So earlier today I came across Elchanan Mossel's Dice Paradox, and I am having some trouble understanding the solution. The question is as follows:

You throw a fair six-sided die until you get 6. What is the expected number of throws (including the throw giving 6) conditioned on the event that all throws gave even numbers?

*Quoted from Jimmy Jin in "Elchanan Mossel’s dice problem"*

In the paper it goes on to state why a common wrong answer is $3$. Then afterwards explains that this problem has the same answer to,

"What is the expected number of times you can roll only $2$’s or $4$’s until you roll any other number?"

I don't understand why this is the case. If the original problem is asking for specifically a $6$, shouldn't that limit many of the possible sequences?

I also attempted to solve the problem using another method, but got an answer different from both $3$ and the correct answer of $1.5$.

I saw that possible sequences could have been something like:

$$\{6\}$$ $$\{2,6\}, \{4,6\}$$ $$\{2,2,6\}, \{2,4,6\}, \{4,2,6\}, \{4,4,6\}$$ $$\vdots$$

To which I set up the following summation and solved using Wolfram Alpha:

$$\text{Expected Value} =\sum_{n=1}^\infty n\left( {\frac{1}{6}} \right)^n 2^{n-1} = 0.375$$ Obviously this is different and probably incorrect, but I can't figure out where the error in the thought process is.

Any help on understanding this would be greatly appreciated.

*A blog post discussing the problem can be found here.*