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.