What's the correct intuition behind the expected number of die rolls until a 6 if all previous are even? [duplicate]

Question

I came across a puzzle on YouTube recently (spoiler).

You roll a fair dice until you get a 6. What is the expected number of rolls, including the roll of 6, conditioned on the event that all previous rolls were even numbers?

Questions:

1. Can you solve it without looking (honor system!)

2. (more important) Can you explain the correct intuition?

The video gives mathematical answer, which is totally fine, but the intuition of the inverse of 1/3 is hard to shake. What's the right way to think about this?

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Edit: Getting a lot of incorrect answers!

The correct answer is NOT 3.

Here's some python code that demonstrates this easily (and the YouTube video has the mathematical proof too):

import random
runs = 1000

number=[]

while True:
    if len(number)>=runs:
        print(f"Expected number is {sum(number)/len(number):.2f}")
        break
    #Start a throw sequence:
    seq=[]
    while True:
        throw = random.randint(1,6)
        if throw%2==1:
            #non conditioned
            break
        if throw==6:
            number.append(len(seq)+1)
            break
        seq.append(throw)

Answer 1

Okay, so you reset every odd roll. Therefore, E = E[6] + E[consecutive 2,4s before a 6]. E[6] is 1. E[2,4] = 1/3*E[2,4] -> E[2,4] = 1/2. So E=1+1/5 = 1.5. Expected Value is 1 1/2 dice rolls.

Now, you asked for the correct intuition as well. When you are resetting every odd roll, you are trying to find the expected value of times you roll a 2 or a 4 before the end runs in a 6, and add that to the expected value when you roll a 6 first. It makes sense that anytime you start with a 2 or a 4 there is a roughly 50% chance you will end the run prematurely, and that there is a higher chance that the final run will be just a 6 roll.

Answer 2

Well, you know that a roll can be either be a $2,4,6$, and it's the same chance for each, so it's $\frac{1}{3}$ for any of them. If you note that it's a geometric random variable, then it's immediately $3$.

Otherwise, let $X$ be the event of rolling a 6. We have that $$E[X] = \frac{1}{3}(E[X|\text{Rolled a 6}]+1) + \frac{2}{3}(E[X|\text{Rolled a 2 or a 4}]+1)$$ but note that $E[X|\text{Rolled a 2 or a 4}] = E[X]$, and $E[X|\text{Rolled a 6}] = 0$, as we stop rolling at that point, so solving this for $E[X]$, we get $E[X] = 3$ as desired.