-th component.

Hence, the conditional probability of getting a $6$ for the first time on the $n$-th roll, given that the $(n-1)$ prior rolls are even, is
$$\left({\small{\frac{1}{3}}}\right)^{n-1}\left({\small{\frac{1}{6}}}\right)$$ so the required conditional expectation is
$$
\sum_{n=1}^\infty 
n\left(\left({\small{\frac{1}{3}}}\right)^{n-1}\left({\small{\frac{1}{6}}}\right)\right)
$$
which is equal to ${\large{\frac{3}{2}}}$.
