Let’s take an imaginary HL deck.
Tell me what position you think Zephrys is in (1-30), and then I turn over 28 cards, which are not Zephrys, leaving only two cards not revealed - the one you picked, and one other.
With two cards left, what are the odds that your card is Zephrys?
(Don’t rush to the 3 door puzzle, this is not quite the same)
1/30, since “my card” is the position you had me guess and doesn’t depend on anything you do after that.
just google “the three door probability problem”, the probability for the chosen card wont change regardless of how you try to maniplulate it
No correct answers yet
2 correct answers so far and one poor attempt to claim the game is rigged 
one in 30 is correct
That’s only because what you think you’re asking isn’t what the text of your post says.
Bro shortly, after drawing 28 not Zef cards, there are 2 cards left, its 50-50.
Guys 1-30 its about when you chose the card or not.
But to draw a Zef, its 50 
wrong. the three door probability problem tells us that when you pick your card at 1/30 that card has the probability 1/30, when you then remove the incorrect choices your card still has the probability 1/30 while the 2nd card ends up with a 29/30 probability
the 1-30 is the base for the probability problem, for your scenario you cant pick a card before removing the other cards
No, because you cant change your pick at those 2 cards that are left.
your chances where determined in those 30 cards and it doesnt matter if he removes 28, your probabillity is still 1/30
And still no correct answers!
The answer is, it depends on how the cards are selected. If I turned over 28 cards at random, then it’s 50/50. Otherwise it’s 1/30.
This is relevant when you consider how cards are selected when being drawn.
incorrect, please read up on some probability theory before you attempt a probability discussion. the key action is that you made someone pick a card, not how you turned over the cards.
But you didnt
Maths and humans will always be troubling.
I deliberately omitted the detail I used to turn over cards. This is why 5/30 != (1-30)+(1/29)…(1/26). You’re not really my target audience
And here’s the actual correct answer.
Suppose the case with a deck of 30 cards.
You draw 28 cards, from the top , one by one. None is Zephrys. Only 2 cards in deck.
So whats the probability of zef being as 29? Its 1/30. Whats the prob of zef being as 30? 1/30.
But here is the question.
What is the prob to draw zef, its 50-50 qs it shares the same prob to be 29 or 30.
that logic only works if you didnt pick a card, by picking a card in belief that it might be zeph you lock that card into a 1/30 probability reagardless of what actions you take afterwards, meaning if you picked cards 29 then card 30 has a 29/30 chance to be zeph
your target audience are people that dont know how probability works? regardless that doesnt change the fact that your logic is wrong.
No you where very specific with your choice of words by saying that you would turn over 28 cards that are not zephrys.
the formulation is the very basis for a three door probability problem