Reverse monty hall problem probability? 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?
 A: 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:

So if the host did reveal one of the good ones would switching make you ?
  more likely to lose because it's the reverse of the monty hall with two 
  goods and a bad. Or, once revealed that one good could have been a good, 
  bad or a rhino for all it matters because it's already eliminated and 
  it's the regular monty hall paradox and your chances increase by 
  switching?

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.
