Suppose you roll a (fair, 6-sided, perfectly ordinary) die repeatedly until you roll a 6. As is well known, the expected (i.e., long-term average over many trials) number of rolls required is 6.
Now suppose we ask this question conditional on never having rolled any odd numbers. That is, suppose you learn that someone has followed the procedure above, and that they rolled only even numbers up to the point where they rolled that 6. What is the expected number of rolls they took?
(That is: What is the expected number of rolls taken to roll a 6, conditional on having got no odd numbers in the process of rolling that 6? The question is not anything like "what's the expected number of rolls next time, given that you didn't get any odd numbers last time?". I've added this paragraph and slightly reworded the question above because it seems from comments that some readers interpreted the question in an unintended way.)
This question is interesting because there is an extremely tempting wrong answer.
Some history: The question seems to have first been asked by Elchanan Mossel while teaching an undergraduate probability course at the University of Pennsylvania in 2014-15. Gil Kalai posted it on his blog as part of his Test Your Intuition series, and then posted a followup. There is interesting discussion in the comments of both. And here is a nice little article about the question, in PDF form.
WARNING: All the links in the paragraph above may take you to places that give away some or all of the solution. If you want to solve the puzzle for yourself, don't follow those links until you've done it.
[See this Meta question for some relevant PSE-specific context, which in particular explains why this is community wiki. An older question has been merged into this one, and it's possible that some details of answers or comments may look odd as a result.]
Answer
The answer is indeed...
11⁄2 rolls
...because the question is equivalent to...
What is 1 more than the average number of consecutive rolls of only 2s and/or 4s before any other roll terminates the series?
One way to see this equivalence is to note that only 2/4 streaks terminated by 6 are considered. (A first-roll 6 terminates a 2/4 streak of length 0.) Yet the average length of those streaks is the same as of streaks terminated by 1, 3 or 5. Collect all such streaks, including those terminated by 6, and the result is all streaks of 2/4, each with the same expected length.
Calculations:
E = expected number of consecutive even rolls up to and including the first 6
= (expected number of consecutive 2s and/or 4s) + (one more roll of 6)
= T + 1
where
T = expected number of consecutive 2s and/or 4s before some other roll
= (probability of 2 or 4) × (1 for this roll + expected subsequent 2s and/or 4s)
= 1⁄3 (1 + T)
so
T = 1⁄2
thus
E = T + 1
= 11⁄2
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