Suppose that you roll a 6-sided fair die repeatedly and record the values in order. Given the fact that the first 6 appears before the first 5, what is the EV of the number of rolls until you get a 6 (including the roll that got the 6)? Answer quickly based off of your intuition (curious to see how intuitive this is).
1,000@100Anonymous You were actually the only person to get it right lol, 3 is correct and verified by Monte Carlo. The intuitive explanation is that each roll has a 1/3 chance of being a 5 or 6, so the EV of the number of rolls until we get a 5 or 6 is 3. We then condition on 6 appearing first. @traders
@spiderduckpig in hindsight, i dont know why i thought 5 was actually correct. checking again, yeah 3 was correct and i got confused again.
What do you think? A variety of responses here. Please don't comment the answer if you know it