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

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?

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)

• What's the convention? Do I delete this question or leave it as a duplicate? May 3 at 20:25
• Do not delete good duplicates! May 3 at 23:18

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.