So I understand the monty hall problem fine but not the reverse.
Premise of question: As I was watching a show on youtube there were 3 people eating burritos and one had a bad filling and one of the people had the option to switch with one of the other two people then all people bite the middle of the burrito.
What I think I understand: if the burritos are distributed at random you can assume a 66% of getting a good burrito. If you stay with your burrito you remain with a 66% chance. If you switch your odds will go down because the best you can do is the same fine burrito or you can get the bad burrito.
What I dont understand: What is your odds if the host doesnt eliminate one of the burritos? Is it the same 66% no matter what cause you dont know any additional information even if the options of switching are equal to or worse than your own hopefully good option?
I don't think your decision matters since no information is added by the elimination/non elimination of a burrito.
Consider the different strategies of switch/stay. $$P(good_i) = 1-P_{bad,i}=2/3$$
If you go with switch: $$P(good|good_i)= 1/2$$ $$P(good|bad_i)= 1$$ $$P(good) = \frac{2}{3}\frac{1}{2}+\frac{1}{3}=2/3$$
If you stay: $$P(good|good_i)= 1$$ $$P(good|bad_i)= 0$$ $$P(good) = \frac{2}{3}+0=2/3$$
So it doesn't matter as expected
Edit:The question is extended to include additional question from asker:
Say the host unwittingly revealed one of the good burritos. I believe you need to take into account the unwittingly part. Say you choose a burrito and the host is known to not know which burrito is which.
$$P(good_i|good_{revealed}) = \frac{P(good_{revealed}|good_i)}{P(good_{revealed})}P(good_i) = \frac{\frac{1}{2}}{\frac{2}{3}}\frac{2}{3}=1/2$$
If the host doesn't know then no information is transmitted and you are neither better off switching nor staying.
Say the host always reveals a good burrito, then we are in standard Monty hall territory. We should stay.